In which of the following reaction `K_(p) gt K_(c)`

Options

A. `N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`
B. `H_(2)(g)+I_(2)(g)hArr 2HI(g)`
C. `PCl_(3)(g)+Cl_(2)(g)hArrPCl_(5)(g)`
D. `2SO_(3)(g)hArr 2SO_(2)(g)+O_(2)(g)`

Answer

To determine in which of the given reactions \( K_p \) is greater than \( K_c \), we can use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c R T^{\Delta N_G} \]

where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta N_G \) is the change in the number of moles of gas, calculated as:

\[ \Delta N_G = \text{(sum of stoichiometric coefficients of products)} - \text{(sum of stoichiometric coefficients of reactants)} \]

For \( K_p \) to be greater than \( K_c \), \( \Delta N_G \) must be positive (\( \Delta N_G > 0 \)).

Now, let's analyze each of the given reactions:

### Option A:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
- Products: 2 (from \( 2NH_3 \))
- Reactants: 1 (from \( N_2 \)) + 3 (from \( 3H_2 \)) = 4
- Calculation of \( \Delta N_G \):
\[ \Delta N_G = 2 - 4 = -2 \]
Since \( \Delta N_G < 0 \), \( K_p < K_c \).

### Option B:
\[ H_2 + I_2 \rightleftharpoons 2HI \]
- Products: 2 (from \( 2HI \))
- Reactants: 1 (from \( H_2 \)) + 1 (from \( I_2 \)) = 2
- Calculation of \( \Delta N_G \):
\[ \Delta N_G = 2 - 2 = 0 \]
Since \( \Delta N_G = 0 \), \( K_p = K_c \).

### Option C:
\[ PCl_3 + Cl_2 \rightleftharpoons PCl_5 \]
- Products: 1 (from \( PCl_5 \))
- Reactants: 1 (from \( PCl_3 \)) + 1 (from \( Cl_2 \)) = 2
- Calculation of \( \Delta N_G \):
\[ \Delta N_G = 1 - 2 = -1 \]
Since \( \Delta N_G < 0 \), \( K_p < K_c \).

### Option D:
\[ 2SO_3 \rightleftharpoons 2SO_2 + O_2 \]
- Products: 2 (from \( 2SO_2 \)) + 1 (from \( O_2 \)) = 3
- Reactants: 2 (from \( 2SO_3 \))
- Calculation of \( \Delta N_G \):
\[ \Delta N_G = 3 - 2 = 1 \]
Since \( \Delta N_G > 0 \), \( K_p > K_c \).

### Conclusion:
The reaction in which \( K_p \) is greater than \( K_c \) is **Option D**.

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