At a certain temperature, $K_c = 4$ for the equilibrium $A(g) \rightleftharpoons B(g)$. If the value of $K_c$ for the reaction $2A(g) \rightleftharpoons 2B(g)$ at the SAME temperature is required, it equals

1. NCERT Table 6.4: if a balanced equation is multiplied by $n$, the new equilibrium constant is $K_c$ raised to the power $n$. 2. The original reaction $A \rightleftharpoons B$ has $K_c = [B]/[A] = 4$. 3. The new reaction is $2A \rightleftharpoons 2B$, i.e. the original multiplied by $n = 2$. 4. So the new equilibrium constant is $K'_c = (K_c)^2 = 4^2 = 16$. 5. Verify directly: $K'_c = [B]^2/[A]^2 = ([B]/[A])^2 = (K_c)^2$. Consistent. 6. Option A would apply if the coefficient were unchanged. Option B is double instead of squared. Option C is $1/K_c^2$ (the squared reversal). _Source: NCERT Class 11 Chemistry, Ch 6, §6.4 + Table 6.4 (multiplied-by-$n$ row), p. 9._