### 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 Process1. **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

Given, 2HI(g) ⇌ H2(g) + I2(g) Equilibrium constant (KC) = 0.0180 Temperature = 698 K Initially…

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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Chapter1: Chemical Foundations

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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

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