The equilibrium constant of the following reactions at `400 K` are given:
`2H_(2)O(g) hArr 2H_(2)(g)+O_(2)(g), K_(1)=3.0xx10^(-13)`
`2CO_(2)(g) hArr 2CO(g)+O_(2)(g), K_(2)=2xx10^(-12)`
Then, the equilibrium constant K for the reaction
`H_(2)(g)+CO_(2)(g) hArr CO(g)+H_(2)O(g)`
is

Options

A. `2.04`
B. `20.5`
C. `0.85`
D. `1.4`

Explanation

To find the equilibrium constant \( K \) for the reaction

\[ H_2(g) + CO_2(g) \rightleftharpoons CO(g) + H_2O(g) \]

we will use the given equilibrium constants \( K_1 \) and \( K_2 \) for the two reactions provided.

### Step 1: Write the expressions for the given reactions and their equilibrium constants.

1. For the reaction

\[ 2H_2O(g) \rightleftharpoons 2H_2(g) + O_2(g) \quad (K_1 = 3.0 \times 10^{-13}) \]

The equilibrium constant \( K_1 \) can be expressed as:

\[ K_1 = \frac{[H_2]^2 [O_2]}{[H_2O]^2} \]

2. For the reaction

\[ 2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g) \quad (K_2 = 2.0 \times 10^{-12}) \]

The equilibrium constant \( K_2 \) can be expressed as:

\[ K_2 = \frac{[CO]^2 [O_2]}{[CO_2]^2} \]

### Step 2: Write the expression for the desired reaction.

For the reaction

\[ H_2(g) + CO_2(g) \rightleftharpoons CO(g) + H_2O(g) \]

The equilibrium constant \( K \) can be expressed as:

\[ K = \frac{[CO][H_2O]}{[H_2][CO_2]} \]

### Step 3: Relate the desired reaction to the given reactions.

To find \( K \), we can manipulate the given reactions.

- The first reaction can be reversed and halved:

\[ H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g) \quad (K = \sqrt{K_1}) \]

Thus,

\[ K = \sqrt{K_1} \]

- The second reaction can also be halved:

\[ CO(g) + \frac{1}{2}O_2(g) \rightleftharpoons CO_2(g) \quad (K = \frac{1}{\sqrt{K_2}}) \]

Thus,

\[ K = \frac{1}{\sqrt{K_2}} \]

### Step 4: Combine the expressions for \( K \).

We can combine these two results:

\[ K = \frac{1}{\sqrt{K_2}} \cdot \sqrt{K_1} \]

### Step 5: Substitute the values of \( K_1 \) and \( K_2 \).

Substituting the values:

\[ K = \frac{1}{\sqrt{2.0 \times 10^{-12}}} \cdot \sqrt{3.0 \times 10^{-13}} \]

Calculating \( \sqrt{K_1} \) and \( \sqrt{K_2} \):

\[ \sqrt{K_1} = \sqrt{3.0 \times 10^{-13}} \approx 5.477 \times 10^{-7} \]
\[ \sqrt{K_2} = \sqrt{2.0 \times 10^{-12}} \approx 4.472 \times 10^{-6} \]

Thus,

\[ K = \frac{5.477 \times 10^{-7}}{4.472 \times 10^{-6}} \approx 0.1225 \]

### Step 6: Final Calculation

Calculating this gives us:

\[ K \approx 2.5 \times 10^{-1} \text{ or } 0.25 \]

### Conclusion

The equilibrium constant \( K \) for the reaction

\[ H_2(g) + CO_2(g) \rightleftharpoons CO(g) + H_2O(g) \]

is approximately \( 0.25 \).