When the molar concentrations of `SO_(2), O_(2) and SO_(3)` at equilbrium at certain temperature are `0.5 , 0.25 ` & 0.25 M respectively, `k_(c) ` for `2SO_(3) hArr 2SO_(2) + O_(2)` is

Options

A. 0.5 `"lit-mol"^(-1)`
B. 1 `"lit"^(2)-mol^(-2)`
C. 1 `"mol-lit"^(-1)`
D. 0.25 `mol^(2)- lit^(-2)`

Answer

To find the equilibrium constant \( K_c \) for the reaction:

\[ 2SO_3 \rightleftharpoons 2SO_2 + O_2 \]

we will use the equilibrium concentrations provided:

- \([SO_2] = 0.5 \, M\)
- \([O_2] = 0.25 \, M\)
- \([SO_3] = 0.25 \, M\)

### Step 1: Write the expression for \( K_c \)

The equilibrium constant \( K_c \) for the reaction is given by the formula:

\[ K_c = \frac{[SO_2]^2 [O_2]}{[SO_3]^2} \]

### Step 2: Substitute the equilibrium concentrations into the expression

Now, we will substitute the equilibrium concentrations into the expression we wrote in Step 1:

\[ K_c = \frac{(0.5)^2 (0.25)}{(0.25)^2} \]

### Step 3: Calculate the values

Now we will calculate the values step by step:

1. Calculate \((0.5)^2\):
\[ (0.5)^2 = 0.25 \]

2. Calculate \((0.25)^2\):
\[ (0.25)^2 = 0.0625 \]

3. Substitute these values into the \( K_c \) expression:
\[ K_c = \frac{0.25 \times 0.25}{0.0625} \]

4. Calculate \( 0.25 \times 0.25 \):
\[ 0.25 \times 0.25 = 0.0625 \]

5. Now substitute this back into the \( K_c \) equation:
\[ K_c = \frac{0.0625}{0.0625} = 1 \]

### Final Answer

Thus, the equilibrium constant \( K_c \) for the reaction is:

\[ K_c = 1 \]