The equilibrium constant, Ke, for the following reaction is 0.0180 at 698 K. 2HI(g) → H₂(g) + 1₂ (9) Calculate the equilibrium concentrations of reactant and products when 0.357 moles of HI(g) are introduced into a 1.00 L vessel at 698 K. [HI] = M [H₂] = [1₂] = = M M

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### Equilibrium Calculation for the Reaction 2HI(g) ⇌ H₂(g) + I₂(g)

The equilibrium constant, \( K_c \), for the following reaction is 0.0180 at 698 K.

\[ 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \]

Calculate the equilibrium concentrations of reactant and products when 0.357 moles of \( HI(g) \) are introduced into a 1.00 L vessel at 698 K.

\[ [HI] = \underline{\hspace{60px}} \, M \]
\[ [H_2] = \underline{\hspace{60px}} \, M \]
\[ [I_2] = \underline{\hspace{60px}} \, M \]

### Explanation of the Equilibrium Process

1. **Initial Concentrations:**
   - Initially, only \( HI \) is present in the reaction vessel.
   - Moles of \( HI \) = 0.357 moles
   - Since the volume of the vessel is 1.00 L,
     \[ [HI]_{initial} = \frac{0.357\, \text{moles}}{1.00\, \text{L}} = 0.357\, M \]
   - \( [H_2]_{initial} = 0\, M \)
   - \( [I_2]_{initial} = 0\, M \)

2. **Change in Concentrations:**
   - As the reaction reaches equilibrium, \( HI \) will dissociate into \( H_2 \) and \( I_2 \).
   - Let the change in concentration of \( HI \) be \( -2x \).
   - Therefore, the change in concentrations of \( H_2 \) and \( I_2 \) will be \( +x \) each.
   - At equilibrium:
     \[ [HI] = 0.357 - 2x \]
     \[ [H_2] = x \]
     \[ [I_2] = x \]

3. **Equilibrium Expression:**
   - The equilibrium constant expression for the reaction is:
     \[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]
   - Substituting the equilibrium concentrations
Transcribed Image Text:### Equilibrium Calculation for the Reaction 2HI(g) ⇌ H₂(g) + I₂(g) The equilibrium constant, \( K_c \), for the following reaction is 0.0180 at 698 K. \[ 2HI(g) \rightleftharpoons H_2(g) + I_2(g) \] Calculate the equilibrium concentrations of reactant and products when 0.357 moles of \( HI(g) \) are introduced into a 1.00 L vessel at 698 K. \[ [HI] = \underline{\hspace{60px}} \, M \] \[ [H_2] = \underline{\hspace{60px}} \, M \] \[ [I_2] = \underline{\hspace{60px}} \, M \] ### Explanation of the Equilibrium Process 1. **Initial Concentrations:** - Initially, only \( HI \) is present in the reaction vessel. - Moles of \( HI \) = 0.357 moles - Since the volume of the vessel is 1.00 L, \[ [HI]_{initial} = \frac{0.357\, \text{moles}}{1.00\, \text{L}} = 0.357\, M \] - \( [H_2]_{initial} = 0\, M \) - \( [I_2]_{initial} = 0\, M \) 2. **Change in Concentrations:** - As the reaction reaches equilibrium, \( HI \) will dissociate into \( H_2 \) and \( I_2 \). - Let the change in concentration of \( HI \) be \( -2x \). - Therefore, the change in concentrations of \( H_2 \) and \( I_2 \) will be \( +x \) each. - At equilibrium: \[ [HI] = 0.357 - 2x \] \[ [H_2] = x \] \[ [I_2] = x \] 3. **Equilibrium Expression:** - The equilibrium constant expression for the reaction is: \[ K_c = \frac{[H_2][I_2]}{[HI]^2} \] - Substituting the equilibrium concentrations
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