The equilibrium constant, Kp, for the following reaction is 10.5 at 350. K. 2CH₂Cl₂ (9) ⇒ CH4 (9) + CCl4 (9) If an equilibrium mixture of the three gases in a 10.8 L container at 350. K contains CH₂Cl2 at a pressure of 0.644 atm and CH4 at a pressure of 0.460 atm, the equilibrium partial pressure of CC14 is atm.
The equilibrium constant, Kp, for the following reaction is 10.5 at 350. K. 2CH₂Cl₂ (9) ⇒ CH4 (9) + CCl4 (9) If an equilibrium mixture of the three gases in a 10.8 L container at 350. K contains CH₂Cl2 at a pressure of 0.644 atm and CH4 at a pressure of 0.460 atm, the equilibrium partial pressure of CC14 is atm.
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![**Equilibrium Constant and Partial Pressures in Reactions**
**The equilibrium constant, \(K_p\), for the following reaction is 10.5 at 350 K.**
\[ 2 \text{CH}_2 \text{Cl}_2 (g) \rightleftharpoons \text{CH}_4 (g) + \text{CCl}_4 (g) \]
If an equilibrium mixture of the three gases in a 10.8 L container at 350 K contains CH\(_2\)Cl\(_2\) at a pressure of 0.644 atm and CH\(_4\) at a pressure of 0.460 atm, the equilibrium partial pressure of CCl\(_4\) is \(\_\_\_\_\) atm.
**Discussion:**
The equation above describes the equilibrium between dichloromethane (CH\(_2\)Cl\(_2\)), methane (CH\(_4\)), and carbon tetrachloride (CCl\(_4\)) gases. The equilibrium constant, \(K_p\), is a measure of the ratio of the concentrations of the products to the reactants at equilibrium, adjusted for the stoichiometry of the reaction.
To find the equilibrium partial pressure of CCl\(_4\), we can use the formula for \(K_p\) that involves partial pressures:
\[
K_p = \frac{P_{\text{CH}_4} \cdot P_{\text{CCl}_4}}{(P_{\text{CH}_2 \text{Cl}_2})^2}
\]
Where:
- \( K_p = 10.5 \)
- \( P_{\text{CH}_2 \text{Cl}_2} = 0.644 \, \text{atm} \)
- \( P_{\text{CH}_4} = 0.460 \, \text{atm} \)
Rearrange the formula to solve for \( P_{\text{CCl}_4} \):
\[
P_{\text{CCl}_4} = \frac{K_p \cdot (P_{\text{CH}_2 \text{Cl}_2})^2}{P_{\text{CH}_4}}
\]
Plug in the known values to find \( P_{\text{CCl}_4} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e7279b6-864f-464c-a490-1592f397f56d%2Fc3d319fb-4e08-4720-b3ce-17b5245386b9%2Fxxmt3pl_processed.png&w=3840&q=75)
Transcribed Image Text:**Equilibrium Constant and Partial Pressures in Reactions**
**The equilibrium constant, \(K_p\), for the following reaction is 10.5 at 350 K.**
\[ 2 \text{CH}_2 \text{Cl}_2 (g) \rightleftharpoons \text{CH}_4 (g) + \text{CCl}_4 (g) \]
If an equilibrium mixture of the three gases in a 10.8 L container at 350 K contains CH\(_2\)Cl\(_2\) at a pressure of 0.644 atm and CH\(_4\) at a pressure of 0.460 atm, the equilibrium partial pressure of CCl\(_4\) is \(\_\_\_\_\) atm.
**Discussion:**
The equation above describes the equilibrium between dichloromethane (CH\(_2\)Cl\(_2\)), methane (CH\(_4\)), and carbon tetrachloride (CCl\(_4\)) gases. The equilibrium constant, \(K_p\), is a measure of the ratio of the concentrations of the products to the reactants at equilibrium, adjusted for the stoichiometry of the reaction.
To find the equilibrium partial pressure of CCl\(_4\), we can use the formula for \(K_p\) that involves partial pressures:
\[
K_p = \frac{P_{\text{CH}_4} \cdot P_{\text{CCl}_4}}{(P_{\text{CH}_2 \text{Cl}_2})^2}
\]
Where:
- \( K_p = 10.5 \)
- \( P_{\text{CH}_2 \text{Cl}_2} = 0.644 \, \text{atm} \)
- \( P_{\text{CH}_4} = 0.460 \, \text{atm} \)
Rearrange the formula to solve for \( P_{\text{CCl}_4} \):
\[
P_{\text{CCl}_4} = \frac{K_p \cdot (P_{\text{CH}_2 \text{Cl}_2})^2}{P_{\text{CH}_4}}
\]
Plug in the known values to find \( P_{\text{CCl}_4} \).
![**Chemical Equilibrium and Reaction Analysis**
**Equilibrium Constant, \( K_c \):**
The equilibrium constant (\( K_c \)) for the following reaction at 298 K is given as 1.80 × 10\(^{-4}\):
\[ \text{NH}_4\text{HS} (s) \rightleftharpoons \text{NH}_3 (g) + \text{H}_2\text{S} (g) \]
**Reaction Details:**
In this decomposition reaction, solid ammonium hydrogen sulfide (NH₄HS) dissociates into ammonia (NH₃) gas and hydrogen sulfide (H₂S) gas.
**Problem Statement:**
Given an equilibrium mixture of the three compounds in a 4.30 L container at 298 K, containing:
- 3.12 mol of NH₄HS
- 0.310 mol of NH₃
**Objective:**
Calculate the number of moles of H₂S present at equilibrium.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e7279b6-864f-464c-a490-1592f397f56d%2Fc3d319fb-4e08-4720-b3ce-17b5245386b9%2Fi5pxjw_processed.png&w=3840&q=75)
Transcribed Image Text:**Chemical Equilibrium and Reaction Analysis**
**Equilibrium Constant, \( K_c \):**
The equilibrium constant (\( K_c \)) for the following reaction at 298 K is given as 1.80 × 10\(^{-4}\):
\[ \text{NH}_4\text{HS} (s) \rightleftharpoons \text{NH}_3 (g) + \text{H}_2\text{S} (g) \]
**Reaction Details:**
In this decomposition reaction, solid ammonium hydrogen sulfide (NH₄HS) dissociates into ammonia (NH₃) gas and hydrogen sulfide (H₂S) gas.
**Problem Statement:**
Given an equilibrium mixture of the three compounds in a 4.30 L container at 298 K, containing:
- 3.12 mol of NH₄HS
- 0.310 mol of NH₃
**Objective:**
Calculate the number of moles of H₂S present at equilibrium.
Expert Solution
Step 1: Defining equilibrium constant!
Answer:
For any reaction value of equilibrium constant KC is equal to the ratio of molar concentration of products and reactants at equilibrium.
For any reaction value of equilibrium constant KP is equal to the ratio of partial pressure of product gases and reactant gases at equilibrium.
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