Chemical Engineering - Chemical Reaction Engineering - Discussion
Discussion Forum : Chemical Reaction Engineering - Section 1 (Q.No. 28)
28.
The rate constant of a chemical reaction increases by 100 times when the temperature is increased from 400 °K to 500 °K. Assuming transition state theory is valid, the value of E/R is
Discussion:
15 comments Page 2 of 2.
Ajay said:
1 decade ago
For transition state theory answer is C. While B is when we use Arrhenius equation.
Surya said:
1 decade ago
Here the answer is wrong the rate follows this equation according to transition state theory:
K = Ko*e^(-E/RT)*T.
Here k increased by 100 times and temperatures are 400 and 500 then we get the answer 8764.
K = Ko*e^(-E/RT)*T.
Here k increased by 100 times and temperatures are 400 and 500 then we get the answer 8764.
V v said:
1 decade ago
Correct answer: C
As they ask to solve by transition state theory.
Equation is: K = Ko*e^(-E/RT)*T.
As they ask to solve by transition state theory.
Equation is: K = Ko*e^(-E/RT)*T.
Shan Rana said:
1 decade ago
Correct option is [B] 9210K.
@Suman has given the right solution.
1/100 = e^(E/R400 - E/R500).
ln 0.01 = E/R (-1/2000).
-4.61 = E/R (-0.0005).
E/R = 4.61/0.0005.
E/R = 9210 K.
@Suman has given the right solution.
1/100 = e^(E/R400 - E/R500).
ln 0.01 = E/R (-1/2000).
-4.61 = E/R (-0.0005).
E/R = 4.61/0.0005.
E/R = 9210 K.
(1)
Suman said:
1 decade ago
k = k0*e^-(E/RT).
K1 = k0*e^-(E/400R).
K1 = k0*e^-(E/500R).
GIVEN THAT,
K2 = 100K1.................(1).
K1/K2 = e^-(E/400R-E/500R).........(2).
Substitute eqn 1 in 2 and take logarithmic on both sides you will get the answer.
K1 = k0*e^-(E/400R).
K1 = k0*e^-(E/500R).
GIVEN THAT,
K2 = 100K1.................(1).
K1/K2 = e^-(E/400R-E/500R).........(2).
Substitute eqn 1 in 2 and take logarithmic on both sides you will get the answer.
(2)
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