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 find the concentration of CO at equilibrium when starting with 1.0 mol L⁻¹ of CO₂, we will follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of CO₂ can be represented as:
\[ 2 \text{CO}_2(g) \rightleftharpoons 2 \text{CO}(g) + \text{O}_2(g) \]

### Step 2: Set up the initial concentrations
Initially, we have:
- \([CO_2] = 1.0 \, \text{mol L}^{-1}\)
- \([CO] = 0 \, \text{mol L}^{-1}\)
- \([O_2] = 0 \, \text{mol L}^{-1}\)

### Step 3: Define the change in concentration
Let \(x\) be the amount of CO₂ that dissociates. According to the stoichiometry of the reaction:
- The change in concentration for CO₂ will be \(-2x\)
- The change in concentration for CO will be \(+2x\)
- The change in concentration for O₂ will be \(+x\)

### Step 4: Write the equilibrium concentrations
At equilibrium, the concentrations will be:
- \([CO_2] = 1.0 - 2x\)
- \([CO] = 2x\)
- \([O_2] = x\)

### Step 5: Write the expression for the equilibrium constant \(K_c\)
The equilibrium constant expression for the reaction is:
\[ K_c = \frac{[CO]^2 [O_2]}{[CO_2]^2} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(2x)^2 (x)}{(1.0 - 2x)^2} \]

### Step 6: Substitute the value of \(K_c\)
Given \(K_c = 2.96 \times 10^{-92}\):
\[ 2.96 \times 10^{-92} = \frac{(2x)^2 (x)}{(1.0 - 2x)^2} \]
This simplifies to:
\[ 2.96 \times 10^{-92} = \frac{4x^3}{(1.0 - 2x)^2} \]

### Step 7: Make an approximation
Since \(K_c\) is very small, we can assume that \(x\) is very small compared to 1. Thus, we can approximate:
\[ 1.0 - 2x \approx 1.0 \]
This gives us:
\[ 2.96 \times 10^{-92} \approx \frac{4x^3}{1.0^2} \]
So:
\[ 4x^3 = 2.96 \times 10^{-92} \]

### Step 8: Solve for \(x\)
Rearranging gives:
\[ x^3 = \frac{2.96 \times 10^{-92}}{4} = 0.74 \times 10^{-92} \]
Taking the cube root:
\[ x = \sqrt[3]{0.74 \times 10^{-92}} \]

### Step 9: Calculate \(x\)
Calculating \(x\):
\[ x \approx 1.96 \times 10^{-31} \, \text{mol L}^{-1} \]

### Step 10: Find the concentration of CO at equilibrium
Since \([CO] = 2x\):
\[ [CO] = 2(1.96 \times 10^{-31}) = 3.92 \times 10^{-31} \, \text{mol L}^{-1} \]

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