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

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### 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} \
Transcribed Image Text:### 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.
Transcribed Image Text:**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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