# Sum: The rate constant of a first order reaction increases from 2 × 10−2 to 4 × 10−2 when the temperature changes - Chemistry

The rate constant of a first order reaction increases from 2 × 10−2 to 4 × 10−2 when the temperature changes from 300 K to 310 K. Calculate the energy of activation (Ea).

(log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)

#### Solution

According to the Arrhenius equation,

k=Ae(Ea/RT)

From this, we get

"log"k_2/k_1=E_a/(2.303R)((T_2-T_1)/(T_1T_2))

We are given that

Initial temperature, T1=300 K

Final temperature, T2=310 K

Rate constant at initial temperature, k1=2×102

Rate constant at final temperature, k2=4×102

Gas constant, R=8.314 J K1 mol1

Substituting the values, we get

"log"((4xx10^(-2))/(2xx10^(-2)))=E_a/(2.303xx8.314)((310-300)/(300xx310))

therefore " activation energy of the reaction, "E_a=(log2xx2.303xx8.314xx300xx310)/10

=535985.94" J mol"^(-1)

=535.98" kJ mol"^(-1)

Concept: Temperature Dependence of the Rate of a Reaction
Is there an error in this question or solution?