The decomposition of N2O5 produces NO2 . if the initial concentration of N2O5 is 2.33mol/l and after 184 min it reduced to 2.08mol/l . Calculate the average rate of reaction. What is the rate of production of NO2 if the reaction is 2N2O5 ------------- 4NO2+O2 ?
Rate of decrease in conc. of N2O5-
Δ (N2O5)/ Δt = - (C2-C1)/ 184 where C2= 2.08 mol/L C1= 2.33 mol/L
= 0.00135 mol L-1 min-1
Average rate of reaction-
Δ x / Δ t = -1/2 Δ [N2O5] / Δt = 1/4 Δ [NO2] / Δ t = Δ O2 /Δt
average rate - Δx/ Δt = -1/2 Δ N2O5 / Δt
= 1/2 x 0.00135 = 6.75 x 10-4 mol/L/min
Rate of production on NO2-
-1/2 Δ [N2O5] / Δt = 1/4 Δ [NO2] / Δ t 1/2 x 0.00135 = 1/4 Δ [NO2] / Δ t Δ [NO2] / Δ t = 4 x 0.00135 / 2 = 2.7 x 10-3 mol/L/min
Δ (N2O5)/ Δt = - (C2-C1)/ 184 where C2= 2.08 mol/L C1= 2.33 mol/L
= 0.00135 mol L-1 min-1
Average rate of reaction-
Δ x / Δ t = -1/2 Δ [N2O5] / Δt = 1/4 Δ [NO2] / Δ t = Δ O2 /Δt
average rate - Δx/ Δt = -1/2 Δ N2O5 / Δt
= 1/2 x 0.00135 = 6.75 x 10-4 mol/L/min
Rate of production on NO2-
-1/2 Δ [N2O5] / Δt = 1/4 Δ [NO2] / Δ t 1/2 x 0.00135 = 1/4 Δ [NO2] / Δ t Δ [NO2] / Δ t = 4 x 0.00135 / 2 = 2.7 x 10-3 mol/L/min