Question

7. A reaction proceeds according to the following equation: NO2 (g) → 2NO (g) + O2 (g) At an early stage of the reaction, [NO] = 0.0021 M and 5.0 minutes later, [NO] = 0.0066 M. What is the average rate of the reaction during this time period, expressed in units of M·s-1?

Answer 1

NO2(g) --------> 2NO(g) + O2(g)

-∆[NO2]/∆t = (1/2)∆[NO]/∆t

∆[NO] = 0.0066M - 0.0021M = 0.0045M

∆t = 5min = 300s

∆[NO]/∆t = 0.0045M/300s

∆[NO]/∆t = 1.5 ×10^-5Ms^-1

-∆[NO2]/∆t = (1/2) × 1.5 ×10^-5Ms^-1 =7.5 ×10^-6Ms^-1

Therefore

Avarage rate of reaction during the time period

= 7.5×10^-6Ms^-1