Answer:
The true statements are given below.
Explanation:
1 In the electron transport chain a series of reactions moves electrons through carriers.
2 Coenzyme Q and cytochrome are the components of the electron transport chain.
During electron transport chain electrons are transported from one electron carrier have high reduction potential to another electron carrier having comparatively low reduction potential than the previous one by a series of redox reactions.
Coenzyme Q is a component of complex 2 whereas cytochrome is an important component of both complex 3 and complex 4.
The electron transport chain is a series of reactions that moves electrons through carriers and is located in the inner mitochondrial matrix in eukaryotic cells and the cytoplasmic membrane in prokaryotic cells. Its products are water and ATP. Coenzyme Q and cytochrome are components of the electron transport chain, but Coenzyme A is not. The electron transport chain operates independently of other metabolic processes.
Explanation:The electron transport chain (ETC) is a series of reactions that moves electrons through carriers. It is located in the inner mitochondrial matrix in eukaryotic cells and the cytoplasmic membrane in prokaryotic cells. The products of the electron transport chain are water and ATP. Coenzyme Q and cytochrome are components of the electron transport chain, but Coenzyme A is not. The electron transport chain operates independently of other metabolic processes.
Acetonitrile (CH3CN) is an important industrial cheical. Among other things, it Is used to make plastic moldings, which have multiple uses, from car parts to Lego bricks. Which statement below about acetonitrile is FALSE? a. Acetonitrile has 16 valence electrons in its Lewis structure. b. Acetonitrile has one triple bond. C. Acetonitrile has one pair of nonbonding electrons. d. All atoms satisfy the octet rule in acetonitrile. e. One carbon atom and the nitrogen atom have nonzero formal charges.
The molecule acetonitrile or CH3CN, indeed has 16 valence electrons and a triple bond follows its Lewis structure. However, the statement that acetonitrile has a pair of nonbonding electrons is incorrect because all 16 valence electrons are used in the molecule's bonds.
Explanation:The molecule in question is acetonitrile, or CH3CN. To analyze these statements, it's crucial to consider the Lewis structure of this molecule: the carbon atom bonded to three hydrogen atoms, that carbon bonded to another carbon atom, and that second carbon bonded to a nitrogen through a triple bond. Hydrogen has 1 valence electron, carbon has 4, and nitrogen has 5.
a. The combined total of valence electrons for acetonitrile is indeed 16 - (3x1 for hydrogen, 2x4 for carbon, and 1x5 for nitrogen).
b. There is a triple bond between the carbon and nitrogen atoms.
c. The entirety of the 16 valence electrons are used in the molecule's bonds, leaving no nonbonding electrons. This statement is false.
d. All atoms in the structure satisfy the octet rule, each having a complete outermost shell.
e. Neither carbon atom nor the nitrogen atom has a nonzero formal charge; all atoms in acetonitrile have a formal charge of zero.
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The false statement about acetonitrile (CH3CN) is that it has one pair of nonbonding electrons. All the outer shell electrons in acetonitrile are involved in bonding, hence, there are no non-bonding electrons in the molecule.
Explanation:Based on your question about acetonitrile (CH3CN), which is an important industrial chemical, the statement that is FALSE is: Acetonitrile has one pair of nonbonding electrons. When drawing the Lewis structure for acetonitrile, we find that all the outer shell electrons are involved in bonding, hence, there are no non-bonding electrons in acetonitrile. The molecule's other properties mentioned in the options are correct: it has 16 valence electrons, a triple bond between the carbon and nitrogen atoms, all atoms satisfy the octet rule, and the carbon attached to the nitrogen, as well as the nitrogen atom itself, have nonzero formal charges due to the triple bond and unequal sharing of electrons.
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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.
What effect will spraying liquid water into the equilibrium given below have if NH3 is far more soluble in water than is N2 or H2?
N2(g) + 3H2(g) 2NH3(g)
A) More NH3(g) will form.
B) More H2(g) will form.
C) Less NH3(g) will form.
D) This will not affect the system.
E) More N2(g) will form.
Answer:
A) More NH3(g) will form.
Explanation:
According to the Le Chatellier Principle if we remove product from a reaction which is in equilibrium, then the effect will to produce more product in order to replace what has been removed.
In this reaction also , on spraying liquid water, Ammonia being water soluble will be consumed. Hence, The reaction will shift in forward direction leading to more formation of Ammonia (NH3).
Hence option A will be correct .
Final answer:
Spraying water into the equilibrium involving N2, H₂, and NH₃ will cause more NH₃ to form because NH₃ is more soluble in water than N₂ or H₂, and removing NH₃ by dissolving it in water shifts the equilibrium towards producing more NH₄. the correct option is A.
Explanation:
The question asks what effect spraying liquid water into the equilibrium N2(g) + 3H2(g) ⇒ 2NH3(g) will have, given that NH₃ is far more soluble in water than N2 or H2. According to Le Chatelier's Principle, if a change is applied to a reaction at equilibrium, the system will adjust to partially counteract that change. In this case, spraying water will remove NH₃ by dissolving it. This reduction in the concentration of NH3 will cause the system to shift towards the product side to replace the removed NH₃, thus more NH₃ will form. Therefore, the correct answer is A) MoreNH₃(g) will form.
A buffer of pH 9.24 is to be prepared from a weak acid and its salt. The best acid from which to prepare the buffer is
(A) phthalic acid, K1 = 1.3 x 10-3 (1st ionization)
(B) hydrohen phthalate, K2 = 3.9 x 10-5
(C) arsenious acid, Ka = 6 x 10-10
(D) formic acid, Ka = 1.8 x 10-5
(E) phenol, Ka = 1.3 x 10-10
Answer:
(C) arsenious acid, Ka = 6 x 10⁻¹⁰
Explanation:
A buffer is prepared by a weak acid and the conjugate base coming from its salt. Its function is to resist abrupt changes in pH when an acid or a base are added. The best working range of a buffer is in the range of pKa ± 1. Let's consider the 5 options and their pKa (pKa = -log Ka).
(A) phthalic acid, K1 = 1.3 x 10⁻³ (1st ionization) pKa = 2.9
(B) hydrogen phthalate, K2 = 3.9 x 10⁻⁵ pKa = 4.4
(C) arsenious acid, Ka = 6 x 10⁻¹⁰ pKa = 9
(D) formic acid, Ka = 1.8 x 10⁻⁵ pKa = 4.7
(E) phenol, Ka = 1.3 x 10⁻¹⁰ pKa = 9.8
The acid whose pKa is closer to the desired pH is arsenious acid. Its working range of pH is 8 - 10. In the second place, phenol could work as a buffer system since the working pH range is 8.8 - 10.8.
How many moles of KOH are in 130.0 mL of a 0.85 M solution of KOH?
Given thatH2(g) + F2(g) -> 2HF(g) => ∆H = -546.6 kJ . mol-12H2(g) + O2(g) -> 2H20(l) => ∆H = -571.6 kJ. mol-1Calculate the value of ∆H for:2F2(g) + 2H20(l) -> 4HF(g) + O2(g)
Answer:
The correct answer is -521.6 KJ/mol
Explanation:
In order to solve the problem, we have to rearrange and to add the reactions with the aim to obtain the requested reaction.
We have:
1) H₂(g) + F₂ (g) → 2HF(g) => ∆H₁ = -546.6 kJ/mol
2) 2H₂(g) + O₂(g) → 2H₂0(l) => ∆H₂ = -571.6 kJ/mol
If we multiply reaction 1 by 2 and add the inverse reaction of 2, we obtain:
2H₂(g) + 2F₂ (g) → 4HF(g) => 4 x ∆H₁ = 4 x (-546.6 kJ/mol)
2H₂0(l) → 2H₂(g) + O₂(g) => (-1) x ∆H₂ = (-1) x (571.6 kJ/mol)
-------------------------------------
2H₂(g) + 2F₂ (g)+ 2H₂0(l) → 4HF(g) + 2H₂(g) + O₂(g)
We eliminate 2H₂(g) is repeated in both members of the reaction, and we obtain: 2F₂(g) + 2H₂0(l) -> 4HF(g) + O₂(g)
Finally, the ΔH of the reaction is calculated as follows:
ΔHtotal= (2 x ΔH₁) + ((-1)ΔH₂
ΔHtotal= (2 x (-546.6 KJ/mol)) + ((-1) x (-571.6 KJ/mol)
ΔHtotal= -521.6 KJ/mol
A certain half-reaction has a standard reduction potential =E0red+0.13V . An engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least 1.10V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell.
Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have? If so, check the "yes" box and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, check the "no" box. yes, there is a minimum. =E0red V no minimum
Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have? If so, check the "yes" box and calculate the maximum. Round your answer to 2 decimal places. If there is no upper limit, check the "no" box. yes, there is a maximum. =E0red V no maximum
By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the cathode of this cell. Note: write the half reaction as it would actually occur at the cathode. e
The minimum standard reduction potential at the cathode in this galvanic cell should be at least 1.23V to provide the needed power. There is no defined maximum limit for this value. A suitable half-reaction for the cathode could involve the reduction of silver ions to silver metal.
Explanation:For a galvanic cell to work, the potential at the cathode must be greater than that at the anode. In this specific case, the reduction potential at the anode, E0anode, is given as +0.13V. The difference in potentials between the cathode and anode corresponds to the provided voltage of the cell, which is 1.10V in this particular scenario.
Using this information, the reduction potential at the cathode can be calculated using E0cell = E0cathode - E0anode formula. Substituting the known values, 1.10V = E0cathode - (+0.13V), we find that E0cathode should be at least 1.23V to provide the necessary power.
There is technically no maximum limit for E0cathode as increasing this value only increases the power provided by the cell. As for a suitable half-reaction, a reduction of silver ions (Ag+) to silver metal (Ag) having a reduction potential of +0.80V could be used:
Ag+ + e- --> Ag
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There is a minimum standard reduction potential for the cathode: 0.97 V. The maximum standard reduction potential is generally assumed to be less than +2.87 V. An example of a balanced cathode half-reaction is 2H⁺ (aq) + 2e⁻ → H₂ (g).
Yes, there is a minimum standard reduction potential for the half-reaction at the cathode. The cell potential, E°cell, is calculated as:
E°cell = E°cathode - E°anodeGiven that E°anode = -0.13 V (since it’s the reduction potential, but we use it as an oxidation potential here), and the cell must provide at least 1.10 V, we have:
1.10 V = E°cathode - (-0.13 V)1.10 V = E°cathode + 0.13 VSolve for E°cathode:
E°cathode ≥ 1.10 V - 0.13 VE°cathode ≥ 0.97 VYes, there is a maximum standard reduction potential. Practically, the maximum standard reduction potential of the cathode is determined by the potential of the reference electrode (often chosen to be +2.87 V for fluorine). Therefore, any standard reduction potential must be less than this value.
One possible balanced equation that can serve as the cathode reaction is the reduction of hydrogen ions:2H⁺ (aq) + 2e⁻ → H₂ (g); E° = 0.00 VThe 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.
A student titrated a 25.00-mL sample of a solution containing an unknown weak, diprotic acid (H2A) with N2OH. If the titration required 17.73 mL of 0.1036 M N2OH to completely neutralize the acid, calculate the concentration (in M) of the weak acid in the sample.
(a) 9.184 x 10 M
(b) 3.674 x 10-2 M
(c) 7.304 x 10-2 M
(d) 7.347 x 10-2 M
(e) 1.469 x 101 M
Answer: The concentration of weak acid is [tex]3.674\times 10^{-2}M[/tex]
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2A[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=2\\M_1=?M\\V_1=25.00mL\\n_2=1\\M_2=0.1036M\\V_2=17.73mL[/tex]
Putting values in above equation, we get:
[tex]2\times M_1\times 25.00=1\times 0.1036\times 17.73\\\\M_1=\frac{1\times 0.1036\times 17.73}{2\times 25.00}=3.674\times 10^{-2}M[/tex]
Hence, the concentration of weak acid is [tex]3.674\times 10^{-2}M[/tex]
The sun produces energy via fusion. One of the fusion reactions that occurs in the sun is 411H→42He+201e How much energy in joules is released by the fusion of 2.01 g of hydrogen-1? Express your answer to three significant figures and include the appropriate units.
Answer: The energy released for the the given amount of hydrogen -1 atom is [tex]1.2474\times 10^{11}J[/tex]
Explanation:
First we have to calculate the mass defect [tex](\Delta m)[/tex].
The given equation follows:
[tex]4_{1}^{1}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+2_0^{1}\textrm{e}[/tex]
To calculate the mass defect, we use the equation:
Mass defect = Sum of mass of product - Sum of mass of reactant
[tex]\Delta m=(2m_{e}+m_{He})-(4m_{H})[/tex]
We know that:
[tex]m_e=0.00054858g/mol\\m_{H}=1.00782g/mol\\m_{He}=4.00260g/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta m=((2\times 0.00054858)+4.00260)-(4\times 1.00782)=-0.027583g=-2.7583\times 10^{-5}kg[/tex]
(Conversion factor: 1 kg = 1000 g )
To calculate the energy released, we use Einstein equation, which is:
[tex]E=\Delta mc^2[/tex]
[tex]E=(-2.7583\times 10^{-5}kg)\times (3\times 10^8m/s)^2[/tex]
[tex]E=-2.4825\times 10^{11}J[/tex]
The energy released for 4 moles of hydrogen atom is [tex]2.4825\times 10^{11}J[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of hydrogen atom = 2.01 g
Molar mass of hydrogen atom = 1 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of hydrogen atom}=\frac{2.01g}{1g/mol}=2.01mol[/tex]
We need to calculate the energy released for the fusion of given amount of hydrogen atom. By applying unitary method, we get:
As, 4 moles of hydrogen atom releases energy of = [tex]2.4825\times 10^{11}J[/tex]
Then, 2.01 moles of hydrogen atom will release energy of = [tex]\frac{2.4825\times 10^{11}}{4}\times 2.01=1.2474\times 10^{11}J[/tex]
Hence, the energy released for the the given amount of hydrogen -1 atom is [tex]1.2474\times 10^{11}J[/tex]
A 0.327-g sample of azulene (C10H8) is burned in a bomb calorimeter and the temperature increases from 25.20 °C to 27.60 °C. The calorimeter contains 1.17×103 g of water and the bomb has a heat capacity of 786 J/°C. Based on this experiment, calculate ΔE for the combustion reaction per mole of azulene burned (kJ/mol). C13H24O4(s) + 17 O2(g) 13 CO2(g) + 12 H2O(l) E =______ kJ/mol.
Explanation:
The given data is as follows.
Molecular weight of azulene = 128 g/mol
Hence, calculate the number of moles as follows.
No. of moles = [tex]\frac{mass}{\text{molecular weight}}[/tex]
= [tex]\frac{0.392 g}{128 g/mol}[/tex]
= 0.0030625 mol of azulene
Also, [tex]-Q_{rxn} = Q_{solution} + Q_{cal}[/tex]
[tex]Q_{rxn} = n \times dE[/tex]
[tex]Q_{solution} = m \times C \times (T_{f} - T_{i})[/tex]
[tex]Q_{cal} = C_{cal} \times (T_{f} - T_{i})[/tex]
Now, putting the given values as follows.
[tex]Q_{solution} = 1.17 \times 10^{3} g \times 10^{3} \times 4.184 J/g^{o}C \times (27.60 - 25.20)^{o}C[/tex]
= 11748.67 J
So, [tex]Q_{cal} = 786 J/^{o}C \times (27.60 - 25.20)^{o}C[/tex]
= 1886.4 J
Therefore, heat of reaction will be calculated as follows.
[tex]-Q_{rxn}[/tex] = (11748.67 + 1886.4) J
= 13635.07 J
As, [tex]Q_{rxn} = n \times dE[/tex]
13635.07 J = [tex]-n \times dE[/tex]
dE = [tex]\frac{13635.07 J}{0.0030625 mol}[/tex]
= 4452267.75 J/mol
or, = 4452.26 kJ/mol (as 1 kJ = 1000 J)
Thus, we can conclude that [tex]\Delta E[/tex] for the given combustion reaction per mole of azulene burned is 4452.26 kJ/mol.
Suppose that the mechanism of hydrolysis of tert- butyl chloride preceded by a different mechanism so that the predicted rate law was: (d[RCl]/dt)=k[RCl][H2O]. From this experiment, would you be able to distinguish between this mechanism (which gives an order creation of reaction =2) and the mechanism proposed above (which gives a Diest order rate law)? Explain your answer.
Answer:
B, because that was the answer on quizlet.
Explanation:
In a galvanic cell,a oxidation occurs at the (name of electrode)b.the cathode is the (sign) electrodec.cations flow in solution toward the (name of electrode)d.electrons flow from the (name of electrode) to (name of electrode)
In a galvanic cell, oxidation occurs at the anode, the cathode is the positive electrode, cations flow toward the cathode, and electrons flow from the anode to the cathode.
Explanation:In a galvanic cell, the following processes occur:
Oxidation occurs at the anode, which is the electrode in the left half-cell in a standard galvanic cell diagram.The cathode is the positive electrode, which is where the reduction occurs, and is located in the right half-cell in a standard galvanic cell diagram.Cations in the solution flow toward the cathode, which has a positive charge due to the reduction process that occurs there. Electrons flow from the anode to the cathode, moving through the external circuit and enabling electrical work to be done.Learn more about Galvanic Cell here:https://brainly.com/question/32505497
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A solution contains 0.021 M Cl? and 0.017 M I?. A solution containing copper (I) ions is added to selectively precipitate one of the ions. At what concentration of copper (I) ion will a precipitate begin to form? What is the identity of the precipitate? Ksp(CuCl) = 1.0 × 10-6, Ksp(CuI) = 5.1 × 10-12.
A) 4 .8 × 10-5 M, CuClB) 3 .0 × 10-10 M, CuIC) 3 .0 × 10-10 M, CuClD) 4 .8 × 10-5 M, CuIE) N o precipitate will form at any concentration of copper (I).
Answer:
A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M
The precipitate formed is CuI
Explanation:
Step 1: Data given
The solution contains 0.021 M Cl- and 0.017 M I-.
Ksp(CuCl) = 1.0 × 10-6
Ksp(CuI) = 5.1 × 10-12.
Step 2: Calculate [Cu+]
Ksp(CuCl) = [Cu+] [Cl-]
1.0 * 10^-6 = [Cu+] [Cl-]
1.0 * 10^-6 = [Cu+] [0.021]
[Cu+] = 1.0 * 10^-6 / 0.021
[Cu+] = 4.76 *10^-5 M
Ksp(CuI) = [Cu] [I]
5.1 * 10^-12 = [Cu+] [I-]
5.1 * 10^-12 =[Cu+] [0.017]
[Cu+] = 5.1 * 10^-12 / 0.017
[Cu+] = 3.0 *10^-10 M
[Cu+]from CuI hast the lowest concentration
A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M
The precipitate formed is CuI
A scientist measures the standard enthalpy change for the following reaction to be -213.5 kJ: CO(g) 3 H2(g)CH4(g) H2O(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(g) is kJ/mol. Submit AnswerRetry Entire Group
Answer:
Approximately [tex]\rm -249.4\; kJ \cdot mol^{-1}[/tex].
Explanation:
[tex]\rm CO\; (g) + 3\; H_2\; (g) \to CH_4\; (g) + H_2 O\; (g)[/tex].
Note that hydrogen gas [tex]\rm H_2\; (g)[/tex] is the most stable allotrope of hydrogen. Since [tex]\rm H_2[/tex] is naturally a gas under standard conditions, the standard enthalpy of formation of [tex]\rm H_2\; (g)[/tex] would be equal to zero. That is:
[tex]\Delta H^{\circ}_f(\rm H_2\; (g)) = 0[/tex]Look up the standard enthalpy of formation for the other species:
[tex]\Delta H^{\circ}_f(\rm CO\; (g)) = -110.5\; kJ \cdot mol^{-1}[/tex], [tex]\Delta H^{\circ}_f(\rm CH_4\; (g)) = -74.6\; kJ \cdot mol^{-1}[/tex].(Source: CRC Handbook of Chemistry and Physics, 84th Edition (2004).)
[tex]\displaystyle \Delta H^{\circ}_\text{reaction} = \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants})[/tex].
In other words, the standard enthalpy change of a reaction is equal to:
the sum of enthalpy change of all products, minusthe sum of enthalpy change of all reactants.In this case,
[tex]\begin{aligned} & \sum \Delta H^{\circ}_f(\text{products}) \\ =& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + \Delta H^{\circ}_f(\mathrm{H_2O\;(g)})\end{aligned}[/tex].
[tex]\begin{aligned} & \sum \Delta H^{\circ}_f(\text{reactants}) \\ =& \Delta H^{\circ}_f(\mathrm{CO\;(g)}) + 3\times \Delta H^{\circ}_f(\mathrm{H_2\;(g)})\end{aligned}[/tex].
Note that the number [tex]3[/tex] in front of [tex]\Delta H^{\circ}_f(\mathrm{H_2\;(g)})[/tex] corresponds to the coefficient of [tex]\rm H_2[/tex] in the chemical equation.
[tex]\begin{aligned}&\Delta H^{\circ}_\text{reaction} \\ =& \sum \Delta H^{\circ}_f(\text{products}) - \sum \Delta H^{\circ}_f(\text{reactants})\\ =& \left(\Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + \Delta H^{\circ}_f(\mathrm{H_2O\;(g)})\right) \\ &- \left(\Delta H^{\circ}_f(\mathrm{CO\;(g)}) + 3\times \Delta H^{\circ}_f(\mathrm{H_2\;(g)})\right) \\ =& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 \times 0 -110.5)\end{aligned}[/tex].
In other words,
[tex]\begin{aligned} & \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) + (-74.6) - (3 \times 0 -110.5) \\=& \Delta H^{\circ}_\text{reaction} = -213.5\; \rm kJ\cdot mol^{-1} \end{aligned}[/tex].
Therefore,
[tex]\begin{aligned}& \Delta H^{\circ}_f(\mathrm{CH_4\;(g)}) \\ =& -213.5 - ((-74.6) - (3 \times 0 -110.5)) \\=& -249.4\; \rm kJ\cdot mol^{-1} \end{aligned}[/tex].
Which one of the following processes produces a decrease in the entropy of the system?
A. boiling water to form steam
B. dissolution of solid KCl in water
C. mixing of two gases into one container
D. freezing water to form ice
E. melting ice to form water.
Explanation:
Entropy -
In a system, the randomness is measured by the term entropy .
Randomness basically refers as a form of energy that can not be used for any work.
The change in entropy is given by amount heat per change in temperature.
When solid is converted to liquid or gas entropy increases,
As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to liquid or gas, the force of attraction between the molecule decreases and hence entropy increases.
So,
The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,
And
If the particles are loosely held , the entropy will increase , i.e. , when gas is converted to liquid or solid .
A. boiling water to form steam ,
change of state from liquid to gas ,
and , hence entropy increases .
B. dissolution of solid KCl in water ,
the number of particles increases ,
and hence , entropy increases .
C. mixing of two gases into one container ,
the number of particles increases ,
and hence , entropy increases .
D. freezing water to form ice ,
change of state from liquid to solid ,
and hence , entropy decreases .
E. melting ice to form water .
change of solid to liquid ,
and hence , entropy increases .
Final answer:
The process that results in a decrease in entropy is freezing water to form ice, as it involves water transitioning from a disordered liquid state to an ordered solid state.
Explanation:
The process that produces a decrease in the entropy of the system is freezing water to form ice (Option D). During freezing, water molecules move from a higher state of disorder in the liquid phase to a more ordered solid phase, resulting in reduced entropy. Freezing involves the transition of water from the liquid phase, where the molecules are more disordered, to the solid phase, where they are arranged in a structured pattern. This structured pattern, characterized by a fixed position of the molecules in the crystal lattice of ice, embodies a lower state of entropy.
Even though freezing results in a decrease in the system's entropy, this does not violate the second law of thermodynamics because the surrounding environment's entropy increases when heat is released during the freezing process, ensuring that the total entropy of the system plus its surroundings does not decrease.
Which species should have the shortest bond length?
A.) N2
B.) O2
C.) SO2
D.) SO3
Answer:
The correct option is: A) N₂
Explanation:
The Bond length of a chemical bond is the length of a chemical bond formed between two given atoms.
Bond length is inversely proportional to the bond order of the chemical bond, which is the total number of bonds between two atoms. Thus as the bond order increases, the bond length decreases.
A) N₂: The nitrogen-nitrogen bond in dinitrogen is a triple bond (N≡N).
Thus the bond order = 3.
B) O₂: The oxygen-oxygen bond in dioxygen is a triple bond (O=O).
Thus the bond order = 2.
C) SO₂: Sulfur dioxide is a resonance stabilized molecule and its resonance hybrid shows that the sulfur-oxygen bond in sulfur dioxide is a partial double bond.
Thus the bond order = 1.5
D) SO₃: Sulfur trioxide is a resonance stabilized molecule and its resonance hybrid shows that the sulfur-oxygen bond in sulfur trioxide is a partial double bond.
Thus the bond order = 1.33
Since the bond order of N₂ is the largest, therefore, the N-N bond length is the shortest.
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
The amount of I3-(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O3^2-(aq) (thiosulfate ion).
The determination is based on the net ionic equation
2S2O3^2-(aq)+I3^-(aq) ---> S4O6^2-(aq)+31^-(aq)
Given that it requires 35.8 mL of 0.350 M Na2S2O3(aq) to titrate a 30.0-mL sample of I3^-(aq) calculate the molarity of I3^-(aq) in the solution
Answer:
molarity of I3^-(aq) in the solution = 0.21 M
Explanation:
We are given the balanced chemical reaction and the volume and molarity of the Na₂S₂O₃ so we can calculate the moles of the thiosulfate that were required, and then can calculate the molarity the I₃⁻ by the definition of molarity.
First lets convert the volume of Na₂S₂O₃ to liters:
35.8 mL x 1 L/1000 mL = 0.0358 L
# moles Na₂S₂O₃ = 0.350 mol/L x 0.0358 L = 0.0125 mol
From the stoichiometry of the reaction we know 2 mol Na₂S₂O₃ s react with 1 mol I₃⁻ , therefore mol of I₃⁻ will be given by
1 mol I₃⁻ / 2 mol Na₂S₂O₃ x 0.0125 mol Na₂S₂O₃ = 0.0063 mol I₃⁻
and its molarity is:
0.0063 mol I₃⁻ / 0.030 L = 0.21 M
1. In the first step of the mechanism for this process, a phenoxide anion is generated. This phenoxide anion goes on to act as a nucleophile via an SN2 mechanism, displacing the chloride on 3-chloro-1,2-propanediol. Why doesn’t the phenoxide anion act as a base to deprotonate one of the alcohols on 3-chloro-1,2-propanediol? Write a brief, specific explanation (1-2 sentences).
Answer:
See the explanation
Explanation:
In this case, in order to get an elimination reaction we need to have a strong base. In this case, the base is the phenoxide ion produced the phenol (see figure 1).
Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).
Finally, the phenoxide will attack the primary carbon attached to the Cl. The C-Cl bond would be broken and the C-O would be produced at the same time to get a substitution (see figure 1).
Final answer:
The phenoxide anion prefers acting as a nucleophile in an SN2 mechanism on 3-chloro-1,2-propanediol because the electronegative chloride creates an electrophilic carbon that is more reactive towards nucleophilic attack than the deprotonation of an alcohol.
Explanation:
The phenoxide anion does not act as a base to deprotonate one of the alcohols on 3-chloro-1,2-propanediol because the presence of the chloride makes that carbon a more electrophilic center, which is highly susceptible to nucleophilic attack.
In an SN2 mechanism, the nucleophile favors attacking an electrophilic carbon, here significantly activated by the chloride leaving group, rather than deprotonating an alcohol which is a less electrophilic and less favorable process.
This is especially true when considering that alcohols are not particularly acidic, and thus their protons are not as easily abstracted by a base as compared to more acidic hydrogens (e.g., hydrogens adjacent to carbonyl groups).
The SN2 reaction is characterized by the simultaneous bond formation by the nucleophile and bond breaking by the leaving group, typically observed in primary alkyl halides. The phenoxide anion is a good nucleophile due to its negative charge, making it highly reactive towards electrophilic carbons, particularly against an atom that bears a good leaving group like chloride.
The Ksp of AgCl is 1.8x10^-10 and the Ksp of AgI is 8.3x10^-17. A solution is .100M in I- and Cl-. When a silver nitrate solution is slowly added to this mixture, what is the molarity of iodide ions when AgCl just starts to precipitate?
A.) 1.0x10^-5M
B.) 9.1x10^-9M
C.) 8.3x10^-7M
D.) 4.6x10^-8M
Answer:
[tex][I^-]=4.6*10^{-8}M[/tex]
Explanation:
The expression of the Ksp:
[tex]Ksp_{AgCl}=[Ag^{+}][Cl^-][/tex]
[tex]Ksp_{AgI}=[Ag^{+}][I^-][/tex]
When the product of the concentrations of both ions equals the Ksp, the salt starts to precipitate.
For the AgCl:
[tex]1.8*10^{-10}M^{2}=[Ag^{+}]*0.1M[/tex]
[tex][Ag^{+}]=1.8*10^{-9}M[/tex]
Initially the concentration of I- was 0.1 M, due to the lower Ksp than the AgCl's, the AgI will precipite before. So, when AgCl starts to precipitate the concentration of I- will be in equilibrium, following the Ksp equation.
[tex]8.3*10^{-17}M^{2}=1.8*10^{-9}M*[I^-][/tex]
[tex][I^-]=4.6*10^{-8}M[/tex]
10.0 g of ammonium nitrate (?Hsoln = 25.7 kJ/mol, molar mass = 80.0 g/mol) dissolves in 100.0 g of water.
What is the change in temperature of the solution?
Assume the specific heat capacity of the solution is 4.2 J
Answer:
The change in temperature = 7.65 °C
Explanation:
Step 1: Data given
10.0 grams of ammonium nitrate dissolves in 100.0 grams of water
Hsoln = 25.7 kJ/mol
Molar mass = 80.04 g/mol
Heat capacity of the solution = 4.2 J
Step 2: Calculate moles ammonium nitrate
Moles = mass / molar mass
Moles = 10.0 grams / 80.04 g/mol
Moles =0.125 moles
Step 3: Calculate q
q = 25.7 kJ/mol * 0.125 moles
q = 3.2125 kJ = 3212.5 J
Step 4: Calculate change in temperature
q = m*c*ΔT
3212.5 J = 100g *4.2 J * ΔT
ΔT= 7.65
The change in temperature = 7.65 °C
Explanation:
The given data is as follows.
Molar mass of ammonium nitrate = 80.0 g/mol
So, we will calculate the number of moles of ammonium nitrate as follows.
No. of moles = [tex]\frac{\text{given mass}}{\text{molar mass}}[/tex]
= [tex]\frac{10.0 g}{80.0 g/mol}[/tex]
= 0.125 mol
Heat released due to solution of ammonium nitrate = [tex]\Delta H \times \text{no. of moles}[/tex]
= [tex]25.7 kJ/mol \times 0.125 mol[/tex]
= 3.2125 KJ
= 3212.5 J (as 1 kJ = 1000 J)
Therefore, calculate the total mass of solution as follows.
mass of solution(m) = (10.0 + 100.0 ) g
= 110.0 g
Hence, heat released will be calculated as follows.
Q = [tex]m \times C \times \Delta T[/tex]
3212.5 J = [tex]110.0 \times 4.2 J \times \Delta T[/tex]
[tex]\Delta T = 6.95^{o}C[/tex]
Thus, we can conclude that the change in temperature of the solution is [tex]6.95^{o}C[/tex].
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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a. What is the frequency of radiation that has a wavelength of 11 μm , about the size of a bacterium? b. What is the wavelength of radiation that has a frequency of 5.30×10^14 s^−1 ?c. Would the radiations in part (a) or part (b) be visible to the human eye?d. What distance does electromagnetic radiation travel in 45.0 μs ?
Answer:
a) ν = 2.7 x 10¹³ s⁻¹
b) λ= 5.7 x 10⁻⁷ m
c) Neither will be visible
d) d= 13,500 m
Explanation:
The relationship ν = c/λ where
ν = frequency of the radiation
c = speed of light = 3 x 10^8 m/s
λ = wavelength of the radiation
will be used to solve parts a), b) and c).
For part d. we know that all electromagnetic radiation travel at the speed of light so d= speed x time
a) ν = 3 x 10^8 m/s / 11 x 10 ⁻⁶ m ( 1 μm = 10⁻6 m )
= 2.7 x 10¹³ s⁻¹
b) λ= c/ν = 3 x 10^8 m/s / 5.30 x 10¹⁴ s⁻¹ = 5.7 x 10⁻⁷ m
c) The visible spectrum range is 380 to 470 nm
in a. converting to nm:
11 x 10⁻⁶ m x 10⁹ nm/m = 11,000 nm
in b. converting to nm:
5.7 x 10⁻⁷ m x 10⁹ nm/ m = 570 nm
Neither of these radiations will be visible to the human eye.
d) d= 3 x 10^8 m/s x 45 x 10⁻⁶ s = 13,500 m
The frequency of the radiation is 2.73 x 10¹³ Hz.
The wavelength of the radiation is 566 nm
The distance traveled by the electromagnetic radiation is 13,500 m.
The given parameters;
wavelength of the radiation, λ = 11 μm = 11 x 10⁻⁶ m.The frequency of the radiation is calculated as follows;
c = fλ
[tex]f = \frac{c}{\lambda} \\\\f = \frac{3\times 10^{8}}{11 \times 10^{-6}} \\\\f = 2.73 \times 10^{13} \ Hz[/tex]
The wavelength of the radiation is calculated as follows;
[tex]\lambda = \frac{c}{f} \\\\\lambda = \frac{3\times 10^{8}}{5.3\times 10^{14}} \\\\\lambda = 5.66 \times 10^{-7} \ m\\\\\lambda = 566 \ \times 10^{-9} \ m\\\\\lambda = 566 \ nm[/tex]
The distance traveled by the electromagnetic radiation is calculated as;
d = vt
d = 3 x 10⁸ x 45 x 10⁻⁶
d = 13,500 m
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In a report to a supervisor, a chemist described an experiment in the following way: "0.0800 mol of H2O2 decomposed into 0.0800 mol of H2O and 0.0400 mol of O2." Express the chemistry and stoichiometry of this reaction by a conventional chemical equation. Do not include physical states. Use the smallest possible whole number coefficients.
Final answer:
The chemical equation representing the decomposition of hydrogen peroxide into water and oxygen gas is 2 H2O2 → 2 H2O + O2. This represents a stoichiometric relationship with a 2:1 ratio of hydrogen peroxide to oxygen and a 1:1 ratio of hydrogen peroxide to water.
Explanation:
The stoichiometry of the chemical reaction where hydrogen peroxide (H2O2) decomposes into water (H2O) and oxygen gas (O2) can be expressed by the following balanced chemical equation:
2 H2O2 → 2 H2O + O2
According to the equation given by the student, 0.0800 mol of H2O2 decomposes to produce 0.0800 mol of H2O and 0.0400 mol of O2. However, based on stoichiometric coefficients, we should expect a 1:1 ratio between H2O2 and H2O, and a 2:1 ratio between H2O2 and O2. Therefore, the decomposition of 0.0800 mol H2O2 should yield 0.0400 mol O2 according to the equation, which aligns with what was provided by the chemist.
What does the principal quantum number determine? Check all that apply. Check all that apply. the possible number of electorns on particular orbital the energy of the electron on the outer shell the shape of the orbital the overall size of an orbital the orientation of the orbital the overall size of an atom the energy of an orbital
Answer:
The principle quantum number determines the overall size and energy of an orbital.
Explanation:
The principle quantum number (n) is an integer that has possible values of 1, 2, 3.... and determines the overall size and energy of an orbital.
Angular momentum quantum number (L) determines the shape of atomic orbital.
The location of the electron in the cell is given by quantum mechanics. The principal quantum number determines the size of the atom and the energy of the orbital.
What is quantum mechanics?The quantum mechanism helps in the determination of the location of electrons in an atom with the help of four quantum numbers.
The principal quantum number is the first quantum number that helps in the analysis of the energy of the shell to which the atom belongs.
The principal number determines the size of an atom and the energy of the orbital. Thus, option F is correct.
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Calculate the standard molar enthalpy of formation, in kJ/mol, of NO(g) from the following data:
N2 + 2NO2 = 2NO2 ΔH 66.4kJ @298k
2NO + 02 =2NO2 ΔH -114.1kJ @298k
Answer:
The standard molar enthalpy of formation of 1 mole of NO gas is 90.25 kJ/mol.
Explanation:
We have :
[tex]N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o_{1} = 66.4 kJ[/tex]
[tex]2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o_{2} = -114.1 kJ[/tex]
To calculate the standard molar enthalpy of formation
[tex]N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?[/tex]...[3]
[1] - [2] = [3] (Hess's law)
[tex]N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?[/tex]
[tex]\Delta H^o_{3} =\Delta H^o_{1} - \Delta H^o_{2} [/tex]
[tex]\Delta H^o_{3}=66.4 kJ - [ -114.1 kJ] = 180.5 kJ[/tex]
According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide.
So, the standard molar enthalpy of formation of 1 mole of NO gas :
[tex]\Delta H^o_{f,NO}=\frac{\Delta H^o_{3}}{2 mol}[/tex]
[tex]\Delta H^o_{f,NO}=\frac{180.5 kJ}{2 mol}=90.25 kJ/mol[/tex]
To calculate the standard molar enthalpy of formation of NO(g), the given reactions are used by reversing the second reaction and adding its enthalpy change to the first reaction's ΔH. Dividing the total by 2 gives 90.25 kJ/mol for the formation of 1 mole of NO(g) from its elements.
Explanation:To calculate the standard molar enthalpy of formation of NO(g), we can use the given enthalpy changes for the reactions provided. The two reactions are:
N2(g) + 2O2(g) → 2NO2(g), ΔH = +66.4 kJ2NO(g) + O2(g) → 2NO2(g), ΔH = -114.1 kJWe want to find the enthalpy change for the formation of NO(g) from its elements, which is N2(g) and O2(g). This is the standard molar enthalpy of formation for NO(g). We can set up an enthalpy diagram to help visualize the process.
Steps to Calculate the Standard Molar Enthalpy of Formation:Reverse the second reaction to find the enthalpy change for breaking down 2NO2 into 2NO and O2, which will be +114.1 kJ (reversing a reaction changes the sign of ΔH).Add this value to the first reaction's ΔH, to get the total enthalpy change for the formation of 1 mole of NO(g) from its elements.Working through the math:
ΔHf[NO] = (66.4 kJ + 114.1 kJ) / 2
ΔHf[NO] = 180.5 kJ / 2
ΔHf[NO] = 90.25 kJ/mol
Therefore, the standard molar enthalpy of formation of NO(g) is 90.25 kJ/mol.
Which of the following statements is INCORRECT? A. It is not possible to know the exact location of an electron and its exact energy simultaneously. B. The energies of an hydrogen atom's electrons are quantized. C. The wave mechanical model correctly predicts all the energy states and the orbitals available to an electron in an hydrogen atom. D. The behavior of an atom's electrons can be described by circular orbits around a nucleus.E. Electrons have both wave and particle properties.
Answer:
D. The behavior of an atom's electrons can be described by circular orbits around a nucleus.
Explanation:
Which of the following statements is INCORRECT?
A. It is not possible to know the exact location of an electron and its exact energy simultaneously. CORRECT. This is known as the Heisenberg's uncertainty principle.
B. The energies of an hydrogen atom's electrons are quantized. CORRECT. According to the modern atomic model, the energy levels are quantized, that is, they have a discrete amount of energy.
C. The wave mechanical model correctly predicts all the energy states and the orbitals available to an electron in an hydrogen atom. CORRECT. The characteristics of the energy levels are explained by Schrödinger's wave equation.
D. The behavior of an atom's electrons can be described by circular orbits around a nucleus. INCORRECT. This was postulated by Bohr's atomic model but it is now considered to be incorrect.
E. Electrons have both wave and particle properties. CORRECT. This is what De Broglie called wave-particle duality.
Statement D is incorrect because Niels Bohr's model of electrons moving in circular orbits around a nucleus has been replaced by the quantum mechanical model, which describes electrons as occupying 'orbitals' rather than specific paths.
Explanation:The statement D, 'The behavior of an atom's electrons can be described by circular orbits around a nucleus', is incorrect. This is a description of a model proposed by Niels Bohr to explain the behavior of electrons in an atom, known as the Bohr model. However, this model has been superseded by the wave mechanical model (also known as the quantum mechanical model) which states that we can't know the exact location of an electron, but we can predict where it's likely to be—an area known as an 'orbital'. Electrons do not travel in defined circular orbits as originally proposed by Bohr.
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Be sure to answer all parts. What is the [H3O+] and the pH of a benzoic acid-benzoate buffer that consists of 0.22 M C6H5COOH and 0.41 M C6H5COONa? (Ka of benzoic acid = 6.3 × 10−5) Be sure to report your answer to the correct number of significant figures.
Answer:
The pH will be 4.5
Explanation:
The mixture of benzoic acid (weak acid) and its salt will make a buffer.
The pH of buffer solution can be calculated using Henderson Hassalbalch's equation, which is
[tex]pH=pKa+log\frac{[salt]}{[acid]}[/tex]
pKa = -logKa
pKa = -log([tex]6.3X10^{-5}[/tex])
pKa = 4.2
[tex]pH=4.2 + log\frac{0.41}{0.22}=4.2+0.27=4.47=4.5[/tex]
A gaseous reaction occurs and comes to equilibrium:
2HI(g) + Cl2(g) ---> 2HCl(g) + I2(g) + energy
Which of the following changes to the system will serve to increase the nuber of moles of I2 present at equilibrium?
A.) Increasing the volume at constant temperature
B.) Decreasing the volume at a constant temperature
C.) Increasing the temperature at a constant volume
D.) Decreasing the temperature at constant volume
Answer:
d) Decreasing the temperature at constant volume
Explanation:
if we assume ideal gas behaviour, since the concentration of the gas is proportional to its pressure at constant temperature
(from the ideal gas law PV =nRT or P= CRT), the equilibrium constant in terms of pressure will be
Kp = (p² HCl *p² I2)/(p² HI *p² CL2)
from Dalton's law : pi = P*xi
Kp = (P²*x² HCl *P* x I2)/(P*x HI *P²x² CL2) = (x² HCl * x I2)/(x HI *x² CL2)
since Kp does not change because the T is constant and it does not depend on pressure → the equilibrium will not change due to changes in pressure caused by reductions or increases in volume at constant pressure and composition
also since the reaction is exothermic → an increase in temperature will displace the equilibrium towards the reactants , and thus decreasing the moles of I2 at equilibrium
this can be seen from van't hoff equation :
d ln (K) /dT= ΔH/RT² , since ΔH>0 → K diminishes with increase in temperature
on the other hand, a decrease in temperature will displace the equilibrium towards the products , and thus increasing the moles of I2 at equilibrium