Equilibrium constants for the following reaction 1200 K are given:
`2H_2O (g) hArr 2H_2(g) +O_2(g) , " " K_1=6.4xx10^(-8)`
`2CO_2(g) hArr 2CO(g) +O_2(g) , " " K_2=1.6xx10^(-6)`
The equilibrium constant for the reaction
`H_2(g) + CO_2(g) hArr CO(g) +H_2O(g)` at 1200 K will be

Options

A. `0.05`
B. 20
C. `0.2`
D. `5.0`

Answer

To find the equilibrium constant for the reaction \( H_2(g) + CO_2(g) \rightleftharpoons CO(g) + H_2O(g) \) at 1200 K, we will use the given equilibrium constants \( K_1 \) and \( K_2 \) for the related reactions.

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

1. \( 2H_2O(g) \rightleftharpoons 2H_2(g) + O_2(g) \) with \( K_1 = 6.4 \times 10^{-8} \)
2. \( 2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g) \) with \( K_2 = 1.6 \times 10^{-6} \)

### Step 2: Reverse the first reaction

To obtain \( H_2(g) \) on the left side, we need to reverse the first reaction:

\[ 2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(g) \]

When we reverse a reaction, the equilibrium constant becomes the reciprocal:

\[ K_1' = \frac{1}{K_1} = \frac{1}{6.4 \times 10^{-8}} = 1.5625 \times 10^{7} \]

### Step 3: Write the second reaction as it is

The second reaction remains unchanged:

\[ 2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g) \quad (K_2 = 1.6 \times 10^{-6}) \]

### Step 4: Add the two reactions

Now, we add the modified first reaction and the second reaction:

\[ 2H_2(g) + O_2(g) + 2CO_2(g) \rightleftharpoons 2H_2O(g) + 2CO(g) + O_2(g) \]

The \( O_2(g) \) cancels out:

\[ 2H_2(g) + 2CO_2(g) \rightleftharpoons 2H_2O(g) + 2CO(g) \]

### Step 5: Divide the entire equation by 2

To match the desired reaction \( H_2(g) + CO_2(g) \rightleftharpoons CO(g) + H_2O(g) \), we divide the entire equation by 2:

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

### Step 6: Calculate the new equilibrium constant

When we divide the reaction by 2, the equilibrium constant is raised to the power of \( \frac{1}{2} \):

\[ K = \sqrt{K_1' \times K_2} \]

Substituting the values:

\[ K = \sqrt{(1.5625 \times 10^{7}) \times (1.6 \times 10^{-6})} \]

Calculating the product:

\[ K = \sqrt{(1.5625 \times 10^{7}) \times (1.6 \times 10^{-6})} = \sqrt{2.5 \times 10^{1}} = \sqrt{25} = 5 \]

### Final Answer

The equilibrium constant for the reaction \( H_2(g) + CO_2(g) \rightleftharpoons CO(g) + H_2O(g) \) at 1200 K is \( 5 \).

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