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
8987°K
9210°K
8764°K
8621°K
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
15 comments Page 1 of 2.

Ramsajan said:   2 years ago
According to transition theory answer - 8764k.
And according to Arrhenius theory, it's 9210k.

Athraa said:   3 years ago
Lnk2/lnk1 = E/R(1/T1 - 1/T2).
*k2 = 100k1.
So:
Ln100 = E/R(1÷400-1÷500).
E/R= 9210.

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.
(3)

Kuldeep verma said:   6 years ago
B is the Wrong answer
Right answer is (C) 8764°K
According to transition theory.

Mnnit said:   6 years ago
The Correct answer is C.

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.
(1)

MUHAMMAD SALEH said:   7 years ago
Here we can use ln k2/k1 = E/R (1/T1-1/T2).

Srinath said:   8 years ago
C is the correct option. I also agree.

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.

Ranjan Yadav said:   9 years ago
Ln((100k1/k1)) = E/R[(500 - 400)/(500 * 400)].
E/R = 9210 KELVIN.


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