For the synthesis of ammonia, `K_c` is 1.2 at `375^@C,N_2(g)+3H_2(g)hArr 2NH_3(g)`
What is `K_p` at this temperature ?

Options

A. `4.1xx10^(-8)`
B. `4.2xx10^(-4)`
C. `1.3xx10^(-3)`
D. `3.4xx10^(3)`

Answer

To find the value of \( K_p \) for the synthesis of ammonia at \( 375^\circ C \), we can use the relationship between \( K_p \) and \( K_c \). Here are the steps to solve the problem:

### Step 1: Write the relationship between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c \times (RT)^{\Delta N_g} \]
where:
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta N_g \) is the change in the number of moles of gas (moles of products - moles of reactants).

### Step 2: Determine \( \Delta N_g \)
For the reaction:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
- Moles of gaseous products = 2 (from \( 2NH_3 \))
- Moles of gaseous reactants = 1 (from \( N_2 \)) + 3 (from \( 3H_2 \)) = 4

Thus,
\[ \Delta N_g = \text{moles of products} - \text{moles of reactants} = 2 - 4 = -2 \]

### Step 3: Convert the temperature to Kelvin
The temperature given is \( 375^\circ C \). To convert this to Kelvin:
\[ T(K) = 375 + 273.15 = 648.15 \approx 648 \text{ K} \]

### Step 4: Substitute values into the equation
Now we can substitute the values into the equation for \( K_p \):
- Given \( K_c = 1.2 \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 648 \, \text{K} \)
- \( \Delta N_g = -2 \)

Substituting these values:
\[ K_p = 1.2 \times (0.0821 \times 648)^{-2} \]

### Step 5: Calculate \( RT \)
First, calculate \( RT \):
\[ RT = 0.0821 \times 648 \approx 53.3 \]

### Step 6: Calculate \( K_p \)
Now substituting back into the equation:
\[ K_p = 1.2 \times (53.3)^{-2} \]
Calculating \( (53.3)^{-2} \):
\[ (53.3)^{-2} \approx 0.000353 \]
Now, multiply by \( 1.2 \):
\[ K_p \approx 1.2 \times 0.000353 \approx 0.0004236 \approx 4.2 \times 10^{-4} \]

### Final Answer
Thus, the value of \( K_p \) at \( 375^\circ C \) is:
\[ K_p \approx 4.2 \times 10^{-4} \]

---