V.CHEMICAL KINETICS III Integrated Rate Laws 1. Example -first order integrated
ID: 573901 • Letter: V
Question
V.CHEMICAL KINETICS III Integrated Rate Laws 1. Example -first order integrated rate law problem SOzCl2(g) SO2 (g) + Cl2(g) is first order in sod2 with k = 2.2 x 10-5 and an initial concentration of 2.4x 102 M (a) What is the concentration of S02CIz after 1.0 hours? (b) How long will it take the concentration of poal to fall to 24 ×103M? (c) How long will it take for 75% of the SOC12 to decompose? (d) What is the half-life of SO2C12? (e) What will be the concentration of [SO2C12] after 4 half-lives?Explanation / Answer
The reaction is SO2Cl2(g)------->SO2(g)+ Cl2(g) is 1st order.
for 1st order, -d(SO2C2/dt), rate of reaction = K*[SO2Cl2], where K is rate constant K= 2.2*10-5t/sec
when the equation is integrated noting that at t=0, [SO2Cl2]=[SO2Cl2]0= 2.4*10-2M and at t=t, [SO2Cl2]= [SO2Cl2]
ln[SO2Cl2]= ln[SO2Cl2]0- Kt
when t= 1hr= 60min= 60*60 sec
ln[SO2Cl2]= ln(2.4*10-2 )-2.2*10-5*3600
[SO2Cl2]=0.0221M=2.21*10-2M
2. given at t=t [SO2Cl2]= 2.4*10-3
ln(2.4*10-3)= ln(2.4*10-2)- 2.2*10-5*t
t= 104663 sec= 104663/3600 hr =29 hrs
the equation ln[SO2Cl2]= ln[SO2Cl2]0- Kt can also be written as
ln [SO2Cl2]/[SO2Cl2]0 = -Kt
X= coonversion= 1- [SO2Cl2]/[SO2Cl2]0
hence -ln(1-X)= Kt
given X= 0.75
hence -ln(1-0.75)=2.2*10-5*t
t=63013.4 seconds = 17.5 hrs
when X=0.5, the time required for 50% conversion is designated as half life
hence -ln(1-0.5)= 2.2*10-5*half life
half life =0.693/(2.2*10-5)= 31500 seconds
after 1st half flife, the concentration is 2.4*10-2/2= 1.2*10-2 M
after second half life, the concentration =1.2*10-2/2= 6*10-3M
after third half life, the concentration =6*10-3/2= 3*10-3M
after 4th half life, the concentration is = 3*10-3/2= 1.5*10-3M
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