Answer:
0.477 V
Explanation:
When a substance is gaining electrons, it's reducing, and when the substance loses electrons, it's oxidizing. In a galvanic cell, one substance oxides giving electrons for the other, which reduces. Then, the substance with higher reduction potential must reduce and the other must oxide.
E°cell = E°red(red) - E°red(oxid)
Where, E°red(red) is the reduction potential of the substance that reduces, and E°red(oxid) is the reduction potential of the substance that oxides. For the value given, Cu⁺² reduces, so:
E°cell = +0.337 - (-0.140)
E°cell = 0.477 V
The standard cell potential for this galvanic cell is [tex]0.477 \ V[/tex].
To calculate the standard cell potential [tex](\( E^\circ_\text{cell} \))[/tex] for the galvanic cell using the given reduction potentials, we use the formula:
[tex]\[ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \][/tex]
Given reduction potentials:
[tex]\( E^\circ_\text{red}(\text{Sn}^{2+} \rightarrow \text{Sn}) = -0.140 \, \text{V} \)[/tex]
[tex]\( E^\circ_\text{red}(\text{Cu}^{2+} \rightarrow \text{Cu}) = +0.337 \, \text{V} \)[/tex]
Since the reduction potential for [tex]\( \text{Cu}^{2+} \)[/tex] is more positive, it acts as the cathode:
[tex]\[ E^\circ_\text{cathode} = +0.337 \, \text{V} \][/tex]
And the reduction potential for [tex]\( \text{Sn}^{2+} \)[/tex] is less positive (more negative), so it acts as the anode:
[tex]\[ E^\circ_\text{anode} = -0.140 \, \text{V} \][/tex]
Now, calculate the standard cell potential:
[tex]\[ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \][/tex]
[tex]\[ E^\circ_\text{cell} = (+0.337 \, \text{V}) - (-0.140 \, \text{V}) \][/tex]
[tex]\[ E^\circ_\text{cell} = +0.337 \, \text{V} + 0.140 \, \text{V} \][/tex]
[tex]\[ E^\circ_\text{cell} = +0.477 \, \text{V} \][/tex]
Explain why the HOH molecule is bent, whereas the HBeH molecule is linear. Water, HOH, has bonding electron pairs and lone pair(s) of electrons. The placement of these electrons forces electron-pair geometry, and the resulting HOH molecule is bent. The HBeH molecule has bonding electron pairs and lone pair(s) of electrons. The bonding pairs of electrons must be as one another as possible, resulting in a linear HBeH molecule.
The question is already answered from one point of view but I'll add BeH2 has sp hybridization and H2O has sp3 hybridization getting a different structure type cause where the electron pairs are located around the central atom, that means that you will see the next structures from these molecules.
Answer:
Water (HoH) has
2
bonding electrons and 2 lone pairs. The placement of these electrons is going to be a tetrahedral electron-pair geometry. The HBeH molecule has 2 bonding electron pairs and 0 lone pairs. The bonding pairs must be as far from one another as possible.
Explanation:
I hope that helps any one in the future who needs it <3
Which of the following describes the correct order for relative solubility of minerals in sedimentary rocks?
a. Evaporate minerals, quartz, and calcite all have the same relative solubility.
b. Evaporate minerals are more soluble than calcite and quartz.
c. Evaporate minerals are more soluble than quartz and less soluble than calcite.
d. Evaporate minerals are less soluble than quartz and calcite.
Answer:
Evaporate minerals are more soluble than calcite and quartz.
Explanation:
Evaporate minerals are the water soluble minerals which at higher concentration precipitate out and crystallized forming rocks.
example of chemicals present are:
chlorides and sulphates.
Quartz is silica (very less soluble, or insoluble)
Calcite is calcium carbonate, again an insoluble salt.
Thus
Evaporate minerals are more soluble than calcite and quartz.
The electrons formed from the aerobic oxidation of glucose are
I. ultimately transferred to O2 after several other transfer reactions.
II. transferred to the coenzymes NAD and FAD.
III. directly transferred to O2 during the citric acid cycle.
IV. transferred to succinate and arachidonic acid.
Answer:
i and ii
Explanation:
In the aerobic oxidation of glucose, the electrons formed are transferred to O2 after several others transfer reactions like passing through coenzymes NAD+ and FAD
The following reaction is found to be at equilibrium at 25 celcius: 2SO3--->O2 + 2SO2 + 198kJ/mol. If the value for Kc at 25 celcius is 8.1, are the products or reactants favored and how much?
A.) Reactants are strongly favored
B.) Reactants are weakly favored
C.) Products are strongly favored
D.) Products are weakly favored
Answer:
D.) Products are weakly favored
Explanation:
For the reaction:
2SO₃ ⇄ O₂ + 2SO₂ + 198kJ/mol
The kc is defined as:
kc = [O₂] [SO₂]² / [SO₃]²
As the kc is 8,1:
8,1 [SO₃]² = [O₂] [SO₂]²
The products are favored 8,1 times. This is a weakly favored because the usual kc are in the order of 1x10⁴. Thus, right answer is:
D.) Products are weakly favored
I hope it helps!
Final answer:
For the reaction 2SO3 → O2 + 2SO2 with an equilibrium constant Kc of 8.1 at 25 °C, the products are weakly favored since Kc is greater than 1 but much less than 10^3. So the correct answer is D.
Explanation:
The equilibrium constant Kc indicates whether the reactants or products are favored in a chemical reaction at equilibrium. For the reaction 2SO3 → O2 + 2SO2, the equilibrium constant Kc is given as 8.1 at 25 °C. Since the value of Kc is greater than 1, this means that the reaction favors the formation of products over reactants. If the Kc value was less than 1, the reactants would be favored.
According to the table provided, values of Kc greater than 103 indicate a strong tendency for the reaction to favor the formation of products, suggesting a strongly favored product side for such large Kc values. In this case, with a Kc of 8.1, which is greater than 1 but significantly less than 103, the products are favored to a lesser extent. Therefore, we could conclude that the products are weakly favored, making option D the correct answer.
When 1 mole of Fe2O3(s) reacts with Al(s) to form Al2O3(s) and Fe(s) according to the following equation, 852 kJ of energy are evolved. Fe2O3(s) + 2Al(s)Al2O3(s) + 2Fe(s) Is this reaction endothermic or exothermic? What is the value of q? kJ g
Answer: The reaction is exothermic in nature and the value of q is 852 kJ
Explanation:
Endothermic reactions are defined as the reactions in which energy of products is more than the energy of the reactants. In these reactions, energy is absorbed by the system. The energy term is written on the reactant side of the reaction.
Exothermic reactions are defined as the reactions in which energy of reactants is more than the energy of the products. In these reactions, energy is released by the system. The energy term is written on the product side of the reaction.
We are given:
Heat released in the reaction = 852 kJ/mol
The chemical equation follows:
[tex]Fe_2O_3(s)+2Al(s)\rightarrow Al_2O_3(s)+2Fe(s)+852kJ[/tex]
As, heat is released during the reaction, it is considered as an exothermic reaction.
To calculate the enthalpy change of the reaction, we use the equation:
[tex]\Delta H_{rxn}=\frac{q}{n}[/tex]
where,
[tex]q[/tex] = amount of heat released
n = number of moles = 1 mole
[tex]\Delta H_{rxn}[/tex] = enthalpy change of the reaction = 852 kJ/mol
Putting values in above equation, we get:
[tex]852kJ/mol=\frac{q}{1mol}\\\\q=(852kJ/mol\times 1mole)=852kJ[/tex]
Hence, the reaction is exothermic in nature and the value of q is 852 kJ
Investigating iron thiocyanate requires several chemicals which must be used with caution. Using knowledge of these chemicals, select whether the following statements are true or false. When handling HNO3, it is important to wash your hands frequently since HNO3 is corrosive and can cause burns. To dilute an acid, always add water to add while slowly stirring. Fe(NO3)3 should be handled with caution because it is an oxidizer and can irritate the skin. KSCN is completely nontoxic If KSCN is strongly heated it will cause the evolution of cyanide gas. KSCN, if combined with a strong base, will also cause the evolution of cyanide gas.
Answer:
1. False
2. False
3. True
4. False
5. False
6. False
Explanation:
1. Because HNO₃ is corrosive when handling it is important to use gloves and also protection glasses. If it touches the skin, then it's important to wash the area affected.
2. The dilute of acid is exothermic, so it releases a huge amount of heat. For precaution, the acid must be added at the water slowly stirring, so it can be controlled.
3. Fe(NO₃)₃ is an oxidizer and in contact with the skin it can cause irritation, so it must be handled carefully and with protection.
4. KSC can irritate the eyes, so it's toxic.
5. KSCN is not inflammable and it's not combustible, so when heated it will only change the state for gas, but it will not cause the evolution of cyanide gas.
6. If KSCN reacts with a strong base, it will dissociate, and the ions SCN⁻, which can't react with the OH⁻ ions of the base. So, it will only cause the evolution of cyanide gas if it reacts with a strong acid.
The answers have been given according to which is True or False. It explores the science of chemicals, especially Nitric Acid.
What is a Chemical Reaction?
Two substances will achieve a reaction has occurs when the molecular or ionic structure of a substance is rearranged to create a new form.
The statements thus are:
1) True.
2) False.
3) True.
4) False
5) False
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Arrange the following ionic compounds in order of decreasing amount of energy released in lattice formation: LiCl , CaO , AlN , and KI ?
Answer:in decreasing pattern; AlN>CaO>LiCl>KI.
Explanation:
The energy required to break apart an ionic solid and convert its component atoms into gaseous ions is known as the LATTICE ENERGY.
LATTICE ENERGY has a trend on the periodic table. The trend shows the relationship between LATTICE ENERGY AND THE ATOMIC RADIUS.
Lattice energy is related to solubility, volatilty and hardness.
Down the group on the periodic chart, the ATOMIC RADIUS INCREASES, THE LATTICE ENERGY DECREASES. Which means that between Lithium ion and potassium ion, Lithium ion will have higher lattice energy than that of potassium.
The trend ACROSS THE PERIOD is that, as we go across the period, the charges on metals increase,consequently INCRESING THE LATTICE ENERGY. Between Aluminum ion with a positive charge of three and Calcium with a positive charge of two, Aluminium ion has higher charge, which means it has more lattice energy than calcium.
The ionic compounds in order of their decreasing amount of energy are AIN>CaO>LiCl>Kl.
The lattice energy is used for the estimation of the strength of the bond that is in ionic compounds.
It should be noted that the higher the charge of the ion, the higher will be the lattice energy. On the other hand, the smaller the radius of ion, the larger will be the lattice energy.
Therefore, the ionic compounds in order of their decreasing amount of energy will be AIN>CaO>LiCl>Kl.
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Use the drop-down arrows to make the following statements, regarding bomb calorimetry, correct.
A bomb calorimeter is a piece of equipment designed to measure [ Select ] ["delta E", "delta H", "delta S", "C"] for reactions.
Reactions carried out in a bomb calorimeter occur at constant [ Select ] ["pressure", "volume", "mass", "temperature"] .
To calculate the heat absorbed or released by a reaction carried out in a bomb calorimeter, we use the equation: [ Select ] ["q = C * delta T", "q = m * C * delta T"]
Answer:
Answers are in the explanation
Explanation:
A bomb calorimeter is a constant volume calorimeter in which medition of change in temperature allows the medition of change in internal energy (ΔU) for chemical reactions or physical changes.
This ΔU can be converted in ΔH thus:
ΔH = ΔU + PV
That means that:
A bomb calorimeter is a piece of equipment designed to measure delta H
As there was said:
Reactions carried out in a bomb calorimeter occur at constant volume
To calculate the heat absorbed or released by a reaction carried out in a bomb calorimeter, we use the equation: q = C * delta T
Where C is the heat capacity of the calorimeter, delta T is the change in temperature and q is the heat produced in the chemical or physical process
I hope it helps!
A bomb calorimeter measures the change in internal energy (delta E) of reactions, which occur at constant volume. The heat absorbed or released is calculated using the formula q = C x delta T.
Explanation:A bomb calorimeter is a piece of equipment designed to measure delta E, or the change in internal energy, for reactions. Reactions carried out in a bomb calorimeter occur at constant volume. The heat absorbed or released by a reaction in this calorimeter is calculated using the equation q = C x delta T, where 'q' is the heat, 'C' is the heat capacity of the calorimeter, and 'delta T' is the change in temperature.
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The freezing point of benzene, C6H6, is 5.500 °C at 1 atmosphere. Kf(benzene) = 5.12 °C/m In a laboratory experiment, students synthesized a new compound and found that when 14.44 grams of the compound were dissolved in 292.3 grams of benzene, the solution began to freeze at 4.571 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound?
Answer:
The molecular weight they determined for this compound is 272.9 g/m
Explanation:
We must apply the colligative property of freezing point depression to solve this:
ΔT° = Kf · molality
Where ΔT° means T° fussion pure solvent − T° fussion of solution
and Kf the cryoscopic constant
The statemente expressed that the compound was also found to be nonvolatile and a non-electrolyte, so we don't have to apply the Van't Hoff factor (i)
By the way, the complete formula is this one:
ΔT° = Kf · molality . i
5.5°C - 4.571°C = 5.12 °C/molal . molality
0.929°C = 5.12 °C/molal . molality
0.929°C / 5.12 molal/°C = molality → 0.181 m
Molality means the mol of solute in 1kg of solvent. But, in this solution we used 292.3 g of benzene, so let's find out our moles of solute, in our mass of solvent.
1000 g ___ 0.181 moles
292.3 g ____ (292.3 . 0.181) /1000 = 0.053 moles
The mass, we used of solute is 14.44 g so, to find out the molar mass we must divide mass (g) / moles
14.44 g /0.053 m = 272.9 g/m
The molecular weight they determined for this compound of benzene which is non volatile is 272.9 g/m
The colligative property of freezing point depression is applied to solve this:
ΔT° = Kf · molality
Where ΔT° means T° fusion pure solvent − T° fusion of solution
and Kf is the cryoscopic constant
The statement expressed that the compound was also found to be nonvolatile and a non-electrolyte, so we don't have to apply the Van't Hoff factor (i)
By the way, the complete formula is this one:
ΔT° = Kf · molality . i
5.5°C - 4.571°C = 5.12 °C/molal . molality
0.929°C = 5.12 °C/molal . molality
0.929°C / 5.12 molal/°C = molality → 0.181 m
Molality means the mol of solute in 1kg of solvent. But, in this solution we used 292.3 g of benzene, so let's find out our moles of solute, in our mass of solvent.
1000 g ___ 0.181 moles
292.3 g ____ (292.3 . 0.181) /1000 = 0.053 moles
The mass,used of solute is 14.44 g so, to find out the molar mass divide mass (g) / moles
14.44 g /0.053 m = 272.9 g/m
Thus, the molar mass of compound is 272.9 g/m.
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Consider the following reaction at constant pressure. Use the information provided below to determine the value of ΔS at 473 K. Predict whether or not this reaction will be spontaneous at this temperature.
4NH3 (g) + 3O2 (g) → 2N2 (g) + 6H2O (g) ΔH = –1267 kJ
Answer:
The reaction will be spontaneous
Explanation:
To determine if the reaction will be spontaneous or not at this temperature, we need to calculate the Gibbs's energy using the following formula:
[tex]\Delta G= \Delta H - T * \Delta S [/tex]
If the Gibbs's energy is negative, the reaction will be spontaneous, but if it's positive it will not.
Calculating the [tex]\Delta G= -1267 - 473 K* \Delta S [/tex] :
[tex]\Delta G= -1267 - 473 K* \Delta S [/tex]
Now, other factor we need to determine is the sign of the S variation. When talking about gases, the more moles you have in your system the more enthropic it is.
In this reaction you go from 7 moles to 8 moles of gas, so you can say that you are going from one enthropy to another higher than the first one. This results in: [tex]\Delta S>0[/tex}
Back to this expression:
[tex]\Delta G= -1267 - 473 K* \Delta S [/tex]
If the variation of S is positive, the Gibbs's energy will be negative always and the reaction will be spontaneous.
Which of the following statements is false? The mean free path of a molecule depends on the size of the molecule. ODalton's law of partial pressures indicates that in gaseous mixtures at low pressure each kind of molecule behaves independently of the others O At a given temperature, for a given gas, every molecule has the same speed Collisions of molecules with the container walls give rise to the gas pressure O At high pressures different gases give different values for the ratio PV/nRT.
Answer:
False statement is
At a given temperature, for a given gas, every molecule has the same speed
Explanation:
While checking each and every statement given
The mean free of a molecule actually depends on the size of the molecule because the mean free path is defined as the average distance between two successive collisions of the gas molecules
If the size of the molecule is more, the average distance between two successive collisions decrease and as a result the mean free path of the molecule decreases
Dalton's law of partial pressures is applicable for only ideal gases which means we are assuming that the size of the molecule of a gas is negligible and there are no intermolecular forces of attraction
These two assumptions gets applied at high temperature and low pressure
So Dalton's law of partial pressures tells us that total pressure of the gas is equal to the sum of the partial pressures of the individual gas components
∴ It explains the independent nature of the gas molecule
At a given temperature, for a given gas, all molecules of the gas do not have same speed but overall the average speed of the gas remains same because speed of each molecule of a gas depends on the collision with other molecules of the gas and as the collisions can't be the same therefore molecules of a gas have different speeds
Actually pressure is generated by the collisions of molecules with the container walls because when the gas molecules collide with the container they generate a force which in turn produce the pressure
At high pressure gas do not tend to behave ideally as there will be intermolecular forces and we will write the ratio of PV/nRT as Z which is the compressibility factor of a gas and it will be different for different gases as different gases has different intermolecular forces of attraction
Which of the following statements could be true regarding polar molecules? Choose one or more: A polar molecule will not contain polar bonds. A polar molecule may have one or more lone pairs. A polar molecule has an uneven distribution of electron density. The bond dipoles in a polar molecule will cancel. A polar molecule will never contain nonpolar bonds.
Answer:
A polar molecule may have one or more lone pairs.
A polar molecule has an uneven distribution of electron density.
Explanation:
A molecule in which the dipole bonds will not cancel is a polar molecule. It depends on the geometry of the molecule, and the direction of the dipole. So, to be polar, the molecule must have at least one polar bond. It may have nonpolar bonds, but the total dipole must be different from 0.
A polar bond is formed between elements that have large differents in electronegativities, such as chlorine and hydrogen. When an atom has a large electronegativity, it has lone pairs of electrons in a bond because it has a small size and a great number of electrons in the valence shell. So, a polar molecule may have on or more lone pairs.
Because of the lone pairs presented, and because of the dipole different from 0, it will be partial charges in the atoms, so a polar molecule has an uneven distribution of electrons density.
A polar molecule is a molecule that has opposite charges at both ends of the molecule.
The true statements about polar molecules are;
A polar molecule has an uneven distribution of electron density. A polar molecule may have one or more lone pairs.To say that a molecule is polar means that it has opposite charges at either end of the molecule.
These charges result from uneven distribution of electrons in the molecule. This uneven distribution of electrons in the molecule is also caused by the large difference in electronegativity between the bonding atoms. This electronegativity difference causes the electron density of the bond to draw closer one of bonding atoms thereby conferring a partial negative charge on that atom and a resulting partial positive charge on the other atom.
Many polar molecules have lone pairs on atoms in the molecule such as NH3, H2O, H2S, etc.
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Which element is reduced in the following reaction? Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O Which element is reduced in the following reaction? Fe2S3 + 12HNO3 2Fe(NO3)3 + 3S + 6NO2 + 6H2O H N O NO2 S
Answer: The nitrogen atom is getting reduced.
Explanation:
Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
The given chemical equation follows:
[tex]Fe_2S_3+12HNO_3\rightarrow 2Fe(NO_3)_3+3S+6NO_2+6H_2O[/tex]
On reactant side:
Oxidation state of iron atom = +3
Oxidation state of nitrogen atom = +5
Oxidation state of sulfur atom = -2
On product side:
Oxidation state of iron atom = +3
Oxidation state of sulfur atom = 0
Oxidation state of nitrogen atom in [tex]NO_2[/tex] = +4
Oxidation state of nitrogen atom in [tex]Fe(NO_3)_3[/tex] = +5
As, the oxidation state of nitrogen atom in [tex]NO_2[/tex] is decreasing from +5 to +4. So, it is getting reduced.
And, oxidation state of sulfur atom is increasing from -2 to 0. So, it is getting oxidized.
Hence, the nitrogen atom is getting reduced.
The element that gets reduced in the chemical reaction is nitrogen.
What is reduction?Reduction in chemistry means when an atom accepts electrons to become more negatively charged.
According to this question, the following chemical reaction is given:
Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O
On reactant side:
Oxidation state of iron atom = +3Oxidation state of nitrogen atom = +5Oxidation state of sulfur atom = -2On product side:
Oxidation state of iron atom = +3Oxidation state of sulfur atom = 0Oxidation state of nitrogen atom in = +4As indicated above, the oxidation number of nitrogen reduces from +5 to +4, hence, it is the element that gets reduced.
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What is electron affinity?energy change associated with the gaining of several of electrons required to complite the outermost orbital of the atom in the gaseous stateenergy change associated with the gaining of a proton by the atom in the gaseous stateenergy change associated with the gaining of an electron by the atom in the gaseous stateenergy change associated with the gaining of several electrons to complete the outermost sublevel of the atom in the gaseous state
Answer:
Energy change associated with the gaining of an electron by the atom in the gaseous state.
Explanation:
What is electron affinity?
energy change associated with the gaining of several electrons required to complete the outermost orbital of the atom in the gaseous state. FALSE. Electron affinity is associated with the gain of only 1 electron.energy change associated with the gaining of a proton by the atom in the gaseous state. FALSE. Protons cannot be gained nor lost.energy change associated with the gaining of an electron by the atom in the gaseous state. TRUE. This is the definition of electron affinity.energy change associated with the gaining of several electrons to complete the outermost sublevel of the atom in the gaseous state. FALSE. Electron affinity is associated with the gain of only 1 electron.The 1995 Nobel Prize in chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning the formation and decomposition of ozone in the stratosphere. Rowland and Molina hypothesized that chlorofluorocarbons (CFCs) in the stratosphere break down upon exposure to UV radiation, producing chlorine atoms. Chlorine was previously identified as a catalyst in the breakdown of ozone into oxygen gas. Using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine with ozone. (1) ClO ( g ) + O 3 ( g ) ⟶ Cl ( g ) + 2 O 2 ( g ) Δ H ∘ rxn = − 122.8 kJ mol (2) 2 O 3 ( g ) ⟶ 3 O 2 ( g ) Δ H ∘ rxn = − 285.3 kJ mol (3) O 3 ( g ) + Cl ( g ) ⟶ ClO ( g ) + O 2 ( g ) Δ H ∘ rxn = ?
Answer:
[tex]\Delta H_{rxn3}=-162.5 kJ/mol[/tex]
Explanation:
The reaction we need to calculate:
[tex]O_3 (g) + Cl (g) \longrightarrow ClO (g) + O_2 (g)[/tex]
1) [tex]O_3 (g) + ClO (g) \longrightarrow Cl (g) +2 O_2 (g)[/tex]
[tex]\Delta H_{rxn}=-122.8 kJ/mol[/tex]
We need the ClO in the products side, so we use the inverse of this reaction:
[tex]Cl (g) +2 O_2 (g) \longrightarrow O_3 (g) + ClO (g) [/tex]
[tex]\Delta H_{rxn1}=122.8 kJ/mol[/tex]
2) [tex]2 O_3 (g) \longrightarrow 3 O_2 (g)[/tex]
[tex]\Delta H_{rxn2}=-285.3 kJ/mol[/tex]
Now we need to combine this two:
[tex]Cl (g) +2 O_2 (g) + 2 O_3 (g)\longrightarrow O_3 (g) + ClO (g) + 3 O_2 (g) [/tex]
[tex]Cl (g) + O_3 (g)\longrightarrow ClO (g) + O_2 (g) [/tex]
The enthalpy of reaction:
[tex]\Delta H_{rxn3}=\Delta H_{rxn1}+ \Delta H_{rxn2}=122.8 kJ/mol-285.3 kJ/mol[/tex]
[tex]\Delta H_{rxn3}=-162.5 kJ/mol[/tex]
Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presence of superheated steam. (a) Given these data, find ΔH o rxn , ΔG o rxn , and ΔS o rxn at 298 K: ΔH o f (kJ/mol) ΔG o f (kJ/mol) S o (J/mol·K) Ethylbenzene, C6H5−CH2CH3 −12.5 119.7 255 Styrene, C6H5−CH=CH2 103.8 202.5 238 ΔH o rxn ΔG o rxn ΔS o rxn kJ kJ J/K (b) At what temperature is the reaction spontaneous? °C (c) What are ΔG o rxn and K at 600.°C? ΔG o rxn K kJ/mol × 10 Enter your answer in scientific notation.
The change in enthalpy, entropy, and free energy were calculated for the dehydrogenation reaction of ethylbenzene into styrene. The reaction was found to be endothermic and results in a decrease in overall disorder. Under the given conditions, the reaction will never be spontaneous.
Explanation:The processes involved in the production of styrene from ethylbenzene are fairly complex and require knowledge of thermodynamics. We'll begin with ΔH°rxn, which is found by subtracting the enthalpy (ΔH) of the reactants from that of the products: ΔH°rxn = [ΔH°f(styrene)] - [ΔH°f(ethylbenzene)] = 103.8 kJ/mol - (-12.5 kJ/mol) = 116.3 kJ/mol. This means the reaction is endothermic, as heat is absorbed.
The change in entropy ΔS°rxn, obtained likewise, is [S°(styrene) - S°(ethylbenzene)] = (238 J/mol·K - 255 J/mol·K) = -17 J/mol·K. This indicates a decrease in disorder in the system.
With these, we can calculate the change in free energy ΔG°rxn at a given temperature (T) using the equation ΔG°rxn = ΔH°rxn - TΔS°rxn. Substituting the known values at 298 K, ΔG°rxn = 116.3 kJ/mol - (298 K)(-17 J/mol·K) = 121.2 kJ/mol, indicating a non-spontaneous reaction.
For the reaction to be spontaneous, ΔG°rxn must be less than zero. Solving for T in the above equation with ΔG°rxn = 0, yields T = ΔH°rxn / ΔS°rxn = 116.3 kJ/mol / -17 J/mol·K ≈ -6840 K. This value is negative, implying the reaction is never spontaneous under the given conditions.
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The standard cell potential (E°cell) for the voltaic cell based on the reaction below is ________ V. Sn2+ (aq) + 2Fe3+ (aq) → 2Fe2+ (aq) + Sn4+ (aq) The standard cell potential (E°cell) for the voltaic cell based on the reaction below is ________ V. Sn2+ (aq) + 2Fe3+ (aq) 2Fe2+ (aq) + Sn4+ (aq) +1.39 +1.21 +0.46 -0.46 +0.617
The standard cell potential (E°cell) for the given electrochemical reaction is approximately +0.9 V. Here option C is correct.
Identifying the Redox Reactions and Potentials
The given balanced equation reveals two processes: an oxidation and a reduction.
Oxidation: [tex]Sn$^{2+}$[/tex] loses two electrons to become [tex]Sn$^{4+}$[/tex]. This can be represented as:
[tex]$$Sn^{2+}(aq) \rightarrow Sn^{4+}(aq) + 2e^-$$[/tex]
Reduction: [tex]Fe$^{3+}$[/tex] gains one electron to become [tex]Fe$^{2+}$[/tex]. This can be represented as:
[tex]$$Fe^{3+}(aq) + e^- \rightarrow Fe^{2+}(aq)$$[/tex]
Next, we need to find the standard reduction potentials (E°) for these half-reactions. Remember, we need the reverse of the oxidation reaction, so look for the potential of:
[tex]$$Sn^{4+}(aq) + 2e^- \rightarrow Sn^{2+}(aq)$$[/tex]
From the provided table, find the value for this and the given reduction reaction:
Oxidation: E° = -0.154 V
Reduction: E° = +0.771 V
Calculating the Cell Potential (E°cell)
The reduction occurs at the cathode (positive) and the oxidation at the anode (negative). Therefore, the cell potential is calculated by:
[tex]$$E^{\circ}_{cell} = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$$[/tex]
Substituting the values:
[tex]$$E^{\circ}_{cell} = (+0.771 \text{ V}) - (-0.154 \text{ V})$$[/tex]
[tex]$$E^{\circ}_{cell} \approx +0.925 \text{ V}$$[/tex]
However, the answer choices only have one decimal place, so we need to round: [tex]$$E^{\circ}_{cell} \approx \boxed{+0.9 \text{ V}}$$[/tex]. Here option C is correct.
Complete question:
The standard cell potential (E°cell) for the voltaic cell based on the reaction below is ________.
Sn2+(aq) + 2Fe3+(aq) --> 2Fe2+(aq) + Sn4+(aq)
Half-reaction E°(V)
Cr3+(aq) + 3e - --> Cr(s) - 0.74
Fe2+ (aq) + 2e - --> Fe(s) - 0.440
Fe3+(aq) + e - --> Fe2+(s) + 0.771
Sn4+(aq) + 2e - --> Sn2(aq) + 0.154
a) +0.46 V
b) +0.617 V
c) +0.9 V
d) -0.46 V
e) +1.21 V
Assume that in each pair of compounds be- low the type of crystal structure is similar. Consider the strength of the attractive force
bonding the ions together in each compound.
In which case would the relative strengths be ranked correctly?
1. MgCl2 < CaBr2
2. LiCl > LiBr
3. KF < KCl
4. NaI > Li
The attraction between ions in a crystal depends on the size of the ions. Since chloride ion is smaller than bromide ion, the statement LiCl > LiBr is correct.
The attraction between oppositely charged ions in the crystal lattice of an ionic compound has a lot to do with the size of the ions.
For a given crystal lattice, smaller ions implies greater attraction and higher lattice energy.
LiCl and LiBr contains the same cation which is the lithium ion so we turn our attention to the anions.
Chloride ion is smaller than bromide ion, the statement LiCl > LiBr is correct based on the strength of the attractive force bonding the ions together in each compound .
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Final answer:
The correct ranking based on ionic bond strength due to charge and size differences is LiCl > LiBr. This considers the Coulombic attraction, where a smaller ionic radius and a higher charge increase the strength of the bond.
Explanation:
The question asks to rank the strength of the attractive force bonding the ions together in each compound pair, correctly. When considering the relative strengths of ionic bonds, factors such as ionic size and charge play significant roles. The strength of an ionic bond increases with the charge of the ions and decreases with the size of the ions. Therefore, compounds with ions of higher charges and smaller sizes have stronger bonds due to the higher Coulombic attraction.
MgCl2 < CaBr2 - Incorrect because Mg2+ has a smaller ionic radius and a higher charge density than Ca2+, leading to stronger bonding in MgCl2.
LiCl > LiBr - Correct since Cl- is smaller than Br-, leading to a stronger Coulombic attraction in LiCl.
KF < KCl - Incorrect because F- is smaller than Cl-, which should result in a stronger attraction in KF due to the smaller ionic radius of F-.
NaI > Li - This comparison is not valid as it compares an ionic compound with an element.
Therefore, the correct ranking of relative strengths based on the charge and size of ions is found in option 2: LiCl > LiBr.
Which of the following metal ions is most likely to exist as a stable eight-coordinate complex with fluoride? Cu2+ Cu+ Fe3+ Ag3+ U4+
Answer:
U⁴⁺
Explanation:
Fluorine stabilizes metals in higher oxidation states with high M:F ratios.
Hard acids prefer to bind to hard bases.
U⁴⁺ is the hardest acid in the list and F⁻ is a hard base, so an eight-coordinate complex of U⁴⁺ is the most stable.
The metal ion that can most adequately exist in the form of 8-coordinate complex would be:
d). U⁴⁺
'Coordination Complex' is characterized as either an ion or a compound that consists of the main metallic atom that is linked through interrelated bonds employing a fixed number of nearby encircling atoms.In the given question, the metal ion U⁴⁺ would most adequately retain in the form of 8 coordinate complexes because it allows metals to sustain in a greater rate of oxidation having a greater ratio of M:F. Since hard bases are defined by stronger acids, U⁴⁺ would be most appropriate as it is a hard acid while F⁻ is a strong base which makes it reliable.
Thus, option d is the correct answer.
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A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in a 25.0 L container at 713 K. At equilibrium, 72.6 g H2S (g) is present? What is the value of K at 713 Kelvin for the following reaction?
Answer: The value of [tex]K_c[/tex] is coming out to be 0.412
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For [tex]Sb_2S_3[/tex]Given mass of [tex]Sb_2S_3[/tex] = 1.00 kg = 1000 g (Conversion factor: 1 kg = 1000 g)
Molar mass of [tex]Sb_2S_3[/tex] = 339.7 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }Sb_2S_3=\frac{1000g}{339.7g/mol}=2.944mol[/tex]
For hydrogen gas:Given mass of hydrogen gas = 10.0 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol[/tex]
For hydrogen sulfide:Given mass of hydrogen sulfide = 72.6 g
Molar mass of hydrogen sulfide = 34 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hydrogen sulfide}=\frac{72.6g}{34g/mol}=2.135mol[/tex]
The chemical equation for the reaction of antimony sulfide and hydrogen gas follows:
[tex]Sb_2S_3(s)+3H_2(g)\rightarrow 2Sb(s)+3H_2S(g)[/tex]
Initial: 2.944 5
At eqllm: 2.944-x 5-3x 2x 3x
We are given:
Equilibrium moles of hydrogen sulfide = 2.135 moles
Calculating for 'x', we get:
[tex]\Rightarrow 3x=2.135\\\\\Rightarrow x=\frac{2.135}{3}=0.712[/tex]
Equilibrium moles of hydrogen gas = (5 - 3x) = (5 - 3(0.712)) = 2.868 moles
Volume of the container = 25.0 L
Molarity of a solution is calculated by using the formula:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex]
The expression of [tex]K_c[/tex] for above equation, we get:
[tex]K_c=\frac{[H_2S]^3}{[H_2]^3}[/tex]
The concentration of solids and liquids are not taken in the expression of equilibrium constant.
[tex]K_c=\frac{(\frac{2.135}{25})^3}{(\frac{2.868}{25})^3}\\\\K_c=0.412[/tex]
Hence, the value of [tex]K_c[/tex] is coming out to be 0.412
The estimated value of K for the reaction Sb2S3 (s) + 3 H2 (g) <--> 2 Sb (s) + 3 H2S (g) at 713K is approximately 1. This estimate is based on initial and equilibrium moles and concentrations of H2 and H2S.
Explanation:The reaction is Sb2S3 (s) + 3 H2 (g) <--> 2 Sb (s) + 3 H2S (g). This type of equation is a chemical equilibrium reaction. Because Sb2S3 and Sb are solids, they are not included in the equilibrium constant expression. Only H2 (g) and H2S (g) are included. To solve the problem, the initial moles of H2 and H2S need to be calculated, and the moles of H2S at equilibrium can be found. After finding the moles, they can be converted into concentrations by dividing by the volume of the container (25.0 L).
Initially, we have 1.00 kg of Sb2S3, which is excess and doesn't appear in the equilibrium equation. So, it can be ignored. Similarly, for hydrogen, 10 g converts to approximately 5 moles. Since no H2S is present initially, let's denote its moles as 0. When the equilibrium is established we know that only 72.6g of H2S is present and that is approximately 3 moles. By stoichiometry, 1 mole of H2 will form 1 mole of H2S. Therefore, 3 moles of H2 have been used to form 3 moles of H2S. Subtracting the used quantity from the total, there is approximately 2 moles of H2 left. Plugging all these into the equilibrium equation, we get K = [H2S]^3 / [H2]^3 which is approximately equal to 1.
This process illustrates the concept of chemical equilibrium and how to calculate the equilibrium constant, K, for a reaction. It emphasizes the importance of stoichiometry, the mole concept, and concentration when dealing with equilibrium reactions.
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The complete question is here:
A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in 25.0L container at 713K. At equilibrium, 72.6 g H2S (g) is present? What is the value of K at 713K for the following reaction? The value of Kp?
Sb2S3 (s) + 3 H2 (g) <--> 2 Sb (s) + 3 H2S (g)
How will you known that you have reached the equivalence point when titrating the NaOH solution with HCl?
Answer:
When the solution (with phenolphthalein) changes to colorless
Explanation:
When titrating with HCl is common to add phenolphthalein as an acid-base indicator.
Phenolphthalein is pink or fucsia when added into a basic solution. On the other hand when it is in acid solutions, is colorless.
So, when titrating, the NaOH solution will be initialy pink due to the phenolphthalein and when reaching the equivalence point, that color will fade out into colorless. This is how you know you hace reached the equivalent point.
On the basis of the Ksp values below, what is the order of the solubility from least soluble to most soluble for these compounds?
AgBr: Ksp = 5.4 x 10-13
Ag2CO3: Ksp = 8.0 x 10-12
AgCl: Ksp = 1.8 x 10-10
Ag2CO3 < AgBr < AgCl
AgBr < Ag2CO3 < AgCl
AgBr < AgCl < Ag2CO3
AgCl < Ag2CO3 < AgBr
Answer:
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Explanation:
The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:
Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).
It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:
Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.
Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that
Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋
Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12
s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶
And for AgCl
AgCl ⇄ Ag⁺ + Cl⁻
Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵
Therefore, AgCl is more soluble than Ag₂CO₃
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Answer:
5.4 x 10^-13
Explanation:
Edmentum says it is correct
If you made a voltaic cell out of this reaction, what half-reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode? If you made a voltaic cell out of this reaction, what half-reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode?
A(aq)+1e−→A−(aq) at the cathode and B(aq)→B+(aq)+1e− at the anode.
A(aq)+1e−→A−(aq) at the anode and B(aq)→B+(aq)+1e− at the cathode
Answer:
A(aq) + 1 e⁻ → A⁻(aq) at the cathode and B(aq) → B⁺(aq) + 1e⁻ at the anode.
Explanation:
In the anode occurs the oxidation, in which the reducing agent loses electrons, and its oxidation number increases.
B(aq) → B⁺(aq) + 1e⁻
In the cathode occurs the reduction, in which the oxidizing agent gains electrons, and its oxidation number decreases.
A(aq) + 1 e⁻ → A⁻(aq)
In a voltaic cell, the anode is where oxidation occurs, and the cathode is where reduction occurs. Electrons flow from the anode (negative electrode) to the cathode (positive electrode). The anode reaction is B(aq) → B+(aq) + 1e−, and the cathode reaction is A(aq) + 1e− → A−(aq).
In a voltaic (galvanic) cell, two half-reactions occur at separate electrodes.
These reactions involve oxidation at the anode and reduction at the cathode.Let's break down the given reactions:
At the anode: The half-reaction is B(aq) → B+(aq) + 1e−. Here, oxidation occurs and electrons are released.At the cathode: The half-reaction is A(aq) + 1e− → A−(aq). Here, reduction occurs and electrons are gained.Electrode Identification
Anode: This is where oxidation takes place (B(aq) → B+(aq) + 1e−). It is the negative electrode.Cathode: This is where reduction takes place (A(aq) + 1e− → A−(aq)). It is the positive electrode.In summary, electrons flow from the anode to the cathode through an external circuit, and a salt bridge maintains charge balance.
Imagine that you have a 7.00 L gas tank and a 3.00 L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 145 atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.
Final answer:
To run out of oxygen and acetylene at the same time, the acetylene tank should be filled to a pressure of 62.14 atm.
Explanation:
To ensure that you run out of oxygen and acetylene at the same time, the number of moles of oxygen and acetylene consumed must be equal. The volume ratio of oxygen and acetylene gases is 7.00 L : 3.00 L, which simplifies to 7 : 3. Therefore, the pressure ratio of oxygen to acetylene should also be 7 : 3. Given that the oxygen tank is filled to a pressure of 145 atm, the acetylene tank should be filled to a pressure of (145 atm) x (3/7) = 62.14 atm.
Which of the following statements is TRUE?
A. Endothermic processes decrease the entropy of the surroundings, at constant T and P.
B. Endothermic processes are never spontaneous.
C. Entropy is not a state function.
D. Exothermic processes are always spontaneous.
E. None of the above are true.
Answer: Option (A) is the correct answer.
Explanation:
Entropy means the degree of randomness present within the molecules of a substance.
And, in an endothermic process heat is absorbed by the reactant molecules. So, when heat is absorbed from the surroundings then there occurs a decrease in heat energy of the surrounding.
As a result, kinetic energy of surrounding elements decreases due to which there is also decrease in the entropy of surroundings.
As spontaneous process is a process that occurs on its own. So, it can be endothermic or exothermic in nature.
A state function is defined as a function that depends on the initial and final states.
Thus, we can conclude that the statement endothermic processes decrease the entropy of the surroundings, at constant T and P, is true.
None of the above statements are true. So the correct statement is E.
A. Endothermic processes decrease the entropy of the surroundings, at constant T and P.
This statement is false. Endothermic processes absorb heat from the surroundings, which can increase the entropy of the system (since disorder increases with heat transfer into the system).
At constant temperature and pressure, the entropy change of the surroundings is related to the heat absorbed or released by the system. For an endothermic process, the surroundings lose heat, which would decrease the entropy of the surroundings, not the system.
B. Endothermic processes are never spontaneous.
This statement is false. The spontaneity of a process is determined by the Gibbs free energy change (G), which depends on both the enthalpy change (H) and the entropy change (S) of the system, according to the equation G = H - TS.
An endothermic process (H > 0) can be spontaneous if the entropy increase of the system is sufficient to overcome the positive enthalpy change, particularly at high temperatures.
C. Entropy is not a state function.
This statement is false. Entropy is indeed a state function, meaning its value depends only on the initial and final states of the system and not on the path taken to go from one state to another.
This is analogous to other state functions like internal energy, enthalpy, and Gibbs free energy.
D. Exothermic processes are always spontaneous.
This statement is false. While exothermic processes (H < 0) release heat, which can contribute to spontaneity, they are not always spontaneous.
The spontaneity of an exothermic reaction also depends on the entropy change of the system.
If the entropy decrease associated with the process is large enough to result in a positive Gibbs free energy change, the process will not be spontaneous.
Therefore, since all the given statements are false, the correct answer is E. None of the above are true.
A 1.0 mol sample of X(g) and a 1.0 mol sample of Q(g) are introduced into an evacuated, rigid 10.0 L container and allowed to reach equilibrium at 50ºC according to the equation above. At equilibrium, which of the following is true about the concentrations of the gases?
X(g) + 2 Q(g) --> R(g) + Z(g)
Kc = 1.3 x 105 at 50ºC
[R] = ½ [Q]
[X] = [Q] = [R] = [Z]
[Q] = ½ [X]
[R] = [Z] > Q
Answer:
[R] = [Z] > Q
Explanation:
Given the equilibrium:
X(g) + 2 Q(g) ⇄ R(g) + Z(g)
The equilibrium constant will be given by:
Kc = 1.3 x 105 at 50ºC = [R] x [Z] / ( [X] x [Q]² )
This value is telling us that at equilibrium the products quantities will be much higher than the reactants.
Looking at the answer choices we see that (a) is false since this is an equilibrium not a areaction that goes to completion;(b) is is false because again this is an equilibrium and it would also imply that Kc is one; (c) is false because in the reaction the ratio is 1X : 2Q.
The last choice is the true statement since the equilibrium constant favors the products. This statement would haven true if Kc had been much less than one.
The concentration of the products are greater than the concentration of the reactants.
The correct answer to the question is [R] = [Z] > Q
Equilibrium constant, K is simply defined as the ratio of the concentration of the products raised to their coefficients to the concentration of the reactants raised to their coefficients.
We can write the equilibrium constant for the reaction as follow:
X(g) + 2Q(g) ⇄ R(g) + Z(g)
K = [R][Z] / [X] [Q]²
From the question given above, we were told that the equilibrium constant has a value of 1.3×10⁵ i.e
K = [R][Z] / [X] [Q]²
1.3×10⁵ = [R][Z] / [X] [Q]²
From the above, we can see that the equilibrium constant has a higher value. This simply means that the concentration of the products is greater than the concentration of the reactants at equilibrium.
Thus, correct answer to the question is:
[R] = [Z] > Q
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What can be said about an Endothermic reaction with a negative entropy change?The reaction isa. spontaneous at all temperatures.b. spontaneous at high temperatures.c. spontaneous at low temperatures.d. spontaneous in the reverse direction at all temperatures.e. nonspontaneous in either direction at all temperatures.What can be said about an Exothermic reaction with a negative entropy change?The reaction isa. spontaneous at all temperatures.b. spontaneous at high temperatures.c. spontaneous at low temperatures.d. spontaneous in the reverse direction at all temperatures.e. nonspontaneous in either direction at all temperatures.
Answer:
1. d. The reaction is spontaneous in the reverse direction at all temperatures.
2. c. The reaction is spontaneous at low temperatures.
Explanation:
The spontaneity of a reaction is associated with the Gibbs free energy (ΔG). When ΔG < 0, the reaction is spontaneous. When ΔG > 0, the reaction is non-spontaneous. ΔG is related to the enthalpy (ΔH) and the entropy (ΔS) through the following expression:
ΔG = ΔH - T. ΔS [1]
where,
T is the absolute temperature (T is always positive)
1. What can be said about an Endothermic reaction with a negative entropy change?
If the reaction is endothermic, ΔH > 0. Let's consider ΔS < 0. According to eq. [1], ΔG is always positive. The reaction is not spontaneous in the forward direction at any temperature. This means that the reaction is spontaneous in the reverse direction at all temperatures.
2. What can be said about an Exothermic reaction with a negative entropy change?
If the reaction is exothermic, ΔH < 0. Let's consider ΔS < 0. According to eq. [1], ΔG will be negative when |ΔH| > |T.ΔS|, that is, at low temperatures.
Final answer:
An endothermic reaction with a negative entropy change is nonspontaneous at all temperatures. An exothermic reaction with a negative entropy change can be spontaneous at low temperatures but not necessarily at high temperatures.
Explanation:
An endothermic reaction with a negative entropy change is one that absorbs energy from its surroundings and results in a decrease in the randomness or disorder of the system. According to thermodynamics, the spontaneity of a reaction depends on the change in free energy (ΔG), which is influenced by enthalpic (ΔH) and entropic (-TΔS) factors.
An endothermic reaction with a negative entropy change would have a positive ΔH and a negative TΔS, making the ΔG positive at all temperatures. Therefore, such a reaction is nonspontaneous at either direction at all temperatures (e).
In contrast, an exothermic reaction with a negative entropy change releases energy and also results in a decrease in the system's disorder. This reaction would have a negative ΔH and a negative TΔS. For this reaction, ΔG could be negative at low temperatures, resulting in spontaneity (spontaneous at low temperatures (c)). At high temperatures, the positive TΔS may outweigh the negative ΔH, resulting in a positive ΔG and nonspontaneity.
Kemmi Major does some experimental work on the combustion of sucrose: C12H22O11(s) 12 O2(g) → 12 CO2(g) 11 H2O(g) She burns a 0.05392 g pellet of sucrose in a bomb calorimeter with excess oxygen. She determines the qrxn to be –916.6 J for the reaction. Calculate the ∆H value for the combustion reaction. (Round the answer to 3 significant digits, units of kJ, pay attention to positive or negative.
Answer: [tex]5.81\times 10^6J/mol[/tex]
Explanation:
Heat of combustion is the amount of heat released when 1 mole of the compound is completely burnt in the presence of oxygen.
[tex]C_{12}H_{22}O_{11}(s)+12O_2\rightarrow 12CO_2(g)+11H_2O(g)[/tex]
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.05392g}{342g/mol}=1.577\times 10^{-4}moles[/tex]
Thus [tex]1.577\times 10^{-4}moles[/tex] of sucrose releases = 916.6 J of heat
1 mole of sucrose releases =[tex]\frac{916.6}{1.577\times 10^{-4}}\times 1=5.81\times 10^6J[/tex] of heat
Thus ∆H value for the combustion reaction is [tex]5.81\times 10^6J/mol[/tex]
The ∆H for the combustion reaction of 0.05392 g of sucrose, given that the qrxn is -916.6 J, is calculated to be approximately -58364.2 kJ. This value is negative indicating an exothermic reaction where heat is released.
Explanation:In this problem, we are tasked with finding the enthalpy change, or ∆H, for the combustion reaction of sucrose. The heat of the reaction, qrxn, is given as -916.6 J for a 0.05392 g pellet of sucrose.
Firstly, we have to convert the grams of sucrose to moles. The molecular weight of sucrose, C12H22O11, is about 342.3 g/mol so the moles of sucrose burned would be 0.05392 g ÷ 342.3 g/mol = 0.000157 mol.
Next, the change in enthalpy per mole of sucrose can be calculated using ∆H = qrxn / moles of sucrose which gives us -916.6 J / 0.000157 mol = -58364152.2293 J/mol. Since the question requires the answer in kJ, we must divide this by 1000 to convert J to kJ. Hence, ∆H = -58364.1523 kJ/mol.
It is important to note that ∆H is negative because the reaction is exothermic, meaning heat is released during the combustion of sucrose. Therefore, the ∆H for the combustion reaction of sucrose is approximately -58364.2 kJ.
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Elimination reactions are favored over nucleophilic substitution reactions:
O at high temperatures.
O when tert-butoxide ion is used.
O when 3° alkyl halides are used as substrates.
O when nucleophiles are used which are strong bases and the substrate is a 2° alkyl halide.
O in all of these cases.
In all of these cases.
Elimination reaction compete with nucleophilic substitution reaction and is favored under conditions of
The heat help to reach the energy required for the dehydrohalogenation reaction with a strong base.
terc-butoxide favor the Hoffman elimination product, terc-butoxide tendency is to break out Zaitseff rule, so in this case the eliminitation in alkyl halides less substituted is favored.
If the substrate of the reaction is an 3rd alkyl halides, the carbocation is more stable so according to Zaitsev's rule in an elimination reaction, the most substituted product will be the most stable, so will be favored.
The nucleophile might to be a strong base because should be strong enough to form a bond with a hydrogen from the alkyl halide and let reaction continues. Regarding to 2° alkyl halide, 3° alkyl halide is favored towards 2° alkyl halide. Being the relative rate of reaction with 3° alkyl halide higher than 2° alkyl halide. That is because provides a greater number of hydrogens for attack by base and hence more favorable for elimination. If the alkyl halide are more branched more stable the alkene.
Elimination reactions are favored over substitution reactions at high temperatures, with tert-butoxide ion, with tertiary alkyl halides, and with strong bases and secondary alkyl halides. Therefore, correct option is: elimination is favored in all these situations.
Elimination reactions are often favored over nucleophilic substitution reactions under certain conditions. The key factors that favor elimination reactions include:
High temperatures: At higher temperatures, elimination (E₂) is favored over substitution (SN₂).Use of tert-butoxide ion: This strong, bulky base favors elimination due to steric hindrance preventing substitution.Use of tertiary (3°) alkyl halides: These substrates are less reactive in SN₂ but highly reactive in E₂.Strong bases with secondary (2°) alkyl halides: Strong bases can favor E₂ over SN₂.Based on these points, elimination reactions are favored in all of these cases.
demonstration of strong electrolytes, weak electrolytes, and nonelectrolytes, Professor Popsnorkle used a lightbulb apparatus that showed how much a solution conducted electricity by the brightness of the lightbulb. When pure water was tested, the bulb did not light. Then Professor Popsnorkle tested the following aqueous solutions. Which one caused the bulb to burn the brightest?
Answer:
Strong electrolyte
Explanation:
Bulb doesn't light when water is used because water doesn't have any appreciable amount of ions for the conduction of electricity.
Strong electrolytes ionize almost completely in solution, thus providing a huge amount of ions to conduct electricity. More the number of ions, more charge is carried and hence greater current and thus leading to bulb glowing brighter.
Weak electrolytes ionize partially in solution, thus providing some amount of ions for carrying charge. This results in a small amount of current going to the bulb and hence the bulb lights up with very low intensity.
Non electrolytes doesn't ionize either and hence is similar to water. These have almost zero conductivity and hence the bulb doesn't light at all.