At T(K) , the equilibrium constant of A_2(g) + B_2(g) leftharpoons...

For the reaction $A_2(g) + B_2(g) \rightleftharpoons C(g)$, the equilibrium constant is $K_1 = 2.7 \times 10^{-5}$.

When a reaction is multiplied by a factor $n$, the new equilibrium constant becomes $K^n$.

The given reaction is multiplied by $\dfrac{1}{3}$ to obtain the reaction $\dfrac{1}{3}A_2(g) + \dfrac{1}{3}B_2(g) \rightleftharpoons \dfrac{1}{3}C(g)$.

The new equilibrium constant $K_2$ is given by:

$K_2 = (K_1)^{1/3}$

$K_2 = (2.7 \times 10^{-5})^{1/3}$

$K_2 = (27 \times 10^{-6})^{1/3}$

$K_2 = 3 \times 10^{-2}$

Answer: $3 \times 10^{-2}$