At 700 K, the equilibrium constant `K_p` for the reaction `2SO_(3(g)) hArr 2SO_(2(g))+O_(2(g))` is `1.80xx10^(-3)` and `kP_(alpha)` is 14, `(R = 8.314 "Jk"^(-1) "mol"^(-1))` . The numerical value in moles per litre of `K_c` for this reaction at the same temperature will be

Explanation

To find the numerical value of \( K_c \) for the reaction \( 2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)} \) at 700 K, we can use the relationship between \( K_p \) and \( K_c \):

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

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

First, we need to determine \( \Delta n \), which is the change in the number of moles of gas during the reaction.

For the reaction:
- Reactants: 2 moles of \( SO_3 \)
- Products: 2 moles of \( SO_2 \) + 1 mole of \( O_2 \)

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

### Step 2: Substitute values into the equation

Now we can substitute \( K_p \), \( R \), \( T \), and \( \Delta n \) into the equation:

Given:
- \( K_p = 1.80 \times 10^{-3} \)
- \( R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \)
- \( T = 700 \, \text{K} \)
- \( \Delta n = 1 \)

The equation becomes:

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

Substituting the values:

\[ 1.80 \times 10^{-3} = K_c \times (8.314 \times 700)^1 \]

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

Now calculate \( RT \):

\[ RT = 8.314 \times 700 = 5819.8 \, \text{J mol}^{-1} \]

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

Now we can rearrange the equation to solve for \( K_c \):

\[ K_c = \frac{K_p}{RT} \]

Substituting the values:

\[ K_c = \frac{1.80 \times 10^{-3}}{5819.8} \]

### Step 5: Calculate \( K_c \)

Now perform the calculation:

\[ K_c = \frac{1.80 \times 10^{-3}}{5819.8} \approx 3.09 \times 10^{-7} \, \text{mol/L} \]

### Final Answer

The numerical value of \( K_c \) for the reaction at 700 K is approximately:

\[ K_c \approx 3.09 \times 10^{-7} \, \text{mol/L} \]

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