At which temperature, P C K K `K_P/K_C` value will be `1/4` for dissociation equilibrium of ammonia `[2 NH_3 ƒ N_2 + 3H_2]`?

Options

A. `6.09 K`
B. `12.18 K`
C. `24.36 K`
D. `18.27K`

Answer

To find the temperature at which the value of \( \frac{K_P}{K_C} \) will be \( \frac{1}{4} \) for the dissociation equilibrium of ammonia given by the reaction:

\[ 2 \text{NH}_3 \rightleftharpoons \text{N}_2 + 3 \text{H}_2 \]

we can follow these steps:

### Step 1: Write the expressions for \( K_P \) and \( K_C \)

The equilibrium constant \( K_P \) in terms of partial pressures is given by:

\[ K_P = \frac{(P_{\text{N}_2})(P_{\text{H}_2})^3}{(P_{\text{NH}_3})^2} \]

The equilibrium constant \( K_C \) in terms of concentrations is given by:

\[ K_C = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} \]

### Step 2: Relate \( K_P \) and \( K_C \)

The relationship between \( K_P \) and \( K_C \) is given by the equation:

\[ K_P = K_C (RT)^{\Delta n} \]

where \( \Delta n \) is the change in the number of moles of gas, which is calculated as:

\[ \Delta n = \text{moles of products} - \text{moles of reactants} = (1 + 3) - 2 = 2 \]

Thus, we can write:

\[ K_P = K_C (RT)^2 \]

### Step 3: Set up the equation for \( \frac{K_P}{K_C} \)

Now, we can express \( \frac{K_P}{K_C} \):

\[ \frac{K_P}{K_C} = (RT)^2 \]

### Step 4: Set the equation equal to \( \frac{1}{4} \)

We are given that \( \frac{K_P}{K_C} = \frac{1}{4} \):

\[ (RT)^2 = \frac{1}{4} \]

### Step 5: Solve for \( RT \)

Taking the square root of both sides gives us:

\[ RT = \frac{1}{2} \]

### Step 6: Solve for temperature \( T \)

Now, we can solve for \( T \):

\[ T = \frac{1}{2R} \]

Using the value of the gas constant \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \):

\[ T = \frac{1}{2 \times 0.0821} \approx 6.09 \, \text{K} \]

### Final Answer

The temperature at which \( \frac{K_P}{K_C} = \frac{1}{4} \) for the dissociation equilibrium of ammonia is approximately **6.09 K**.

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