Question

14.44 The first-order rate constant for the decomposition of N205, 2N205(g)-→ 4 NO2(g) + O2(g), at 70°C is 6.82 × 10-3 s-1. Suppose we start with 0.250 mol of N205(g) 1S in a volume of 2.0 L. (a) How many moles of N2O5 will re- main after 10.0 min? (b) How many minutes will it take for the quantity of N205 to drop to 0.100 mol? (c) What is the half-life, in minutes, of N2Os at 70 °C?

Answer 1

Answer:

Given reaction is

2N2O5(g) --------> 4NO2(g)+O2(g)

It is first order. Rate constant k=6.82x10^-3 s^-1

Given initial moles of N2O5=0.25 mol and volume=2L

Then [N2O5]o=moles/volume=0.25 mol/2L=0.125 mol/L

(a) [N2O5]=? at time=10 min=10x60 sec=600 s.

First order rate law

ln[N2O5]=ln[N2O5]o - kt

ln[N2O5]=ln(0.125) - (6.82x10^-3s^-1 x 600 s)=-2.079-4.092

ln[N2O5]=-6.171

[N2O5]=e^-6.171=0.00209 M=0.00209 mol/L

Moles of [N2O5]=0.00209 mol per 1L and 0.00418 mol per 2L solution.

(b) Moles of N2O5=0.1mol then [N2O5]=0.1 mol/2L=0.05mol/L at time t

ln[N2O5]=ln[N2O5]o - kt

(6.82x10^-3 s^-1)t=ln(0.125)-ln(0.05)

t=(0.916)/(6.82x10^-3 s^-1)=134.35 s

time=134.35 /60=2.24 min. (Since 1 min =60 s)

(c) Half life of first order t1/2=0.693/k

t1/2=0.693/(6.82x10^-3 s)=101.61 s=101.61/60

t1/2=1.69 min.