The following equilibrium constants were determined at `1120 K :`
`2CO(g)hArrC(s)+CO_(2)(g), , K_(p1)=10^(-14)atm^(-1)`
`CO(g)+Cl_(2)(g)hArrCOCl_(2)(g), , K_(p2)=6xx10^(-3)atm^(-1)`
What is the equilibrium constant `K_(c)` for the foollowing reaction at `1120 K:`
`C(s)+CO_(2)(g)+2Cl_(2)(g)hArr2COCl_(2)(g)`

Options

A. `3.31xx10^(11)M^(-1)`
B. `5.5xx10^(10)M^(-1)`
C. `5.51xx10^(6)M^(-1)`
D. None of these

Answer

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

\[ C(s) + CO_2(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g) \]

Given the following equilibrium constants at \( 1120 \, K \):

1. \( 2CO(g) \rightleftharpoons C(s) + CO_2(g) \), \( K_{p1} = 10^{-14} \, atm^{-1} \)
2. \( CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g) \), \( K_{p2} = 6 \times 10^{-3} \, atm^{-1} \)

### Step 1: Reverse the first reaction

The first reaction needs to be reversed to match the desired equation. When we reverse a reaction, the equilibrium constant is inverted.

\[ C(s) + CO_2(g) \rightleftharpoons 2CO(g) \quad K_{p1}' = \frac{1}{K_{p1}} = \frac{1}{10^{-14}} = 10^{14} \, atm \]

### Step 2: Multiply the second reaction by 2

Next, we multiply the second reaction by 2 to match the stoichiometry of the desired equation.

\[ 2CO(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g) \quad K_{p2}' = (K_{p2})^2 = (6 \times 10^{-3})^2 = 36 \times 10^{-6} \, atm^{-2} \]

### Step 3: Add the two modified reactions

Now we add the modified reactions together:

1. \( C(s) + CO_2(g) \rightleftharpoons 2CO(g) \) \( (K_{p1}' = 10^{14}) \)
2. \( 2CO(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g) \) \( (K_{p2}' = 36 \times 10^{-6}) \)

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

\[ C(s) + CO_2(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g) \]

### Step 4: Calculate the overall \( K_p \)

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

\[ K_p = K_{p1}' \times K_{p2}' = 10^{14} \times 36 \times 10^{-6} = 36 \times 10^{8} \, atm \]

### Step 5: Convert \( K_p \) to \( K_c \)

To find \( K_c \), we use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c (RT)^{\Delta n} \]

Where:
- \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \)
- \( T = 1120 \, K \)
- \( \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - (1 + 2) = -1 \)

Thus, we have:

\[ K_c = \frac{K_p}{(RT)^{\Delta n}} = K_p \cdot (RT)^{1} \]

Calculating \( RT \):

\[ RT = 0.0821 \, L \cdot atm/(K \cdot mol) \times 1120 \, K = 92.352 \, L \cdot atm/mol \]

Now substituting:

\[ K_c = 36 \times 10^{8} \cdot 92.352 \]

Calculating \( K_c \):

\[ K_c \approx 3.32 \times 10^{10} \, L/mol \]

### Final Answer

The equilibrium constant \( K_c \) for the reaction at \( 1120 \, K \) is approximately:

\[ K_c \approx 3.32 \times 10^{10} \, L/mol \]

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