Question

Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of Cl2 loss is 4.60 × 10-2 M/s, what is the rate of formation of NOCl? 2 NO(g) + Cl2(g) → 2 NOCl(g)

Answer 1

Answer: 0.0920 M/s

Explanation: For every 1 mol of Cl2 used up, 2 mol of NOCl is produced. Thus, the rate of formation of NOCl is double the rate of Cl2 loss, which is 2*(4.60*10^-2) = 9.20*10^-2 M/s, or 0.0920 M/s.