Question

question 25 (4 points)
the equilibrium constant for reaction 1 is k. the equilibrium constant for reaction 2 is?

1. so₂ (g) + (1/2) o₂ (g) so₃ (g)
2. 2so₃ (g) 2so₂ (g) + o₂

a) 1/k²
b) 2k
c) 1/2k

Explanation:

Step1: Write K for Reaction 1

For reaction 1: $\ce{SO_{2}(g) + \frac{1}{2}O_{2}(g)
ightleftharpoons SO_{3}(g)}$, the equilibrium constant $K = \frac{[\ce{SO_{3}}]}{[\ce{SO_{2}}][\ce{O_{2}}]^{\frac{1}{2}}}$.

Step2: Write K' for Reaction 2

For reaction 2: $\ce{2SO_{3}(g)
ightleftharpoons 2SO_{2}(g) + O_{2}(g)}$, the equilibrium constant $K' = \frac{[\ce{SO_{2}}]^{2}[\ce{O_{2}}]}{[\ce{SO_{3}}]^{2}}$.

Step3: Relate K and K'

Notice that reaction 2 is -2 times reaction 1 (reverse and double). From reaction 1, square both sides: $K^{2} = \frac{[\ce{SO_{3}}]^{2}}{[\ce{SO_{2}}]^{2}[\ce{O_{2}}]}$. Then take reciprocal: $\frac{1}{K^{2}} = \frac{[\ce{SO_{2}}]^{2}[\ce{O_{2}}]}{[\ce{SO_{3}}]^{2}}$, which is equal to $K'$. So $K' = \frac{1}{K^{2}}$.

Answer:

a) $1/K^{2}$