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.
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