The rate constant of a decomposition reaction becomes ten times its initial value when the temperature is increased from $200 \text{ K}$ to $250 \text{ K}$ . Calculate the activation energy of the reaction.
(Given: $\mathrm{R} = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ and $2.303 \times 8.314 = 19.15$ )

Question

The rate constant of a decomposition reaction becomes ten times its initial value when the temperature is increased from

200 \text{ K}

to

250 \text{ K}

. Calculate the activation energy of the reaction.
(Given:

\mathrm{R} = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}

and

2.303 \times 8.314 = 19.15

)

Quantrex Academy · Accepted Answer

Correct answer: B. 19.15 kJ mol^ -1

Using the Arrhenius equation: ( (k_2)/(k_1) ) = (E_a)/(2.303 R) ( (1)/(T_1) - (1)/(T_2) ) Given k_2 = 10 k_1 , T_1 = 200 K , T_2 = 250 K , and 2.303 R = 19.15 J K^ -1 mol^ -1 . Substituting the values: (10) = (E_a)/(19.15) ( (1)/(200) - (1)/(250) ) 1 = (E_a)/(19.15) ( (250 - 200)/(200 × 250) ) 1 = (E_a)/(19.15) ( (50)/(50000) ) 1 = (E_a)/(19.15) ( (1)/(1000) ) E_a = 19.15 × 1000 J mol^ -1 E_a = 19.15 kJ mol^ -1 Answer: 19.15 kJ mol^ -1

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