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Chemistry Question on Equilibrium

The equilibrium constant for the reaction
SO3 (g)⇌SO2 (g)+12O2 (g)\text{SO}_3 \, (g) \rightleftharpoons \text{SO}_2 \, (g) + \frac{1}{2} \text{O}_2 \, (g)
is KC=4.9×10−2K_C = 4.9 \times 10^{-2}. The value of KCK_C for the reaction given below is
2SO2 (g)+O2 (g)⇌2SO3 (g)2 \text{SO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2 \text{SO}_3 \, (g) is

A

4.9

B

41.6

C

49

D

416

Answer

416

Explanation

Solution

Given the equilibrium constant for the reaction:
SO3 (g)⇌SO2 (g)+12O2 (g)\text{SO}_3 \, (g) \rightleftharpoons \text{SO}_2 \, (g) + \frac{1}{2} \text{O}_2 \, (g)
is Kc=4.9×10−2K_c = 4.9 \times 10^{-2}.
For the reaction:
2SO2 (g)+O2 (g)⇌2SO3 (g)2\text{SO}_2 \, (g) + \text{O}_2 \, (g) \rightleftharpoons 2\text{SO}_3 \, (g)
we need the equilibrium constant Kc′K_c'.
Since the second reaction is the reverse and doubled version of the original reaction, we calculate Kc′K_c' as:
Kc′=(1Kc)2=(14.9×10−2)2K_c' = \left( \frac{1}{K_c} \right)^2 = \left( \frac{1}{4.9 \times 10^{-2}} \right)^2
Kc′=416.49K_c' = 416.49