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 1 of 2.
Rafi said:
5 years ago
According to transition theory;
K = ko.T.e^-E/RT
And from this we get;
Ln(k2/k1) = ln(T2/T1) + E/R {1/T1 -1/T2};
As K2 = 100K1 T1=400k T2=500k;
Ln(100k1/k1) = ln(500/400) + E/R {1/400 - 1/500};
Solving this we get E/R= 8764K.
K = ko.T.e^-E/RT
And from this we get;
Ln(k2/k1) = ln(T2/T1) + E/R {1/T1 -1/T2};
As K2 = 100K1 T1=400k T2=500k;
Ln(100k1/k1) = ln(500/400) + E/R {1/400 - 1/500};
Solving this we get E/R= 8764K.
(3)
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)
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)
Vishal Verma said:
7 years ago
Option (C) is correct when it is given to transition state theory.
Option (B) is correct if we have given to assume Arrhenius theory.
Option (B) is correct if we have given to assume Arrhenius theory.
(1)
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.
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.
Ajay said:
1 decade ago
For transition state theory answer is C. While B is when we use Arrhenius equation.
Ranjan Yadav said:
9 years ago
Ln((100k1/k1)) = E/R[(500 - 400)/(500 * 400)].
E/R = 9210 KELVIN.
E/R = 9210 KELVIN.
Parth said:
8 years ago
Correct Answer is 8764 k. Here we are calculating according to transition state theory, So formula is k = k0e-(E/RT)*T.
Srinath said:
8 years ago
C is the correct option. I also agree.
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