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
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5baa0102-60f7-4f8d-865b-4da67ad08e15%2F523b97d0-9f1a-4ad8-9d4b-0bebda8f0160%2Fudrtmox_processed.jpeg&w=3840&q=75)
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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