Consider the following equilibria: 2 SO3 (g) <=> 2 SO2 (g) + O2 (g) ----- Kc = 2

Question

Consider the following equilibria:
2 SO3 (g) <=> 2 SO2 (g) + O2 (g) ----- Kc = 2.3 x 10-7
2 NO3 (g) <=> 2 NO2 (g) + O2 (g) ------ Kc = 1.4 x 10-3
Calculate the equilibrium constant for the reaction
SO2(g) + NO3(g) <=> SO3(g) + NO2(g).

a. 6.1

Explanation / Answer

2 SO3 (g) 2 SO2 (g) + O2 (g) ----- Kc = 2.3 x 10-7 now, 2 SO2 (g) + O2 (g) ------>2 SO3 (g) , Kc1 = 1/( 2.3 x 10-7) also, 2 NO3 (g) 2 NO2 (g) + O2 (g) ------ Kc2 = 1.4 x 10-3 Adding the above equations, 2SO2(g) + 2NO3(g) 2SO3(g) + 2NO2(g). ----- Kc = Kc1 x Kc2 SO2(g) + NO3(g) SO3(g) + NO2(g). we have , Kc = (kc)^2 = 1.6 x 10^-4 Option - B

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