### Chemical Equilibrium and Le Chatelier's Principle#### Problem Statement:For the reaction:\[ \text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{HBr}(g) \]Given the equilibrium constant \( K = 7.2 \times 10^{-4} \) at 1362 K, and \( \Delta H^\circ \) is positive. A vessel is charged with the following partial pressures at 1362 K:- \(\text{HBr}\): 48.2 Pa- \(\text{H}_2\): 1385 Pa- \(\text{Br}_2\): 3370 Pa#### Question:Will the reaction proceed to the left or right to reach equilibrium?#### Options:- \( \bigcirc \) left- \( \bigcirc \) rightTo solve this, we need to perform an equilibrium calculation using the reaction quotient \( Q \) and compare it to the equilibrium constant \( K \).**Calculation Steps:**1. **Calculate the Reaction Quotient (Q):**\[ Q = \frac{{\text{(Pressure of HBr)}^2}}{{\text{(Pressure of H}_2) \times \text{(Pressure of Br}_2)}} \]Given:- Pressure of HBr \( = 48.2 \) Pa- Pressure of H\(_2\) \( = 1385 \) Pa- Pressure of Br\(_2\) \( = 3370 \) Pa\[ Q = \frac{(48.2)^2}{(1385) \times (3370)} \]2. **Compare \( Q \) and \( K \):**- If \( Q < K \), the reaction will proceed to the right.- If \( Q > K \), the reaction will proceed to the left.- If \( Q = K \), the system is at equilibrium.Let's substitute the values and calculate \( Q \):\[ Q = \frac{48.2 \times 48.2}{1385 \times 3370} \]\[ Q = \frac{2323.24}{4665450} \approx 4.98 \times 10^{-4} \]Given that \( K = 7.2 \times 10^{-4} \ **Calculate the pressure (in pascals) of each species at equilibrium.**\[ P_{H_2} = \boxed{\hspace{10cm}} \text{Pa} \]\[ P_{Br_2} = \boxed{\hspace{10cm}} \text{Pa} \]\[ P_{HBr} = \boxed{\hspace{10cm}} \text{Pa} \]This set of equations is provided to calculate the equilibrium pressures of different chemical species in a reaction. The pressures of \( \text{H}_2 \) (hydrogen gas), \( \text{Br}_2 \) (bromine gas), and \( \text{HBr} \) (hydrogen bromide) are sought in units of pascals (Pa). Each equation includes a text box for the respective pressure values to be inputted after calculations.

The question will be answered by using simple concept of chemical equilibrium.

For H, (g) + Br, (g) =2 HBr(g), K = 7.2 x 10-4 at 1362 K and AH° is positive. A vessel is charged with 48.2 Pa HBr, 1385 Pa H,, and 3370 Pa Br, at 1362 K. Will the reaction proceed to the left or right to reach equilibrium? left O right

For H, (g) + Br, (g) =2 HBr(g), K = 7.2 x 10-4 at 1362 K and AH° is positive. A vessel is charged with 48.2 Pa HBr, 1385 Pa H,, and 3370 Pa Br, at 1362 K. Will the reaction proceed to the left or right to reach equilibrium? left O right

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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Question

### Chemical Equilibrium and Le Chatelier's Principle

#### Problem Statement:
For the reaction:

\[ \text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{HBr}(g) \]

Given the equilibrium constant \( K = 7.2 \times 10^{-4} \) at 1362 K, and \( \Delta H^\circ \) is positive. A vessel is charged with the following partial pressures at 1362 K:
- \(\text{HBr}\): 48.2 Pa
- \(\text{H}_2\): 1385 Pa
- \(\text{Br}_2\): 3370 Pa

#### Question:
Will the reaction proceed to the left or right to reach equilibrium?

#### Options:
- \( \bigcirc \) left
- \( \bigcirc \) right

To solve this, we need to perform an equilibrium calculation using the reaction quotient \( Q \) and compare it to the equilibrium constant \( K \).

**Calculation Steps:**

1. **Calculate the Reaction Quotient (Q):**

\[ Q = \frac{{\text{(Pressure of HBr)}^2}}{{\text{(Pressure of H}_2) \times \text{(Pressure of Br}_2)}} \]

Given:
- Pressure of HBr \( = 48.2 \) Pa
- Pressure of H\(_2\) \( = 1385 \) Pa
- Pressure of Br\(_2\) \( = 3370 \) Pa

\[ Q = \frac{(48.2)^2}{(1385) \times (3370)} \]

2. **Compare \( Q \) and \( K \):**

- If \( Q < K \), the reaction will proceed to the right.
- If \( Q > K \), the reaction will proceed to the left.
- If \( Q = K \), the system is at equilibrium.

Let's substitute the values and calculate \( Q \):

\[ Q = \frac{48.2 \times 48.2}{1385 \times 3370} \]
\[ Q = \frac{2323.24}{4665450} \approx 4.98 \times 10^{-4} \]

Given that \( K = 7.2 \times 10^{-4} \

**Calculate the pressure (in pascals) of each species at equilibrium.**

\[ P_{H_2} = \boxed{\hspace{10cm}} \text{Pa} \]

\[ P_{Br_2} = \boxed{\hspace{10cm}} \text{Pa} \]

\[ P_{HBr} = \boxed{\hspace{10cm}} \text{Pa} \]

This set of equations is provided to calculate the equilibrium pressures of different chemical species in a reaction. The pressures of \( \text{H}_2 \) (hydrogen gas), \( \text{Br}_2 \) (bromine gas), and \( \text{HBr} \) (hydrogen bromide) are sought in units of pascals (Pa). Each equation includes a text box for the respective pressure values to be inputted after calculations.

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