At $$T$$(K), the equilibrium constant of $$A_2(g) + B_2(g) \rightleftharpoons C(g)$$ is $$2.7 \times 10^{-5}$$. What is the equilibrium constant for $$\frac{1}{3}A_2(g) + \frac{1}{3}B_2(g) \rightleftharpoons \frac{1}{3}C(g)$$ at the same temperature?

The equilibrium constant $$K_1$$ for the reaction
$$A_2(g)+B_2(g)\;\rightleftharpoons\;C(g)$$ is given as $$K_1 = 2.7 \times 10^{-5}$$.

If every stoichiometric coefficient of a balanced chemical equation is multiplied by a factor $$n$$, the new equilibrium constant becomes $$K^{\,n}$$.
Conversely, if every coefficient is divided by $$n$$, the new equilibrium constant is the $$n^{\text{th}}$$ root of the old one.

The required reaction is obtained by dividing all coefficients by $$3$$:
$$\tfrac13 A_2(g)+\tfrac13 B_2(g)\;\rightleftharpoons\;\tfrac13 C(g)$$.
Hence its equilibrium constant $$K_2$$ is
$$K_2 \;=\;(K_1)^{1/3} \;=\;(2.7 \times 10^{-5})^{1/3}.$$

Write the number and the power of ten separately:
$$K_2 = (2.7)^{1/3}\,\times\,10^{-5/3}.$$ Compute each factor:

• Cube root of $$2.7$$:
$$ (2.7)^{1/3} \approx 1.4 $$ (because $$1.4^3 = 2.744$$).
• Cube root of $$10^{-5}$$:
$$ 10^{-5/3} = 10^{-1.6667}. $$

Convert $$10^{-1.6667}$$ to decimal using $$10^{-1} = 0.1$$ and $$10^{-0.6667}\approx0.215$$:
$$10^{-1.6667} \approx 0.1 \times 0.215 \approx 0.0215.$$

Multiply the two factors:
$$K_2 \approx 1.4 \times 0.0215 \approx 0.030 \;=\;3.0 \times 10^{-2}.$$

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

Option D which is: $$3 \times 10^{-2}$$