Kp = 2.4 x 104 at 1073 K %D 2H2S(g) = 2H2(g) + S2(g) a. A reaction mixture contains 0.122 atm of H2, 0.055 atm of S2, and 0.445 atm of H2S. Is the reaction mixture at equilibrium? Show your work mathematically (calculate Q), If it's not, in what direction will reaction proceed? E.00348 PH2S Q>K so Shifts to left b. What is the value of Kp at 1073 K for the reaction written as: H2(g) + ½S2(g) = H2S(g)? c. What is the value of Kc for the reaction as originally written? .09348
Kp = 2.4 x 104 at 1073 K %D 2H2S(g) = 2H2(g) + S2(g) a. A reaction mixture contains 0.122 atm of H2, 0.055 atm of S2, and 0.445 atm of H2S. Is the reaction mixture at equilibrium? Show your work mathematically (calculate Q), If it's not, in what direction will reaction proceed? E.00348 PH2S Q>K so Shifts to left b. What is the value of Kp at 1073 K for the reaction written as: H2(g) + ½S2(g) = H2S(g)? c. What is the value of Kc for the reaction as originally written? .09348
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![### Equilibrium and Reaction Quotient Analysis
#### Consider the Reaction:
\[ \text{2H}_2\text{S(g)} \rightleftharpoons \text{2H}_2\text{(g)} + \text{S}_2\text{(g)} \]
\[ K_p = 2.4 \times 10^{-4} \text{ at 1073 K} \]
#### a. Initial Concentrations and Reaction Quotient, \( Q \)
A reaction mixture contains:
- 0.122 atm of H\(_2\)
- 0.055 atm of S\(_2\)
- 0.445 atm of H\(_2\)S
Is the reaction mixture at equilibrium? If not, in which direction will the reaction proceed?
Calculate \( Q \) using the following expression:
\[
Q = \frac{P_{H_2}^2 \cdot P_{S_2}}{P_{H_2S}^2} = \frac{(0.112)^2 \cdot (0.055)}{(0.445)^2} = 0.00348
\]
Since \( Q > K \), the reaction will shift to the left.
#### b. Calculating \( K_p \) for Modified Reaction
For the reaction:
\[ \text{H}_2\text{(g)} + \frac{1}{2}\text{S}_2\text{(g)} \rightleftharpoons \text{H}_2\text{S(g)} \]
Calculate \( K_p \):
\[
K_p = \frac{P_{H_2S}}{P_{H_2} \cdot \left(\frac{1}{2}P_{S_2}\right)} = \frac{(0.445)}{(0.112) \cdot \left(\frac{1}{2} \times 0.055\right)} = 1.4 \times 10^2
\]
#### c. Calculating \( K_c \)
Determine \( K_c \) for the original reaction:
\[ K_c = 0.00348 \]
This transcription, suitable for an educational website, illustrates the methodologies for computing reaction quotients and equilibrium constants, helping learners understand the dynamic nature of reversible reactions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F055086f6-4d26-404b-b10d-4e78791428d4%2Fd61bd066-eba4-4f3b-a9b3-5a4dcfab3ebb%2Fcn4qkkq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Equilibrium and Reaction Quotient Analysis
#### Consider the Reaction:
\[ \text{2H}_2\text{S(g)} \rightleftharpoons \text{2H}_2\text{(g)} + \text{S}_2\text{(g)} \]
\[ K_p = 2.4 \times 10^{-4} \text{ at 1073 K} \]
#### a. Initial Concentrations and Reaction Quotient, \( Q \)
A reaction mixture contains:
- 0.122 atm of H\(_2\)
- 0.055 atm of S\(_2\)
- 0.445 atm of H\(_2\)S
Is the reaction mixture at equilibrium? If not, in which direction will the reaction proceed?
Calculate \( Q \) using the following expression:
\[
Q = \frac{P_{H_2}^2 \cdot P_{S_2}}{P_{H_2S}^2} = \frac{(0.112)^2 \cdot (0.055)}{(0.445)^2} = 0.00348
\]
Since \( Q > K \), the reaction will shift to the left.
#### b. Calculating \( K_p \) for Modified Reaction
For the reaction:
\[ \text{H}_2\text{(g)} + \frac{1}{2}\text{S}_2\text{(g)} \rightleftharpoons \text{H}_2\text{S(g)} \]
Calculate \( K_p \):
\[
K_p = \frac{P_{H_2S}}{P_{H_2} \cdot \left(\frac{1}{2}P_{S_2}\right)} = \frac{(0.445)}{(0.112) \cdot \left(\frac{1}{2} \times 0.055\right)} = 1.4 \times 10^2
\]
#### c. Calculating \( K_c \)
Determine \( K_c \) for the original reaction:
\[ K_c = 0.00348 \]
This transcription, suitable for an educational website, illustrates the methodologies for computing reaction quotients and equilibrium constants, helping learners understand the dynamic nature of reversible reactions.
Transcribed Image Text:**Reaction Equilibrium Changes**
The reaction \(2\text{H}_2\text{S (g)} \rightleftharpoons 2\text{H}_2\text{ (g)} + \text{S}_2\text{ (g)}\) is endothermic. Describe the effect of the following changes on the equilibrium:
i. **Decreasing the temperature:**
*Effect:* Shifts to the right.
ii. **Adding a catalyst:**
*Effect:* No effect.
iii. **Removing \( \text{H}_2\text{ (g)} \):**
*Effect:* Shifts right.
iv. **Adding \(\text{H}_2\text{S (g)}\):**
*Effect:* Shifts right.
v. **Decreasing the volume of the reaction container** (Hint: what happens to the pressure?):
*Effect:* Shifts to the left, favoring side with more moles.
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