At 700 K , the equilibrium constant , `K_p` for the reaction `2SO_3(g) hArr 2SO_2(g) +O_2 (g)` is `1.8 xx10^(-3)` atm. The value of `K_c` for the above reaction at the same temperature in moles per litre would be

Options

A. `1.1 xx 10^(-7)`
B. `3.1 xx 10^(-5)`
C. `6.2 xx 10^(-7)`
D. `9.3 xx 10^(-7)`

Answer

To find the value of \( K_c \) for the reaction

\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]

given that \( K_p = 1.8 \times 10^{-3} \) atm at 700 K, we can use the relationship between \( K_p \) and \( K_c \):

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

where:
- \( R \) is the universal gas constant, \( 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T \) is the temperature in Kelvin
- \( \Delta n \) is the change in the number of moles of gas, calculated as the moles of products minus the moles of reactants.

### Step 1: Calculate \( \Delta n \)

For the reaction:

- Moles of products: \( 2 \, \text{(SO}_2\text{)} + 1 \, \text{(O}_2\text{)} = 3 \)
- Moles of reactants: \( 2 \, \text{(SO}_3\text{)} = 2 \)

Thus,

\[ \Delta n = \text{Moles of products} - \text{Moles of reactants} = 3 - 2 = 1 \]

### Step 2: Substitute values into the equation

Now we can substitute the known values into the equation:

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

Substituting \( K_p = 1.8 \times 10^{-3} \), \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \), \( T = 700 \, \text{K} \), and \( \Delta n = 1 \):

\[ 1.8 \times 10^{-3} = K_c \times (0.0821 \times 700)^1 \]

### Step 3: Calculate \( RT \)

Calculating \( RT \):

\[ RT = 0.0821 \times 700 = 57.47 \, \text{L atm mol}^{-1} \]

### Step 4: Rearranging the equation to solve for \( K_c \)

Now rearranging the equation to solve for \( K_c \):

\[ K_c = \frac{K_p}{RT} = \frac{1.8 \times 10^{-3}}{57.47} \]

### Step 5: Perform the division

Calculating \( K_c \):

\[ K_c = \frac{1.8 \times 10^{-3}}{57.47} \approx 3.13 \times 10^{-5} \, \text{mol/L} \]

### Final Answer

Thus, the value of \( K_c \) for the reaction at 700 K is approximately:

\[ K_c \approx 3.1 \times 10^{-5} \, \text{mol/L} \]

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