The value of K for `H_(2)(g)+CO_(2)(g)hArr H_(2)O(g)+CO(g)` is 1.80 at `1000^(@)C`. If 1.0 mole of each `H_(2) and CO_(2)` placed in 1 litre flask, the final equilibrium concentration of CO at `1000^(@)C` will be

Options

A. 0.295 M
B. 0.385 M
C. 0.572 M
D. 0.473 M

Answer

To solve the problem, we need to find the final equilibrium concentration of CO when the reaction \( H_2(g) + CO_2(g) \rightleftharpoons H_2O(g) + CO(g) \) reaches equilibrium at \( 1000^\circ C \) with a given equilibrium constant \( K = 1.80 \).

### Step-by-Step Solution:

1. **Write the balanced equation**:
\[ H_2(g) + CO_2(g) \rightleftharpoons H_2O(g) + CO(g) \]

2. **Set up the initial conditions**:
- Initially, we have 1 mole of \( H_2 \) and 1 mole of \( CO_2 \) in a 1-liter flask.
- Therefore, the initial concentrations are:
\[ [H_2] = 1 \, M, \quad [CO_2] = 1 \, M, \quad [H_2O] = 0 \, M, \quad [CO] = 0 \, M \]

3. **Define the change in concentrations**:
- Let \( x \) be the amount of \( H_2 \) and \( CO_2 \) that reacts to form products at equilibrium.
- At equilibrium, the concentrations will be:
\[ [H_2] = 1 - x, \quad [CO_2] = 1 - x, \quad [H_2O] = x, \quad [CO] = x \]

4. **Write the expression for the equilibrium constant \( K \)**:
\[ K = \frac{[H_2O][CO]}{[H_2][CO_2]} \]
Substituting the equilibrium concentrations into the expression:
\[ 1.80 = \frac{x \cdot x}{(1 - x)(1 - x)} = \frac{x^2}{(1 - x)^2} \]

5. **Rearranging the equation**:
\[ 1.80(1 - x)^2 = x^2 \]
Expanding and rearranging gives:
\[ 1.80(1 - 2x + x^2) = x^2 \]
\[ 1.80 - 3.60x + 1.80x^2 = x^2 \]
\[ 0 = 0.80x^2 - 3.60x + 1.80 \]

6. **Solving the quadratic equation**:
\[ 0.80x^2 - 3.60x + 1.80 = 0 \]
Dividing through by 0.80:
\[ x^2 - 4.5x + 2.25 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{4.5 \pm \sqrt{(-4.5)^2 - 4 \cdot 1 \cdot 2.25}}{2 \cdot 1} \]
\[ x = \frac{4.5 \pm \sqrt{20.25 - 9}}{2} \]
\[ x = \frac{4.5 \pm \sqrt{11.25}}{2} \]
\[ x = \frac{4.5 \pm 3.35}{2} \]
This gives two possible values for \( x \):
\[ x \approx 3.925 \quad \text{(not valid since it exceeds initial moles)} \quad \text{and} \quad x \approx 0.572 \]

7. **Final equilibrium concentration of CO**:
Since \( x \) represents the concentration of \( CO \) at equilibrium:
\[ [CO] = x = 0.572 \, M \]

### Conclusion:
The final equilibrium concentration of CO at \( 1000^\circ C \) is \( 0.572 \, M \).