Question

The rate constant of a reaction is 1.2*10-5 the order of reaction is

Answer 1

Answer:

We are given that:

k1 = 1.2 x 10-3 sec-1

T1 = 30 + 273 = 303 K

k2 = 2.1 x 10-3 sec-1

T2 = 40 + 273 = 313 K

Substituting these values the equation:

Log (k2 / k1)

= Ea / 2.303 R {(T2 – T1) / T1 T2}

We will get:

Log (2.1 x 10-3 sec-1/ 1.2 x 10-3 sec-1)

= Ea / 2.303 x 8.314 {(313 – 303) / 303 x 313}

Log (2.1 / 1.2) = Ea / 2.303 x 8.314 x (10 / 303 x 313)

Ea = 44126.3 J mol-1

Ea = 44.13 kJ mol-1

Hence, the energy of activation of the reaction is 44.13 kJ mol-1

Answer 2

Answer:

We are given that:

k1 = 1.2 x 10-3 sec-1

T1 = 30 + 273 = 303 K

k2 = 2.1 x 10-3 sec-1

T2 = 40 + 273 = 313 K

Substituting these values the equation:

Log (k2 / k1)

= Ea / 2.303 R {(T2 – T1) / T1 T2}

We will get:

Log (2.1 x 10-3 sec-1/ 1.2 x 10-3 sec-1)

= Ea / 2.303 x 8.314 {(313 – 303) / 303 x 313}

Log (2.1 / 1.2) = Ea / 2.303 x 8.314 x (10 / 303 x 313)

Ea = 44126.3 J mol-1

Ea = 44.13 kJ mol-1

Hence, the energy of activation of the reaction is 44.13 kJ mol-1