Be sure to answer all parts. The equilibrium constant Ke for the reaction H₂(g) + Br₂(g) 52HBr(g) is 2.180 × 106 at 730°C. Starting with 1.20 moles of HBr in a 13.2-L reaction vessel, calculate the concentrations of H₂, Br2, and HBr at equilibrium. [H₂] = [Br₂] = [HBr] = M M M

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**Equilibrium Constant Calculation for the Reaction \(H_2(g) + Br_2(g) \leftrightarrow 2HBr(g)\)**

Be sure to answer all parts.

The equilibrium constant \( K_c \) for the reaction

\[ H_2(g) + Br_2(g) \leftrightarrow 2HBr(g) \]

is \( 2.180 \times 10^6 \) at 730°C. Starting with 1.20 moles of HBr in a 13.2-L reaction vessel, calculate the concentrations of \( H_2 \), \( Br_2 \), and \( HBr \) at equilibrium.

**Initial Data:**
- Initial moles of \(HBr\): 1.20 moles
- Volume of reaction vessel: 13.2 L

**Equilibrium Condition:**
Let the change in concentration of \( H_2 \) and \( Br_2 \) at equilibrium be \( + x \) M, since they are produced from the dissociation of \( HBr \). Consequently, the change in concentration of \( HBr \) would be \( -2x \) M (due to the stoichiometric coefficient of 2).

Initial Concentrations:
- \([H_2]\) = 0 M (since initially no \(H_2\) is present)
- \([Br_2]\) = 0 M (since initially no \(Br_2\) is present)
- \([HBr]\) = \(\frac{1.20 \text{ moles}}{13.2 \text{ L}}\) = 0.0909 M

Change in Concentration:
- \(\Delta[H_2]\) = \( + x \)
- \(\Delta[Br_2]\) = \( + x \)
- \(\Delta[HBr]\) = \( -2x \)

Equilibrium Concentrations:
- \([H_2]\) = \( x \)
- \([Br_2]\) = \( x \)
- \([HBr]\) = \( 0.0909 - 2x \)

Given the equilibrium constant expression for the reaction:

\[ K_c = \frac{[HBr]^2}{[H_2][Br_2]} \]

Substitute the equilibrium concentrations:

\[ 2.180 \
Transcribed Image Text:**Equilibrium Constant Calculation for the Reaction \(H_2(g) + Br_2(g) \leftrightarrow 2HBr(g)\)** Be sure to answer all parts. The equilibrium constant \( K_c \) for the reaction \[ H_2(g) + Br_2(g) \leftrightarrow 2HBr(g) \] is \( 2.180 \times 10^6 \) at 730°C. Starting with 1.20 moles of HBr in a 13.2-L reaction vessel, calculate the concentrations of \( H_2 \), \( Br_2 \), and \( HBr \) at equilibrium. **Initial Data:** - Initial moles of \(HBr\): 1.20 moles - Volume of reaction vessel: 13.2 L **Equilibrium Condition:** Let the change in concentration of \( H_2 \) and \( Br_2 \) at equilibrium be \( + x \) M, since they are produced from the dissociation of \( HBr \). Consequently, the change in concentration of \( HBr \) would be \( -2x \) M (due to the stoichiometric coefficient of 2). Initial Concentrations: - \([H_2]\) = 0 M (since initially no \(H_2\) is present) - \([Br_2]\) = 0 M (since initially no \(Br_2\) is present) - \([HBr]\) = \(\frac{1.20 \text{ moles}}{13.2 \text{ L}}\) = 0.0909 M Change in Concentration: - \(\Delta[H_2]\) = \( + x \) - \(\Delta[Br_2]\) = \( + x \) - \(\Delta[HBr]\) = \( -2x \) Equilibrium Concentrations: - \([H_2]\) = \( x \) - \([Br_2]\) = \( x \) - \([HBr]\) = \( 0.0909 - 2x \) Given the equilibrium constant expression for the reaction: \[ K_c = \frac{[HBr]^2}{[H_2][Br_2]} \] Substitute the equilibrium concentrations: \[ 2.180 \
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