2·The rearrangement of Cyclopropane to Propene is a first order reaction with a

Question

2·The rearrangement of Cyclopropane to Propene is a first order reaction with a rate constant of 6.7 x 10*s1. If the initial concentration of Cyclopropane is 0.050M a. b. c. d. e. What is the molarity of Cyclohexane after 30 minutes? How many minutes does it take for the Cyclohexane concentration to drop to 0.010M? How many minutes does it take for 65% of Cyclohexane to react? What is the half-life (in minutes)? How many minutes will it take for the concentration of Cyclohexane to drop to 12.5% of its initial value? What is the concentration?

Explanation / Answer

for 1st order reaction , -dCA/dt= KCA, where CA= concentration of propane in the present case and K is rate constant and -dCA/dt =rate of reaggangement,

When the equation is intergrated noting that at t=0, CA= CAO and t= t, CA= CA

the equation becomes lnCA= lnCAO-Kt (1)

1. given K= 6.7*10-4/s, CAO= 0.05M ( initial concentration) t= 30 min= 30*60 seconds = 1800 seconds

hence from eq.1 , lnCA= ln(0.05) -6.7*10-4* 1800 , CA= 0.015M

2. gvien CA=0.01M and CAO=0.05M, from eq.1, ln(0.01)= ln(0.05)- 6.7*10-4*t

t= 2402 seconds= 2402/60 min=40 min

3. Conversion XA= 1-CA/CAO

hence from Eq,1, ln(CA/CAO)= -Kt or- ln (1-XA)= -Kt

t= -ln(1-XA)/K, given XA=0.65, t= -ln(1-0.65)/ (6.7*10-4)= 1567 seconds

4. half life is the time required for the concentration to drop to 50% of its initial value, CA= CAO/2 at half life

hence from Eq.1, ln(CAO/2) =lnCAO-K* half life

half life =0.693/K= 0.693/6.7*10-4 =1034 seconds

5. given CA= 0.125CAO

from Eq.1, ln (0.125CAO)= lnCAO-6.7*10-4t

t= 3104 seconds , CA= 0.125*0.05=0.00625M

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