Given equilibrium constants for the following reaction at `1200^(@)C`
`C(g)+CO_(2)(g)hArr 2CO(g), K_(p_(1))=1.47xx10^(3)`
`CO(g)+1//2O_(2)(g)hArr COCl_(2)(g), K_(P_(2))=2xx10^(6)`
Calculate `K_(P)` for the reaction :
`C(g)+CO_(2)+O_(2)(g)hArr 2COCl_(2)(g)`

Options

A. `5.8xx10^(15)`
B. `5.88xx10^(10)`
C. `3.175xx10^(10)`
D. `4.2xx10^(12)`

Answer

To solve the problem, we need to calculate the equilibrium constant \( K_P \) for the reaction:

\[ C(g) + CO_2(g) + O_2(g) \rightleftharpoons 2COCl_2(g) \]

using the given equilibrium constants for the two reactions:

1. \( C(g) + CO_2(g) \rightleftharpoons 2CO(g) \) with \( K_{P_1} = 1.47 \times 10^3 \)
2. \( CO(g) + \frac{1}{2} O_2(g) \rightleftharpoons COCl_2(g) \) with \( K_{P_2} = 2 \times 10^6 \)

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

We have:
- Reaction 1:
\[ C(g) + CO_2(g) \rightleftharpoons 2CO(g) \]
\[ K_{P_1} = 1.47 \times 10^3 \]

- Reaction 2:
\[ CO(g) + \frac{1}{2} O_2(g) \rightleftharpoons COCl_2(g) \]
\[ K_{P_2} = 2 \times 10^6 \]

### Step 2: Modify Reaction 2 for our needs

To use Reaction 2 in our desired reaction, we need to multiply it by 2 to match the stoichiometry of the final reaction:

\[ 2CO(g) + O_2(g) \rightleftharpoons 2COCl_2(g) \]

When we multiply a reaction by a coefficient, the equilibrium constant is raised to that power. Therefore, the new equilibrium constant \( K_{P_2'} \) for this modified reaction will be:

\[ K_{P_2'} = (K_{P_2})^2 = (2 \times 10^6)^2 = 4 \times 10^{12} \]

### Step 3: Combine the modified reactions

Now we can add Reaction 1 and the modified Reaction 2:

1. \( C(g) + CO_2(g) \rightleftharpoons 2CO(g) \) (from Reaction 1)
2. \( 2CO(g) + O_2(g) \rightleftharpoons 2COCl_2(g) \) (modified Reaction 2)

When we add these reactions, the \( 2CO(g) \) cancels out:

\[ C(g) + CO_2(g) + O_2(g) \rightleftharpoons 2COCl_2(g) \]

### Step 4: Calculate the overall equilibrium constant

The overall equilibrium constant \( K_P \) for the final reaction is the product of the equilibrium constants of the individual reactions:

\[ K_P = K_{P_1} \times K_{P_2'} \]

Substituting the values we have:

\[ K_P = (1.47 \times 10^3) \times (4 \times 10^{12}) \]

### Step 5: Perform the multiplication

Calculating the multiplication:

1. \( 1.47 \times 4 = 5.88 \)
2. For the powers of ten: \( 10^3 \times 10^{12} = 10^{15} \)

Thus, we have:

\[ K_P = 5.88 \times 10^{15} \]

### Final Answer

The equilibrium constant \( K_P \) for the reaction \( C(g) + CO_2(g) + O_2(g) \rightleftharpoons 2COCl_2(g) \) is:

\[ K_P \approx 5.88 \times 10^{15} \]