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
A. 8987°K
B. 9210°K
C. 8764°K
D. 8621°K
Answer: Option B
This Question Belongs to Chemical Engineering >> Chemical Reaction Engineering
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A. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\text{k}} + {{\text{k}}_{\text{g}}}$$
B. $${{\text{k}}_{\text{e}}}{\text{ff}} = \frac{{{\text{k}} + {{\text{k}}_{\text{g}}}}}{2}$$
C. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\left( {{\text{k}}{{\text{k}}_{\text{g}}}} \right)^{\frac{1}{2}}}$$
D. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$
The half life period of a first order reaction is given by (where, K = rate constant. )
A. 1.5 K
B. 2.5 K
C. $$\frac{{0.693}}{{\text{K}}}$$
D. 6.93 K
Catalyst is a substance, which __________ chemical reaction.
A. Increases the speed of a
B. Decreases the speed of a
C. Can either increase or decrease the speed of a
D. Alters the value of equilibrium constant in a reversible
A. $$ \propto {\text{CA}}$$
B. $$ \propto \frac{1}{{{\text{CA}}}}$$
C. Independent of temperature
D. None of these
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.
The answer given here i.e. 9210K is correct for Arrhenious law but by solving using Transition State theory we will end up with 8764K.
The equation for transition state theory is K = T*e^(E/RT)
so solving the same equation for two different point we end up with the following given equation:
ln(K2/K1) = ln(T2/T1) + E/R(1/T1 - 1/T2)
Putting all the values given in the question we end up with the answer option C i.e. 8764K
Thank You