Question

A gas mixture was prepared at 3940 K with total pressure 2.67 atm and a mole fraction of 5.93×10-4 of N and 1.88×10-4 of O2. The elementary reaction N(g) + O2(g) NO(g) + O(g) has a second-order rate constant of 1.55×1010 L mol-1 s-1 at this temperature.

Calculate the initial rate of the reaction under these conditions. ________ mol L-1 s-1

Answer 1

initial rate of reaction = 0.118 mol L^-1 s^-1

Explanation

Total pressure = 2.67 atm

mole fraction N = 5.93 x 10^-4

partial pressure of N = (Total pressure) * (mole fraction N)

partial pressure of N = (2.67 atm) * (5.93 x 10^-4)

partial pressure of N = 1.58331 x 10^-3 atm

concentration of N = (partial pressure of N) / [(R * T)]

concentration of N = (1.58331 x 10^-3 atm) / [(0.0821 atm-L/mol-K) * (3940 K)]

concentration of N = 4.89727 x 10^-6 M

Similarly, concentration of O₂ = 1.55259 x 10^-6 M

Rate = k[N][O₂]

Rate = (1.55 x 10¹⁰ Lmol^-1s^-1) * (4.89727 x 10^-6 mol.L^-1) * (1.55259 x 10^-6 mol.L^-1)

Rate = 0.118 mol L^-1 s^-1