If the following reaction at 698 K has equilibrium concentration values of 0.584 M HI, 0.0793 M H₂, and 0.0793 M 12, what is the value of the equilibrium constant? 2H1 H₂(g) + 1₂ (g) 2 (0.584m) → 0.0793 M + 0.0793 M L -3

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**Equilibrium Constant Calculation for the Reaction 2HI ⇌ H₂ + I₂**

**Problem Statement:**
At 698 K, the equilibrium concentrations for the reaction 2HI (g) ⇌ H₂ (g) + I₂ (g) are given as:
- [HI] = 0.584 M
- [H₂] = 0.0793 M
- [I₂] = 0.0793 M

**Objective:**
Calculate the equilibrium constant (Kc) for this reaction.

**Solution:**

1. **Balanced Reaction:**
   \[
   2HI (g) \rightarrow H₂ (g) + I₂ (g)
   \]

2. **Expression for the Equilibrium Constant (Kc):**
   \[
   Kc = \frac{[H₂][I₂]}{[HI]^2}
   \]

3. **Substitute the Equilibrium Concentrations into the Expression:**
   - [HI] = 0.584 M
   - [H₂] = 0.0793 M
   - [I₂] = 0.0793 M
   
   \[
   Kc = \frac{(0.0793)(0.0793)}{(0.584)^2}
   \]

4. **Calculate:**
   \[
   Kc = \frac{0.0793 \times 0.0793}{0.584^2}
   \]
   \[
   Kc = \frac{0.00628049}{0.341056} \approx 5.80 \times 10^{-3}
   \]

**Conclusion:**
The equilibrium constant \( Kc \) for the reaction at 698 K is approximately \( 5.80 \times 10^{-3} \).
Transcribed Image Text:**Equilibrium Constant Calculation for the Reaction 2HI ⇌ H₂ + I₂** **Problem Statement:** At 698 K, the equilibrium concentrations for the reaction 2HI (g) ⇌ H₂ (g) + I₂ (g) are given as: - [HI] = 0.584 M - [H₂] = 0.0793 M - [I₂] = 0.0793 M **Objective:** Calculate the equilibrium constant (Kc) for this reaction. **Solution:** 1. **Balanced Reaction:** \[ 2HI (g) \rightarrow H₂ (g) + I₂ (g) \] 2. **Expression for the Equilibrium Constant (Kc):** \[ Kc = \frac{[H₂][I₂]}{[HI]^2} \] 3. **Substitute the Equilibrium Concentrations into the Expression:** - [HI] = 0.584 M - [H₂] = 0.0793 M - [I₂] = 0.0793 M \[ Kc = \frac{(0.0793)(0.0793)}{(0.584)^2} \] 4. **Calculate:** \[ Kc = \frac{0.0793 \times 0.0793}{0.584^2} \] \[ Kc = \frac{0.00628049}{0.341056} \approx 5.80 \times 10^{-3} \] **Conclusion:** The equilibrium constant \( Kc \) for the reaction at 698 K is approximately \( 5.80 \times 10^{-3} \).
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