Kc is approximately 0.00362, and Kp is approximately 2.167 based on molar concentrations and gas collected data.
To accurately calculate the equilibrium constants [tex]\( K_c \)[/tex] and [tex]\( K_p \)[/tex] for the given reaction, we need to follow the steps correctly:
[tex]\[ \text{CH}_4(\text{g}) + \text{H}_2\text{O}(\text{g}) \leftrightarrow \text{CO}(\text{g}) + 3\text{H}_2(\text{g}) \][/tex]
1. Molar Masses:
- [tex]\(\text{CH}_4\):[/tex] 16.04 g/mol
- [tex]\(\text{H}_2\text{O}\)[/tex] : 18.02 g/mol
- [tex]\(\text{CO}\):[/tex] 28.01 g/mol
- [tex]\(\text{H}_2\):[/tex] 2.02 g/mol
2. Calculate Moles of Each Substance:
[tex]\(\text{CH}_4\): \( \frac{42.3 \, \text{g}}{16.04 \, \text{g/mol}} = 2.638 \, \text{moles} \)\\ \(\text{H}_2\text{O}\): \( \frac{49.2 \, \text{g}}{18.02 \, \text{g/mol}} = 2.730 \, \text{moles} \)\\\(\text{CO}\): \( \frac{8.32 \, \text{g}}{28.01 \, \text{g/mol}} = 0.297 \, \text{moles} \)\\\(\text{H}_2\): \( \frac{2.63 \, \text{g}}{2.02 \, \text{g/mol}} = 1.301 \, \text{moles} \)[/tex]
3. Calculate Concentrations:
[tex]- \([ \text{CH}_4 ] = \frac{2.638 \, \text{moles}}{5.00 \, \text{L}} = 0.528 \text{M} \)\\\\- \([ \text{H}_2\text{O} ] = \frac{2.730 \, \text{moles}}{5.00 \, \text{L}} = 0.546 \, \text{M} \) \\\\- \([ \text{CO} ] = \frac{0.297 \, \text{moles}}{5.00 \, \text{L}} = 0.0594 \, \text{M} \)\\\\- \([ \text{H}_2 ] = \frac{1.301 \, \text{moles}}{5.00 \, \text{L}} = 0.260 \, \text{M} \)[/tex]
4. Calculate [tex]K_c[/tex]
[tex]\[ K_c = \frac{[ \text{CO} ][ \text{H}_2 ]^3}{[ \text{CH}_4 ][ \text{H}_2\text{O} ]} \][/tex]
[tex]\[ K_c = \frac{(0.0594) \times (0.260)^3}{(0.528) \times (0.546)} \][/tex]
[tex]\[ K_c = \frac{0.0594 \times 0.017576}{0.288288} \][/tex]
[tex]\[ K_c = \frac{0.0010431744}{0.288288} \][/tex]
[tex]\[ K_c \approx 0.00362 \][/tex]
5. Calculate [tex]K_p[/tex]
[tex]\[ K_p = K_c \left( RT \right)^{\Delta n} \][/tex]
- [tex]\(\Delta n = \text{moles of products} - \text{moles of reactants} = (1 + 3) - (1 + 1) = 2\)[/tex]
- Assuming [tex]\( T = 298 \, \text{K} \)[/tex] (standard temperature) and [tex]( R = 0.0821 \, \text{L atm/(mol K)} \):[/tex]
[tex]\[ K_p = 0.00362 \left( 0.0821 \times 298 \right)^2 \][/tex]
[tex]\[ K_p = 0.00362 \left( 24.4658 \right)^2 \][/tex]
[tex]\[ K_p = 0.00362 \times 598.5877 \][/tex]
[tex]\[ K_p \approx 2.167 \][/tex]
Conclusion:
[tex]\( K_c \approx 0.00362 \)\\\\ \( K_p \approx 2.167 \)[/tex]
In summary, Kc is approximately 0.00362, and Kp is approximately 2.167
If 4.8 moles of X and 3.4 moles of Y react according to the reaction below, how many moles of the excess reactant will be left over at the end of the reaction?
3X + 2Y “yields”/ X3Y2
1.7 mol Y left over
1.6 mol X left over
0.2 mol Y left over
0.1 mol X left over
Answer : The correct option is, 0.2 mole Y left over .
Explanation : Given,
Moles of X = 4.8 mole
Moles of Y = 3.4 mole
The balanced chemical reaction is,
[tex]3X+2Y\rightarrow X_3Y_2[/tex]
From the balanced reaction, we conclude that
As, 3 moles of X react with 2 moles of Y
So, 4.8 moles of X react with [tex]\frac{2}{3}\times 4.8=3.2[/tex] moles of Y
From this we conclude that, the reactant Y is an excess reagent and X is a limiting reagent.
The moles of excess reagent left over at the end of the reaction = Given moles of X - Required moles of X
The moles of excess reagent left over at the end of the reaction = 3.4 - 3.2 = 0.2 mole
Therefore, the moles of excess reagent left over at the end of the reaction is, 0.2 mole Y left over.
Answer:
The correct answer is : '0.2 mol Y left over'.
Explanation:
[tex]3X + 2Y \rightarrow X_3Y_2[/tex]
Moles of X = 4.8 moles
Moles of Y = 3.4 moles
According to reaction, 3 moles of X react with 2 moles of Y .
Then 4.8 moles of X react with :
[tex]\frac{2}{3}\times 4.8=3.2 [/tex]moles of Y
Moles of Y reacted = 3.2 moles
Moles of Y left unreacted = 3.4 moles - 3.2 moles = 0.2 moles
As we can see that X is in limiting amount and y is present in an excessive amount.And the left over amount of Y is 0.2 moles.
How many moles of co2 are produced when 5.20 mol of ethane are burned in an excess of oxygen?
Why do sea and ocean levels recede (more coast land is exposed) when the planet goes through a major ice age?
The amount of water that evaporates from earth is
Answer:
The amount of water that evaporates from the earth is approximately equal to the amount that falls as precipitation.
Explanation:
Water evaporation is critical to climate because it is directly related to precipitation formation. Water that evaporates from rivers, lakes, oceans and even our bodies helps to form rain. This occurs when the temperature cools. Under these climatic conditions, water vapor returns to its liquid form (condensation) and falls through rainfall. The amount of water evaporated is basically equal to the amount of water that comes back to land in precipitation.
How many moles (of molecules or formula units) are in each sample? part a 71.66 g cf2cl2?
The activation energy for the gas phase isomerization of cyclopropane is 272 kJ. (CH2)3CH3CH=CH2 The rate constant at 718 K is 2.30×10-5 /s. The rate constant will be /s at 753 K.
What is the percent by mass of potassium in K3Fe(CN)6?
Answer:
The percentage of potassium in the given complex is 35.54 %.
Explanation:
Mass of potassium in [tex]K_3Fe(CN)_6[/tex] = 3 × 39.10 g mol=117.3 g/mol
Molar mass of [tex]K_3Fe(CN)_6[/tex] =329.15 g/mol
Percentage of potassium (K) in the the complex:
[tex]\% K=\frac{\text{mass of potassium}}{\text{molar mass of complex}}\times 100[/tex]
[tex]\%K=\frac{117.3 g/mol}{329.15 g/mol}\times 100=35.54\%[/tex]
The percentage of potassium in the given complex is 35.54 %.
Look up the boiling points of anisole and d-limonene. which one do you expect to elute first in gas chromotograpjhy
Final answer:
In gas chromatography, compounds elute based on their boiling points, with those having lower boiling points eluting first. Since anisole has a lower boiling point than d-limonene, anisole is expected to elute first.
Explanation:
The question asks which compound, anisole or d-limonene, would elute first in gas chromatography (GC) based on their boiling points. In gas chromatography, compounds generally elute in order of increasing boiling points because compounds with lower boiling points have lower retention times on the GC column. Also, the elution order correlates with the strength of intermolecular forces (IMFs) affecting the compounds; compounds with stronger IMFs tend to have higher boiling points and adhere more to the stationary phase, thus eluting later. Although the specific boiling points of anisole and d-limonene are not provided in this answer, it is known that anisole has a boiling point of about 154°C, and d-limonene has a boiling point around 176°C. Therefore, one would expect anisole to elute first in gas chromatography due to its lower boiling point compared to d-limonene.
A leaf falls into a shallow lake and is rapidly buried in the sediment the sediment change choose to rock over millions of years which type of fossil would most likely be form
The data in the table below were obtained for the reaction: 2clo2 (aq) + 2 oh- (aq) --> clo3- (aq) + clo2- (aq) + h2o (l) experiment [clo2] (m) [oh-] (m) initial rate (m/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828 what is the order of the reaction with respect to clo2?
The order of the reaction with respect to clo2 is 1.
Explanation:The order of the reaction with respect to clo2 can be determined by comparing the initial rates of different experiments and analyzing the effect of changing the concentration of clo2 on the reaction rate. By comparing the rates of Experiment 1 and Experiment 2, we can see that when the concentration of clo2 is halved, the rate of the reaction is also halved. This indicates that the reaction rate is directly proportional to the concentration of clo2. Therefore, the order of the reaction with respect to clo2 is 1.
Consider the reaction caso4(s)⇌ca2+(aq)+so2−4(aq) at 25 ∘c the equilibrium constant is kc=2.4×10−5 for this reaction.if excess caso4(s) is mixed with water at 25 ∘c to produce a saturated solution of caso4, what is the equilibrium concentration of ca2+?
A chemist mixes oxygen gas and hydrogen gas to form water, which is composed of one oxygen and two hydrogen atoms per molecule. What has occurred? A physical change B chemical change C combustion D precipitation
Answer: The formation of water is a chemical change.
Explanation:
Physical change is defined as the change in which change in shape and size takes place. The chemical composition of a substance remains the same. No new substance is formed during this.
For Example: Melting of ice
Chemical change is defined as the change in which change in chemical composition takes place. A new substance is formed in this.
For Example: Formation of water molecule.
The chemical equation for the formation of water molecule follows:
[tex]2H_2+O_2\righatarrow 2H_2O[/tex]
Hence, the formation of water is a chemical change.
Draw the products for the proton transfer reaction between sodium hydride and ethanol
Final answer:
Sodium hydride donates a hydride ion to ethanol, resulting in the formation of hydrogen gas and the ethoxide ion in a sodium ethoxide complex.
Explanation:
The proton transfer reaction between sodium hydride (NaH) and ethanol (CH3CH2OH) involves sodium hydride acting as a base, donating a hydride ion (H-) to the proton (H+) of the ethanol. This reaction results in the formation of hydrogen gas (H2) and the ethoxide ion (CH3CH2O-), which remains in the solution complexed with the sodium ion (Na+). The balanced equation for this reaction is NaH + CH3CH2OH → H2 + Na+ + CH3CH2O-. This reaction utilizes the hydride ion from the sodium hydride as a nucleophile that abstracts a proton from the ethanol, leading to the evolution of hydrogen gas.
Calculate the vapor pressure at 50°c of a coolant solution that is 54.0:46.0 ethylene glycol-to-water by volume. at 50.0°c, the density of water is 0.9880 g/ml, and its vapor pressure is 92 torr. the vapor pressure of ethylene glycol is less than 1 torr at 50.0°c.
molecular or formula mass P4
When the metal sample reacts with acid, the gas evolved will be collected over water; the gas is said to be "wet". what is the composition of "wet" gas? how can partial pressure of hydrogen gas be obtained from the total pressure of wet gas? how will you obtain the neccessay data ro determine the pressure of hydrogen gas?
can someone help me
Rubbing alcohol evaporates from your hand quickly, leaving a cooling sensation. Because evaporation is an example of a physical property, how do the molecules of gas compare to the molecules as a liquid? 1. The gas particles have a stronger attraction between them and move slower than the liquid. 2. The gas and liquid particles have the same structure and identity but different motion and kinetic energy. 3. The bonds inside the molecule are broken, and atoms move closer together as evaporation occurs. 4. The bonds are broken, and atoms spread apart as it changes from liquid to gas.
Final answer:
Rubbing alcohol molecules as a gas have the same structure as in the liquid but with more kinetic energy and less intermolecular attraction, which upon evaporation causes a cooling effect through evaporative cooling.
Explanation:
When rubbing alcohol evaporates from your hand, it leaves a cooling sensation because the molecules in the liquid state require a certain threshold of kinetic energy to overcome intermolecular forces and escape into a gas state. The correct statement regarding how the molecules of gas compare to the molecules as a liquid is:
The gas and liquid particles have the same structure and identity but different motion and kinetic energy.In the gaseous state, particles move faster and are further apart compared to when they are in the liquid state, where particles are closer together and have stronger intermolecular attractions. This process of evaporation involves evaporative cooling, where the molecules with higher kinetic energy escape, leaving behind those with lower kinetic energy, which results in a decrease in temperature.
For each bond, show the direction of polarity by selecting the correct partial charges. si-p si-s s-p the most polar bond is
The biggest elctronegative difference is between silicon and sulfur. So Si-S will be most polar bond.
The polarity between the two atoms is determined by their relative difference in electronegativity.
The Electronegativity of ,
Silicon= 1.9
Phosphorus= 2.19
Sulfur= 2.58
The direction of polarity,
[tex]\rm \bold{ \delta^+Si\rightarrow \delta^-P}\\\rm \bold{ \delta^+Si\rightarrow \delta^-S}\\\rm \bold{ \delta^+P\rightarrow \delta^-S}[/tex]
Since, the biggest elctronegative difference is between silicon and sulfur (Si-S).
Hence we can say that Si-S will be most polar bond.
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If the molar mass of the compound in problem 1 is 110 grams/mole, what is the molecular formula? With work?
Final answer:
To find the molecular formula, calculate the empirical formula mass, divide the molar mass by this number to find a multiplier, and apply the multiplier to the subscripts in the empirical formula.
Explanation:
Finding the Molecular Formula
To determine the molecular formula of a compound for which we know the molar mass is 110 grams/mole, and have the empirical formula from problem 1, we follow these steps:
Calculate the empirical formula mass by summing the atomic masses of each element in the empirical formula.Divide the given molar mass (110 g/mol) by the empirical formula mass. This will give us a factor by which we multiply the subscripts in the empirical formula to get the molecular formula.If our factor is close to 1, the empirical formula and the molecular formula are the same. If it's a whole number or a simple fraction, multiply each subscript in the empirical formula to get the molecular formula.For example, assuming our empirical formula is CH₂ (not given in the question), we would get the empirical formula mass as 12 (for C) + 2×1 (for H) = 14 g/mol. Then, 110 g/mol divided by 14 g/mol gives us approximately 7.86, which we round to 8 since it should be a whole number. Multiplying the subscripts in CH₂ by 8 gives us C₈H₁₆ as the molecular formula.
mass of 0.432 moles of C8H9O4?
Explanation:
It is known that number of moles present in a substance is equal to the mass divided by molar mass.
Mathematically, No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
As it is given that number of moles are 0.432 moles and molar mass of [tex]C_{8}H_{9}O_{4}[/tex] is 169.15 g/mol.
Hence, calculate the mass of [tex]C_{8}H_{9}O_{4}[/tex] is as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
0.432 mol = [tex]\frac{mass}{169.15 g/mol}[/tex]
mass = 73.072 g
Thus, we can conclude that the mass of 0.432 moles of [tex]C_{8}H_{9}O_{4}[/tex] is 73.072 g.
The activation energy for the gas phase decomposition of 2-bromopropane is 212 kJ.
CH3CHBrCH3CH3CH=CH2 + HBr
The rate constant at 683 K is 6.06×10-4 /s. The rate constant will be 5.06×10-3 /s at
The activation energy can be determined using the Arrhenius equation. To find the activation energy at a different temperature, rearrange the equation and plug in the given rate constant. The frequency factor is approximately 1.529 * 10^9 /s.
Explanation:The activation energy can be determined using the Arrhenius equation:
k = Ae^(-Ea/RT)
where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
To find the activation energy at a different temperature, we can rearrange the equation as:
Ea = -ln(k/A) * RT
Using the given rate constant at 683 K and the activation energy of 212 kJ, we can calculate the frequency factor as follows:
A = k * e^(Ea/RT)
Plugging in the values:
A = (6.06×10^(-4) /s) * e^(212000 J / (8.3145 J/mol*K * 683 K))
A = 1.529 * 10^9 /s
Therefore, the frequency factor is approximately 1.529 * 10^9 /s.
The activation energy for a reaction is the minimum amount of energy required for the reaction to occur. To find the activation energy at a different temperature, we can use the Arrhenius equation. The activation energy will be 2.71 kJ/mol at the desired condition.
Explanation:The activation energy for a reaction is the minimum amount of energy required for the reaction to occur. It represents the energy barrier that the reactants must overcome before they can form products. In the given question, the activation energy for the gas phase decomposition of 2-bromopropane is 212 kJ.
To find the activation energy at a different temperature, we can use the Arrhenius equation:
k = A * e^(-Ea/RT)
where:
- k is the rate constant
- A is the frequency factor
- Ea is the activation energy
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
By rearranging the equation, we can solve for the activation energy at a different temperature:
Ea2 = -ln(k2/k1) * (R/T2 - R/T1)
Substituting the given values:
Ea2 = -ln(5.06x10^-3 / 6.06x10^-4) * (8.314 J/(mol·K) / 683 K - 8.314 J/(mol·K) / 298 K)
Ea2 = -ln(8.333) * (0.0122 - 0.0278) = -ln(8.333) * (-0.0156) = 0.1739 * 0.0156 = 0.00271 J/mol = 2.71 kJ/mol
Therefore, the activation energy will be 2.71 kJ/mol at the desired condition.
for the complete combustion of 5.6dm3 of a gaseuos hydrocarbon CxHy. 28.0dm3 of oxygen gas were used, 16.8dm3 of CO2 gas and 18.0gof liquid water were produced all gases measurement were made at stp. Determine the chemical formula of the hydrocarbon
What are isotopes? What are some examples of common stable isotopes?
Xas shown in table 15.2, kp for the equilibrium n21g2 + 3 h21g2 δ 2 nh31g2 is 4.51 * 10-5 at 450 °c. for each of the mixtures listed here, indicate whether the mixture is at equilibrium at 450 °c. if it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) 98 atm nh3, 45 atm n2, 55 atm h2 (b) 57 atm nh3, 143 atm n2, no h2 (c) 13 atm nh3, 27 atm n2, 82 atm h
A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea = 85.6 kj/mol?
This question involves the application of the Arrhenius equation in chemistry, particularly in calculating the temperature dependence of reaction rates. Given the rate constant and the activation energy at a certain temperature, one can find the rate constant at a different temperature
Explanation:The question pertains to the use of the Arrhenius equation, which calculates the temperature dependence of reaction rates. The equation is k = A * e^(-Ea/RT), where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
The given reaction has a rate constant (k) of 1.35 x 10² s-1 at 25.0°C (or 298.15K). The activation energy (Ea) is given as 85.6 kJ/mol. To find the rate constant at another temperature, rearrange the Arrhenius equation to solve for A, substitute the given values for Ea, k, R and T to find A. Then, plug the calculated A, given Ea, the new temperature in Kelvin (75.0°C or 348.15K), and R into the Arrhenius equation to solve for the new rate constant, k.
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What are ionic compounds typically composed of ?
A. A metal anion and a nonmetal cation
B. Two metal anions
C. A metal cation and non metal anion
D.Two nonmetal cations
Ionic compounds are generally formed from a metal cation (positively charged ion) and a nonmetal anion (negatively charged ion), so the correct answer to your question is option C.
Explanation:Ionic compounds are typically composed of a metal cation and nonmetal anion. This means the correct answer to your question is option C. A cation is a positively charged ion, and in this context, it is typically formed by an element from the left side of the periodic table, or a metal. An anion, on the other hand, is a negatively charged ion, usually formed by an element from the right side of the periodic table, or a nonmetal. When these ions combine, they create an ionic compound, such as NaCl (sodium chloride), where sodium is the metal cation and chloride is the nonmetal anion.
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Which acid is the best choice to create a buffer with ph= 7.66?
The base-dissociation constant of ethylamine (c2h5nh2) is 6.4 ??? 10???4 at 25.0 ??
c. the [h ] in a 1.2 ??? 10-2 m solution of ethylamine is ________ m.
[H⁺]=3.608.10⁻¹²
Further explanationWeak acid ionization reaction occurs partially (not ionizing perfectly as in strong acids)
The ionization reaction of a weak acid is an equilibrium reaction
HA (aq) ---> H⁺ (aq) + A⁻ (aq)
The equilibrium constant for acid ionization is called the acid ionization constant, which is symbolized by Ka
The values for the weak acid reactions above:
[tex]\rm Ka=\dfrac{[H][A^-]}{[HA]}[/tex]
The greater the Ka, the stronger the acid, which means the reaction to the right is also greater
Where Kb is the base ionization constant
LOH (aq) ---> L⁺ (aq) + OH⁻ (aq)
[tex]\rm Kb=\dfrac{[L][OH^-]}{[LOH]}[/tex]
Kb of Ethylamine (C₂H₅NH₂) : 6.4.10⁻⁴
The ethylamine ionization reactions occur in water as follows:
C₂H₅NH₂ + H₂O ⇒ C₂H₅NH₃⁺ + OH⁻
with a Kb value:
[tex]\rm Kb=\dfrac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]}[/tex]
for example x = number of moles / concentration that reacts
Initial concentration of Ethylamine (C₂H₅NH₂) : 1.2.10⁻²
Concentration at equilibrium = 1.2.10⁻² -x
Initial concentration of C₂H₅NH₃ = 0
Concentration at equilibrium = x
Initial concentration OH⁻ = 0
Concentration at equilibrium = x
so the value of Kb =
[tex]\rm Kb=\dfrac{[x][x]}{[1.2.10^{-2}-x]}\\\\assumption\:x=so\:small\:then\\\\6.4.10^{-4}=\dfrac{x^2}{1.2.10^{-2}}\\\\x^2=7.68.10^{-6}\\\\x=2.771.10^{-3}[/tex]
x = [OH⁻] = 2.771.10⁻³
Ka x Kb = [H⁺] [OH-]
a water equilibrium constant value (Kw) of 1.10⁻¹⁴ at 25 °C
Ka x Kb = [H +] [OH-] = 1.10⁻¹⁴
1.10⁻¹⁴ = [H⁺] . 2.771.10⁻³
[H⁺]=3.608.10⁻¹²
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The balanced equation for a hypothetical reaction is A + 5B + 6C → 3D + 3E. What is the rate law for this reaction?
Answer: [tex]Rate=k[A]^1[B]^5[C]^6[/tex]
Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.
Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.
[tex]A+5B+6C\rightarrow 3D+3E[/tex]
[tex]Rate=k[A]^1[B]^5[C]^6[/tex]
k= rate constant
1 = order with respect to A
5 = order with respect to B
6 = order with respect to C
Thus rate law is [tex]Rate=k[A]^1[B]^5[C]^6[/tex]