Question

The first-order rate constant for the decomposition of N2O5, 2N2O5(g)→4NO2(g)+O2(g) at 70∘C is 6.82×10^−3 s−1. Suppose we start with 2.60×10^−2 mol of N2O5(g) in a volume of 2.4 L . How many moles of N2O5 will remain after 4.0 min ? How many minutes will it take for the quantity of N2O5 to drop to 1.9×10^−2 mol ? What is the half-life of N2O5 at 70∘C?

Answer 1

for first order reaction

-ln (1-XA)= Kt

where XA = conversion and K= rate constant = 6.82/1000 sec-1

XA= 1-CA/CAO, where CAO= initial concentration of N2O5= 2.6*10^-2/ 2.4=1.083*10^-2 M, CA = concentration at any time t

-ln(CA/CAO)= 6.82*10^-3* 4*60=1.6368

CA/CAO= exp(-1.6368)=0.1946

CA= 0.1946*1.08*10-2=0.002102 M

moles of N2O5 remaning= 0.002102*2.4=0.005044

b) -ln(1.9*10-2/ 2.6*10-2)= 6.82*10-3*t

t =45.9 minutes

c) half lifr is time requied to drop to fifty percent of its initial value and

half lfie =0.693/K= 0.693/6.82*10-3=101.6 minutes