The equilibrium constant `K_C` for `3C_2 H_2 (g) hArr C_6 H_6 (g)`
is `4L^2 "mol"^(-2)` .If the equilibrium concentration of benzene is 0.5 mol `L^(-1)`, that of accetylene in mol `L^(-1)` must be

Options

A. 0.025
B. 0.25
C. 0.05
D. 0.5

Explanation

To solve the problem, we need to use the equilibrium constant expression for the given reaction:

**Step 1: Write the balanced chemical equation and the expression for \( K_C \)**

The reaction is:
\[ 3C_2H_2 (g) \rightleftharpoons C_6H_6 (g) \]

The equilibrium constant \( K_C \) is defined as:
\[ K_C = \frac{[C_6H_6]}{[C_2H_2]^3} \]

**Step 2: Substitute the known values into the \( K_C \) expression**

We know:
- \( K_C = 4 \, L^2 \, mol^{-2} \)
- The equilibrium concentration of benzene \( [C_6H_6] = 0.5 \, mol \, L^{-1} \)

Substituting these values into the expression gives:
\[ 4 = \frac{0.5}{[C_2H_2]^3} \]

**Step 3: Rearrange the equation to solve for \( [C_2H_2]^3 \)**

Rearranging the equation to isolate \( [C_2H_2]^3 \):
\[ [C_2H_2]^3 = \frac{0.5}{4} \]

**Step 4: Calculate \( [C_2H_2]^3 \)**

Calculating the right side:
\[ [C_2H_2]^3 = \frac{0.5}{4} = 0.125 \]

**Step 5: Take the cube root to find \( [C_2H_2] \)**

Now, take the cube root of both sides to find \( [C_2H_2] \):
\[ [C_2H_2] = (0.125)^{1/3} \]

Calculating the cube root:
\[ [C_2H_2] = 0.5 \, mol \, L^{-1} \]

Thus, the equilibrium concentration of acetylene \( C_2H_2 \) is \( 0.5 \, mol \, L^{-1} \).

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