Equilibrium Constant Calculation for Scaled Reaction Equation

To determine the equilibrium constant for the modified reaction, we apply the properties of equilibrium constants under scaling of stoichiometric coefficients.

For the given reaction: $\text{A}_2(\text{g}) + \text{B}_2(\text{g}) \rightleftharpoons \text{C}(\text{g})$

The equilibrium constant expression is: $K = \frac{[\text{C}]}{[\text{A}_2][\text{B}_2]} = 2.7 \times 10^{-5}$

The target reaction is obtained by multiplying the original chemical equation by a stoichiometric factor of $n = \frac{1}{3}$: $\frac{1}{3}\text{A}_2(\text{g}) + \frac{1}{3}\text{B}_2(\text{g}) \rightleftharpoons \frac{1}{3}\text{C}(\text{g})$

The equilibrium constant for the scaled reaction, $K'$, is given by: $K' = \frac{[\text{C}]^{1/3}}{[\text{A}_2]^{1/3}[\text{B}_2]^{1/3}} = \left(\frac{[\text{C}]}{[\text{A}_2][\text{B}_2]}\right)^{1/3} = K^{1/3}$

Substituting the given value of $K$: $K' = (2.7 \times 10^{-5})^{1/3}$

Rewriting $2.7 \times 10^{-5}$ in standard exponential form for an easy cube root calculation: $2.7 \times 10^{-5} = 27 \times 10^{-6}$

Now, taking the cube root: $K' = (27 \times 10^{-6})^{1/3} = (27)^{1/3} \times (10^{-6})^{1/3} = 3 \times 10^{-2}$

Thus, the equilibrium constant for the target reaction is $3 \times 10^{-2}$.

Correct Option: D