For the reaction `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)` at 298K, enthalpy and entropy changes are -92.4 kJ and -198.2 `JK^(-1)` respectively. Calculate the equilibrium constant of the reaction `(R=8.314JK^(-1)mol^(-1))`.

Explanation

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

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

Given:
- Enthalpy change (\( \Delta H \)) = -92.4 kJ
- Entropy change (\( \Delta S \)) = -198.2 J/K
- Temperature (\( T \)) = 298 K
- Gas constant (\( R \)) = 8.314 J/(K·mol)

### Step 1: Convert Enthalpy Change to Joules
Since \( \Delta H \) is given in kJ, we need to convert it to Joules for consistency with \( \Delta S \).

\[ \Delta H = -92.4 \, \text{kJ} \times 1000 \, \text{J/kJ} = -92400 \, \text{J} \]

### Step 2: Calculate Gibbs Free Energy Change (\( \Delta G \))
Using the formula:

\[ \Delta G = \Delta H - T \Delta S \]

Substituting the values:

\[ \Delta G = -92400 \, \text{J} - 298 \, \text{K} \times (-198.2 \, \text{J/K}) \]

Calculating \( T \Delta S \):

\[ T \Delta S = 298 \times (-198.2) = -59043.6 \, \text{J} \]

Now substituting back into the equation for \( \Delta G \):

\[ \Delta G = -92400 + 59043.6 = -33356.4 \, \text{J} \]

### Step 3: Calculate the Equilibrium Constant (\( K_p \))
Using the relationship between \( \Delta G \) and \( K_p \):

\[ \Delta G = -RT \ln K_p \]

Rearranging gives:

\[ \ln K_p = -\frac{\Delta G}{RT} \]

Substituting the values:

\[ \ln K_p = -\frac{-33356.4 \, \text{J}}{8.314 \, \text{J/(K·mol)} \times 298 \, \text{K}} \]

Calculating the denominator:

\[ RT = 8.314 \times 298 = 2477.572 \, \text{J/mol} \]

Now substituting into the equation for \( \ln K_p \):

\[ \ln K_p = \frac{33356.4}{2477.572} \approx 13.45 \]

### Step 4: Calculate \( K_p \)
To find \( K_p \), we take the exponential of both sides:

\[ K_p = e^{\ln K_p} = e^{13.45} \approx 694000 \]

### Final Answer
The equilibrium constant \( K_p \) for the reaction at 298 K is approximately:

\[ K_p \approx 694000 \]