Question

The rate constant for the 1st order decomposition of N2O5 in the reaction 2N2O5 (g)  4NO2 (g) + O2 (g) is k=3.38x 10-5 s-1 at 25oC.

a. What is the half-life of N2O5?
b. What will be the total pressure, initially 88.3 kPa for the pure N2O5 vapor, (1) 10s and

(2) 10 min after initiation of the reaction?
(Hint: partial pressure is proportional to concentration).

Answer 1

Based on the rate constant value the reaction is 1^st order. For 1^st order reaction

-dCA/dt= KCA

CA= PA/RT

-dPA/dt= KPA

When integrated –ln(PA/PAO)= Kt, PA= partial pressure at time t and PAO= initial pressure, K= rate constant

When PA= 0.5PAO, when the pressure is 50% of the initial pressure, the time is half life

-ln( 0.5)= K*half life, half life =0.693/(3.38*10^-5)= 20503 sec

At 10seconds, -ln(PA/88.3)= 10*3.38*10^-5= 0.000338

PA=88.27 Kpa

From PA= PAO- a/deltan*(Pi-Po)

The reaction is 2N2O5(g) ----à4NO2(g)+O2(g)

Change in no of moles =4+1-2= 3

PI= total pressure at any time, PO= total pressure , detlan= change in no of moles during the reaction = 3

88.27= 88.3- (2/3)*(PI-88.3)

88.27= 88.3-0.666PI+58.9

0.666PI= 88.3+58.9-88.27, PI= 88.4

After 10min= 10*60sec= 600sec

-ln(PA/88.3)= 600*3.38*10^-5 =0.02028

PA= 86.52 Kpa

Hence 86.52= 88.3- (2/3)*(PI-88.3)

0.666PI= 88.3+58.9-86.52 , PI= 91.11 Kpa