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Sum: The rate constant of a first order reaction increases from 2 × 10−2 to 4 × 10−2 when the temperature changes - CBSE (Science) Class 12 - Chemistry

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Question

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)`

 

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Solution Sum: The rate constant of a first order reaction increases from 2 × 10−2 to 4 × 10−2 when the temperature changes Concept: Temperature Dependence of the Rate of a Reaction.
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