The gas-phase equilibrium of the decomposition of NO, shown here, has equilibriu

Question

The gas-phase equilibrium of the decomposition of NO, shown here, has equilibrium constant, Kp, value of 0.00024 at 200 degree C. If a closed vessel was filled with nitrogen monoxide and the initial pressure of NO was 0.05 atm at 200 degree C, what would be the final partial pressure of NO, O_2, N_2? 2NO(g) O_2(g) + N_2(g) For the reaction: 2NO(g) + O_2 2NO_2(g) Determine the rate law: rate = k[NO]^m [O_2]^n Determine also the rate constant: k = (rate)/[NO]^m [O_2]^n Some formulas: K = [products]^p/[reactants]^t Delta G degree rxn = [coefficient * Delta G degree formation] products - [coefficient * Delta G degree formation] reactants

Explanation / Answer

order w.r.t NO

(0.3/0.6)^n = (0.048/0.192)

n = 2

order w.r.t O2

(0.3/1.2)^n =(0.192/0.768)

n = 1

rate = k [NO]^2[O2]^1

K = 0.048/(0.3^2*0.3) = 1.78

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