Question

two different reactions at 300k having same pre exponential factor the activation energy of the 1st reaction is more than that of 2nd reaction by 5•75kJ then find k2÷k1​

Answer 1

According to the Arrhenius equation, K=Ae

−Ea/RT

Where A is proportionality constant, Ea is activation energy, R is universal gas constant and T is the temperature in kelvin.

Taking log both sides,

We get, logk = logA - Ea/2.303RT

So, for 1st reaction , logk₁ = logA - Ea₁/2.303RT

For 2nd reaction, logk₂ = logA - Ea₂/2.303RT

Subtracting equation (2) from equation (1),

logk₁ - logk₂ = (Ea₁ - Ea₂)/2.303RT

Here, (Ea₁ - Ea₂) = 41.9kJ

R = 8.3 J/K/mol

And T = 600K

log(

k₂

k₁

)=

2.303×8.3×600

(41.9×1000)

= 3.66

k₂

k₁

= antilog(3.66)

=0.002