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Chemistry Question on Equilibrium Constant

Given that the equilibrium constant for the reaction 2SO2(g)+O2(g)<=>2SO3(g) {2SO2(g) + O2(g) <=> 2SO3(g)} has a value of278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature? SO3(g)<=>SO2(g)+12O2(g) {SO3(g) <=> SO2(g) + \frac{1}{2} O2(g)}

A

1.8×10−31.8 \times 10^{-3}

B

3.6×10−33.6 \times 10^{-3}

C

6.0×10−26.0 \times 10^{-2}

D

1.3×10−51.3 \times 10^{-5}

Answer

6.0×10−26.0 \times 10^{-2}

Explanation

Solution

2SO2(g)+O2(g)<=>2SO3(g) {2SO2(g) + O2(g) <=> 2SO3(g)} K=278\quad K = 278 (given) SO3(g)<=>SO2(g)+12O2(g) {SO3(g) <=> SO2(g) + \frac{1}{2} O2(g)} K′=(1K)\quad K' = \left(\sqrt{\frac{1}{K}}\right) =1278=35.97×10−4=6×10−2= \sqrt{\frac{1}{278}} = \sqrt{35.97\times10^{-4}} = 6\times10^{-2}