Question

A reaction takes place in 3 steps with rate constant k1, k2, k3. The overall rate rate constant is k= k1*(k2)^1/2 /k3. If the activation energies are 40, 30 ,20 kj for step 1,2,3 respectively then find the overall activation energy of the reaction. Please tell me the solution of this answer as soon as possible.

Answer 1

Answer : The overall activation energy of the reaction is, 35 KJ

Solution : Given,

Activation energy for step 1 = 40 KJ

Activation energy for step 2 = 30 KJ

Activation energy for step 3 = 20 KJ

As we know that

$K=Ae^{\frac{-Ea}{RT}}$

where,

K = rate constant

A = Pre-exponential factor

Ea = activation energy

R = gas constant

T = temperature

Taking 'ln' on both the sides, we get

$lnK=\frac{-Ea}{RT}$

The overall rate constant is,

$K=\frac{k_1\times (k_2)^{1/2}}{k_3}$

Similarly, we are taking 'ln' on the sides in this expression, we get

$Ea=Ea_1+\frac{1}{2}Ea_2-Ea_3$

Now put all the given values in this expression, we get

$Ea=40KJ+\frac{1}{2}\times 30KJ-20KJ=35KJ$

Therefore, the overall activation energy of the reaction is, 35 KJ

Answer 2

Answer:35

Explanation:Given k=k1*(k2)1/2/k3

Taking log both the side

lnk=lnk1+lnk2(1/2)-lnk3

We know that k=Ae-Ea/RT;lnk=lnA-Ea/RT

let every step have same arrhenius constant(i.e.A) and Ea1,Ea2 and Ea3 are the corresponding activation energy for 1st,2nd and the third step respectively.

So, k=lnA-Ea/RT=lnA-Ea1/RT..+(1/2)*[lnA-Ea2/RT]-lnA+Ea3/RT

So if we have value of A then we can get the value of Ea using the equation.