The rate constant for the first order decompoistion of a certain reaction is described by the equation logk(s1)=14.341.25104kt (a) what is the energy of activation for the reaction? (b) at what temperature will its half-life periof be 256min ?
Answers
Explanation:
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Class 12
CHEMISTRY
CHEMICAL KINETICS
The rate constant for the first order decompoistion of a certain reaction is described by the equation `log k (s^(-1)) = 14.34 - (1.25 xx 10^(4)K)/(T)` (a) What is the energy of activation for the reaction? (b) At what temperature will its half-life periof be `256 min`?
Updated On: 1-6-2020
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2.2 k+
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Answer
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Text Solution
Answer
T = 669 K
Solution
(a)
k = Ae^(-E_(a)//RT)
<br> or
log k = log A - (E_(a))/(2.303 RT)
<br>
= 14.34 - (1.25 xx 10^(4))/(T)K
<br> On careful comparison, we get <br>
(E_(a))/(2.303 R) = 1.25 xx 10^(4)
<br> or
(E_(a))/(2.303 xx 8.314) = 1.25 xx 10^(4)
<br> or
E_(a) = 2.3933 xx 10^(5)
<br> (b)
k = (0.693)/("Half time") = (0.693)/(256 xx 60) = 4.51 xx 10^(-5)s^(-1)
<br>
k = Ae^(-E_(a)//RT)
<br> or
log k = log A - (E_(a))/(2.303 RT) = 14.34 - (1.25 xx 10^(4))/(T)
<br> or
log (4.51 xx 10^(-5)) = 14.34 - (1.25 xx 10^(4))/(T)K
<br> or
T = 669 K