Equilibrium constant for the following reactions have been determined at `823K`.
`CaO(s)+H_(2)(g)hArrCO(s)+H_(2)O(g) K_(1)=60`
`CaO(s)+CO(g)hArrCO(S)+CO_(2)(g) K_(2)=400`
Using this information, calculate, equilibrium constant (at the same temperature) for `:-` `CO_(2)(g)+H_(2)(g)hArrCO(s)+H_(2)O(g)K_(3)=?`
`CO(g)+H_(2)O(g)hArrCO_(2)(g)+H_(2)(g)K_(4)=?`

Options

A. `K_(3)=0.15,K_(4)=6.66`
B. `K_(3)=1.5,K_(4)=66.6`
C. `K_(3)=15,K_(4)=666`
D. None

Explanation

To solve the problem, we need to find the equilibrium constants \( K_3 \) and \( K_4 \) for the given reactions using the provided equilibrium constants \( K_1 \) and \( K_2 \).

### Given Reactions and Equilibrium Constants:
1. \( \text{CaO}(s) + \text{H}_2(g) \rightleftharpoons \text{CO}(s) + \text{H}_2O(g) \) with \( K_1 = 60 \)
2. \( \text{CaO}(s) + \text{CO}(g) \rightleftharpoons \text{CO}(s) + \text{CO}_2(g) \) with \( K_2 = 400 \)

### Required Reactions:
1. \( \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(s) + \text{H}_2O(g) \) (to find \( K_3 \))
2. \( \text{CO}(g) + \text{H}_2O(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g) \) (to find \( K_4 \))

### Step 1: Finding \( K_3 \)
To find \( K_3 \), we can manipulate the given reactions.

1. Start with the first reaction:
\[ \text{CaO}(s) + \text{H}_2(g) \rightleftharpoons \text{CO}(s) + \text{H}_2O(g) \quad (K_1 = 60) \]

2. Now, take the second reaction and reverse it:
\[ \text{CO}(s) + \text{CO}_2(g) \rightleftharpoons \text{CaO}(s) + \text{CO}(g) \quad (K_2^{-1} = \frac{1}{400}) \]

3. Adding these two reactions gives:
\[ \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(s) + \text{H}_2O(g) \]

4. The equilibrium constant for the overall reaction is:
\[ K_3 = K_1 \times K_2^{-1} = K_1 \div K_2 = \frac{60}{400} = 0.15 \]

### Step 2: Finding \( K_4 \)
To find \( K_4 \), we can again manipulate the given reactions.

1. Start with the second reaction:
\[ \text{CaO}(s) + \text{CO}(g) \rightleftharpoons \text{CO}(s) + \text{CO}_2(g) \quad (K_2 = 400) \]

2. Now, take the first reaction and reverse it:
\[ \text{CO}(s) + \text{H}_2O(g) \rightleftharpoons \text{CaO}(s) + \text{H}_2(g) \quad (K_1^{-1} = \frac{1}{60}) \]

3. Adding these two reactions gives:
\[ \text{CO}(g) + \text{H}_2O(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g) \]

4. The equilibrium constant for the overall reaction is:
\[ K_4 = K_2 \times K_1^{-1} = K_2 \div K_1 = \frac{400}{60} \approx 6.67 \]

### Final Results:
- \( K_3 = 0.15 \)
- \( K_4 \approx 6.67 \)

### Summary:
- \( K_3 = 0.15 \)
- \( K_4 \approx 6.67 \)