19) 2 H2S(g) 2 H2(g) + S2(g) with Kc = 1.00x10-6 If the initial concentrations for the above reaction are 2 mole H2S, 1mole of H2, and 1mole of S2 in a 2L container, based on the reaction quotient what direction is the reaction going? a) Forward b) Reverse c) at equilibrium d) not enough information to determine P Type here to search
19) 2 H2S(g) 2 H2(g) + S2(g) with Kc = 1.00x10-6 If the initial concentrations for the above reaction are 2 mole H2S, 1mole of H2, and 1mole of S2 in a 2L container, based on the reaction quotient what direction is the reaction going? a) Forward b) Reverse c) at equilibrium d) not enough information to determine P Type here to search
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Chemical Equilibrium Problem**
Consider the following chemical equilibrium reaction:
\[ 2 \text{H}_2\text{S(g)} \leftrightharpoons 2 \text{H}_2\text{(g)} + \text{S}_2\text{(g)} \quad \text{with} \; K_c = 1.00 \times 10^{-6} \]
Given the initial concentrations of the reactants and products:
- 2 moles of \(\text{H}_2\text{S}\)
- 1 mole of \(\text{H}_2\)
- 1 mole of \(\text{S}_2\)
The reaction takes place in a 2L container. Based on the reaction quotient (Q), determine the direction in which the reaction will proceed:
a) Forward \
b) Reverse \
c) At equilibrium \
d) Not enough information to determine
**Answer:**
First, calculate the initial concentrations:
- Concentration of \( \text{H}_2\text{S} \): \( \frac{2 \text{ moles}}{2 \text{ L}} = 1 \text{ M} \)
- Concentration of \( \text{H}_2 \): \( \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M} \)
- Concentration of \( \text{S}_2 \): \( \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M} \)
Next, calculate the reaction quotient (Q):
\[ Q = \frac{[\text{H}_2]^2 [\text{S}_2]}{[\text{H}_2\text{S}]^2} = \frac{(0.5)^2 (0.5)}{(1)^2} = \frac{0.125}{1} = 0.125 \]
Compare \( Q \) with \( K_c \):
- \( Q > K_c \) (0.125 > \( 1.00 \times 10^{-6} \))
Since \( Q > K_c \), the reaction will proceed in the reverse direction to reach equilibrium.
Thus, the correct answer is: **b) Reverse**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2397022e-dfa6-4642-9d62-ec00c66800a3%2Fd80e08cc-d2e3-4658-93e1-b9e78da5fa50%2Fqd3s2gl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Equilibrium Problem**
Consider the following chemical equilibrium reaction:
\[ 2 \text{H}_2\text{S(g)} \leftrightharpoons 2 \text{H}_2\text{(g)} + \text{S}_2\text{(g)} \quad \text{with} \; K_c = 1.00 \times 10^{-6} \]
Given the initial concentrations of the reactants and products:
- 2 moles of \(\text{H}_2\text{S}\)
- 1 mole of \(\text{H}_2\)
- 1 mole of \(\text{S}_2\)
The reaction takes place in a 2L container. Based on the reaction quotient (Q), determine the direction in which the reaction will proceed:
a) Forward \
b) Reverse \
c) At equilibrium \
d) Not enough information to determine
**Answer:**
First, calculate the initial concentrations:
- Concentration of \( \text{H}_2\text{S} \): \( \frac{2 \text{ moles}}{2 \text{ L}} = 1 \text{ M} \)
- Concentration of \( \text{H}_2 \): \( \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M} \)
- Concentration of \( \text{S}_2 \): \( \frac{1 \text{ mole}}{2 \text{ L}} = 0.5 \text{ M} \)
Next, calculate the reaction quotient (Q):
\[ Q = \frac{[\text{H}_2]^2 [\text{S}_2]}{[\text{H}_2\text{S}]^2} = \frac{(0.5)^2 (0.5)}{(1)^2} = \frac{0.125}{1} = 0.125 \]
Compare \( Q \) with \( K_c \):
- \( Q > K_c \) (0.125 > \( 1.00 \times 10^{-6} \))
Since \( Q > K_c \), the reaction will proceed in the reverse direction to reach equilibrium.
Thus, the correct answer is: **b) Reverse**
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