Assume that you are studying the first-order conversions of a reactant X to prod
ID: 1006227 • Letter: A
Question
Assume that you are studying the first-order conversions of a reactant X to products in a reaction vessel with a constant volume of 1.000 L. At 1:00 p.m., you start the reaction at 25 °C with 1.000 mol of X. At 2:00 p.m., you find that 0.600 mol of X remains, and you immediately increase the temperature of the reaction mixture to 35 °C. At 3:00 p.m., you discover that 0.200 mol of X is still present. You want to finish the reaction by 4:00 p.m. but need to continue it until only 0.010 mol of X remains, so you decide to increase the temperature once again. What is the minimum tempera- ture required to convert all but 0.010 mol of X to products by 4:00 p.m.?
Please explain and show calculations
Explanation / Answer
Integrated rate expression for First order reaction is,
ln([A]/[A0]) = -kt .................(1)
Where,
[A0] : Initial concentration of reactant.
[A] : Final concentration of reactant after time t
k = first order rate constant in time-1 unit.
Codition -1) At T = 25 oC = 25 + 273.15 = 298.15 K
Let rate constant be k1 and we have,
[A0] = 1.000 M , [A] = 0.600 M t = 1 hr (1:00 PM to 2:00 PM)
Using rate expression eq. (1) we get,
ln(0.600/1.000) = -K1 x 1
-0.511 = -K1
K1 = 0.511 hr-1. ................(At T1 = 298.15 K)
Condition-2) At T = 35 oC = 35 + 273.15 = 308.15 K
[A0] = 0.600 M , [A] = 0.200 M t = 1 hr (2:00 PM to 3:00 PM)
Using rate expression eq. (1) we get,
ln(0.200/0.600) = -K2 x 1
-1.099 = -K2
K2 = 1.099 hr-1. .............(At T2 = 308.15 K)
Temperature and rate constants are related through Activation energy Ea which is given by Arrhenius equation.
k = A e(-Ea/RT). ..........(2a)
A = constant
For K1 and K2 at T1 and T2 we can write.
K2/K1 = e (Ea/R)[(T2-T1) / T1T2]
on taking natural log.
ln(K2/K1) = (Ea/R)[(T2 - T1)/T1T2] ..............(2b)
Let us put resppective values and find value of Ea.
ln(1.099/0.511) = Ea/8.314 x [(308.15 - 298.15/308.15 x 298.15)
0.765 = Ea/8.314 x (1.09 x 10-4)
Ea = (0.765 x 8.314) / (1.09 x 10-4)
Ea = 58434 J
Now
Condition-3)
At T3 = ?
[A0] = 0.200 M, [A] = 0.010 M, t = 1hr ( 3:00 PM to 4:00 PM)
Let us find required rate constant K3 at this new T3.
Using eq. (1)
ln(0.010/0.200) = -K3 x 1
-2.996 = -K3
K3 = 2.996
With this we have,
Ea = 58434 J, K2 = 1.099, K3 = 2.996, T2 = 308.15, T3 = ?
Using eq. (2a) we can get leq. like (2b) as,
ln(K3/K2) = (Ea/R) [(1/T2) - (1/T3)]
ln(2.996/1.099) = (58434/8.314)[(1/308.15) - (1/T3)]
calculating it we get,
1.003 = 7028.4 [(3.245 x 10-3) - (1/T3)]
1.003/7028.4 = [(3.245 x 10-3) - (1/T3)]
1.427 x 10-4 = [(3.245 x 10-3) - (1/T3)]
1/T3 = 3.245 x 10-3 -1.427 x 10-4.
1/T3 = 3.102 x 10-3.
Reciprocal gives
T3 = 322.34 K
T3 = 322.34 - 273.15 oC
T3 = 49.19 oC
The temperature to be maintained is 322.34 K or 49.19 oC.
======================xxxxxxxxxxxxxxxxxxxx=======================