Consider the following reaction where K. = 1.29×10-² at 600 K. COCI»(g) CO(g) + Cl2(g) A reaction mixture was found to contain 8.79×10-² moles of COCI,(g), 2.04×10*² moles of CO(g), and 3.67×10² moles of Cl,(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qe, equals The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.

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### Equilibrium Concentration Calculation for COCl₂ Decomposition Reaction

#### Reaction Details
Consider the following reaction where the equilibrium constant (Kₐ) is 1.29×10⁻² at 600 K:

\[ \text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g) \]

A reaction mixture was found to contain:
- 8.79×10⁻² moles of \(\text{COCl}_2(g)\)
- 2.04×10⁻² moles of \(\text{CO}(g)\)
- 3.67×10⁻² moles of \(\text{Cl}_2(g)\)

in a 1.00 liter container.

#### Equilibrium Check and Quotient Calculation

**The main question is:** 
- Is the reaction at equilibrium? If not, in which direction must it proceed to achieve equilibrium?

First, we need to calculate the reaction quotient, \( Q_c \).

The reaction quotient, \( Q_c \), is calculated using the formula:

\[ Q_c = \frac{[\text{CO}][\text{Cl₂}]}{[\text{COCl₂}]} \]

Given the concentrations:
- \([\text{COCl₂}] = 8.79 \times 10^{-2} \text{ moles/L}\)
- \([\text{CO}] = 2.04 \times 10^{-2} \text{ moles/L}\)
- \([\text{Cl₂}] = 3.67 \times 10^{-2} \text{ moles/L}\)

Now substituting these values into the formula for \( Q_c \):

\[ Q_c = \frac{(2.04 \times 10^{-2})(3.67 \times 10^{-2})}{8.79 \times 10^{-2}} \]

Perform the calculation:

\[ Q_c = \frac{7.4868 \times 10^{-4}}{8.79 \times 10^{-2}} \]
\[ Q_c = 8.52 \times 10^{-3} \]

#### Comparing \( Q_c \) to \( K_c \)
- \( Q_c = 8.52 \times 10^{-3} \)
- \( K_c = 1.29 \times
Transcribed Image Text:### Equilibrium Concentration Calculation for COCl₂ Decomposition Reaction #### Reaction Details Consider the following reaction where the equilibrium constant (Kₐ) is 1.29×10⁻² at 600 K: \[ \text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g) \] A reaction mixture was found to contain: - 8.79×10⁻² moles of \(\text{COCl}_2(g)\) - 2.04×10⁻² moles of \(\text{CO}(g)\) - 3.67×10⁻² moles of \(\text{Cl}_2(g)\) in a 1.00 liter container. #### Equilibrium Check and Quotient Calculation **The main question is:** - Is the reaction at equilibrium? If not, in which direction must it proceed to achieve equilibrium? First, we need to calculate the reaction quotient, \( Q_c \). The reaction quotient, \( Q_c \), is calculated using the formula: \[ Q_c = \frac{[\text{CO}][\text{Cl₂}]}{[\text{COCl₂}]} \] Given the concentrations: - \([\text{COCl₂}] = 8.79 \times 10^{-2} \text{ moles/L}\) - \([\text{CO}] = 2.04 \times 10^{-2} \text{ moles/L}\) - \([\text{Cl₂}] = 3.67 \times 10^{-2} \text{ moles/L}\) Now substituting these values into the formula for \( Q_c \): \[ Q_c = \frac{(2.04 \times 10^{-2})(3.67 \times 10^{-2})}{8.79 \times 10^{-2}} \] Perform the calculation: \[ Q_c = \frac{7.4868 \times 10^{-4}}{8.79 \times 10^{-2}} \] \[ Q_c = 8.52 \times 10^{-3} \] #### Comparing \( Q_c \) to \( K_c \) - \( Q_c = 8.52 \times 10^{-3} \) - \( K_c = 1.29 \times
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