What is the concentration of CO in equilibrium at `25^(@)C` in a sample of a gas originally containing 1.0 mol `L^(-1)` of `CO_(2)` ? For dissociation of `CO_(2)` at `25^(@)C, K_(c)=2.96xx10^(-92)`.

Options

A. `3.9xx10^(-3)"mole litre"^(-1)`
B. `1.2xx10^(-28)"mole litre"^(-1)`
C. `5.6xx10^(-32)"mol litre"^(-1)`
D. `1.1xx10^(-33)"mol litre"^(-1)`

Explanation

To solve the problem, we need to find the concentration of CO at equilibrium when starting with a concentration of 1.0 mol L⁻¹ of CO₂, given that the dissociation of CO₂ at 25°C has a Kc value of 2.96 x 10⁻⁹².

### Step-by-Step Solution:

1. **Write the balanced chemical equation for the dissociation of CO₂:**
\[ 2 \text{CO}_2(g) \rightleftharpoons 2 \text{CO}(g) + \text{O}_2(g) \]

2. **Define the initial concentration of CO₂:**
The initial concentration of CO₂ is given as:
\[ [\text{CO}_2] = 1.0 \text{ mol L}^{-1} \]

3. **Set up the change in concentration at equilibrium:**
Let \( x \) be the amount of CO₂ that dissociates at equilibrium. According to the stoichiometry of the reaction:
- The concentration of CO₂ at equilibrium will be:
\[ [\text{CO}_2] = 1.0 - x \]
- The concentration of CO at equilibrium will be:
\[ [\text{CO}] = 2x \]
- The concentration of O₂ at equilibrium will be:
\[ [\text{O}_2] = x \]

4. **Write the expression for the equilibrium constant \( K_c \):**
The expression for \( K_c \) for the reaction is:
\[ K_c = \frac{[\text{CO}]^2 [\text{O}_2]}{[\text{CO}_2]^2} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(2x)^2 (x)}{(1.0 - x)^2} \]

5. **Substitute the value of \( K_c \):**
Given \( K_c = 2.96 \times 10^{-92} \):
\[ 2.96 \times 10^{-92} = \frac{4x^3}{(1.0 - x)^2} \]

6. **Assume \( x \) is very small compared to 1.0:**
Since \( K_c \) is very small, we can assume that \( x \) is negligible compared to 1.0:
\[ 1.0 - x \approx 1.0 \]
Thus, the equation simplifies to:
\[ 2.96 \times 10^{-92} = 4x^3 \]

7. **Solve for \( x \):**
Rearranging gives:
\[ x^3 = \frac{2.96 \times 10^{-92}}{4} \]
\[ x^3 = 7.4 \times 10^{-93} \]
Taking the cube root:
\[ x = (7.4 \times 10^{-93})^{1/3} \approx 1.96 \times 10^{-31} \text{ mol L}^{-1} \]

8. **Calculate the concentration of CO at equilibrium:**
Since the concentration of CO is given by \( 2x \):
\[ [\text{CO}] = 2x = 2 \times 1.96 \times 10^{-31} \approx 3.92 \times 10^{-31} \text{ mol L}^{-1} \]

### Final Answer:
The concentration of CO at equilibrium is approximately:
\[ [\text{CO}] \approx 3.92 \times 10^{-31} \text{ mol L}^{-1} \]