Question

For a reaction of the type, , the rate of reaction (- rx) is given by (k1 + k2)cx (k1 + k2 + k3)cx k1 cv k2 cx (k1 - k2)cx

Answer 1

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