Two rate constants of a first‑order reaction are measured as k_1=3.0×10^-4,s^-1 at T_1=300,K and k_2=1.2×10^-3,s^-1 at T_2=350,K. Using the Arrhenius equation, calculate the activation energy E_a (in kJ mol^-1) and the pre‑exponential factor A.

Two rate constants of a first‑order reaction are measured as $k_{1} = 3.0 \times 1 0^{- 4}, s^{- 1}$ at $T_{1} = 300, K$ and $k_{2} = 1.2 \times 1 0^{- 3}, s^{- 1}$ at $T_{2} = 350, K$ . Using the Arrhenius equation, calculate the activation energy $E_{a}$ (in kJ mol $^{- 1}$ ) and the pre‑exponential factor $A$ .

hard 5 marks Class 12 · Chemistry · Chemical Kinetics

Answer

Ea = 24.2 kJ/mol; pre‑exponential factor

A = 4.9 s^{- 1}

The correct answer is Ea = 24.2 kJ/mol; pre‑exponential factor $A = 4.9 s^{- 1}$ — Thus $E_{a}$ is $24.2$  kJ mol $^{- 1}$ and $A$ is $4.9$  s $^{- 1}$ .

Step-by-step solution

Write the Arrhenius equation for each temperature: $k_{1} = A e^{- E_{a} / (R T_{1})}$ and $k_{2} = A e^{- E_{a} / (R T_{2})}$ .
Take the natural logarithm of both equations: $l n k_{1} = l n A - \frac{E _{a}}{R T _{1}}$ and $l n k_{2} = l n A - \frac{E _{a}}{R T _{2}}$ .
Subtract the two equations to eliminate $l n A$ : $l n \frac{k _{1}}{k _{2}} = - \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$ .
Insert the given values: $l n (3.0 \times 1 0^{- 4} /1.2 \times 1 0^{- 3}) = l n (0.25) = - 1.3863$ , $T_{1} = 300$  K, $T_{2} = 350$  K, $R = 8.314$  J mol $^{- 1}$  K $^{- 1}$ .
Calculate the temperature difference: $\frac{1}{T _{1}} - \frac{1}{T _{2}} = 0.00047619, K^{- 1}$ .
Solve for $E_{a}$ : $E_{a} = - R \frac{l n ( k _{1} / k _{2} )}{( 1/ T _{1} - 1/ T _{2} )} = 8.314 \times 1.3863/0.00047619 \approx 24224$  J mol $^{- 1}$ , i.e., $24.2$  kJ mol $^{- 1}$ .
Now determine $A$ using one rate constant, e.g., $k_{1}$ : $A = k_{1} e^{E_{a} / (R T_{1})} = 3.0 \times 1 0^{- 4} \times e^{24224/ (8.314 \times 300)} \approx 4.9$  s $^{- 1}$ .
Thus $E_{a}$ is $24.2$  kJ mol $^{- 1}$ and $A$ is $4.9$  s $^{- 1}$ .

Common mistake: Reversing the sign of the temperature difference term

Concept tested: Arrhenius activation energy calculation

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked

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Two rate constants of a first‑order reaction are measured as k1​=3.0×10−4,s−1 at T1​=300,K and k2​=1.2×10−3,s−1 at T2​=350,K. Using the Arrhenius equation, calculate the activation energy Ea​ (in kJ mol−1) and the pre‑exponential factor A.

Step-by-step solution