At 300 K, the rate constant for the reaction $A \to B$ is $k_{1} = 3.0 \times 1 0^{- 3} s^{- 1}$ . If the activation energy is $120 kJ mol^{- 1}$ , calculate the rate constant at 350 K. Use $R = 8.314 J mol^{- 1} K^{- 1}$ and the Arrhenius equation $k = A e^{- E_{a} / (R T)}$ . Assume $A$ remains constant.

hard 5 marks Class 11 · Chemistry · Organic Chemistry Some Basic Principles and Techniques

Answer

k_{2} \approx 2.9 s^{- 1}

The correct answer is $k_{2} \approx 2.9 s^{- 1}$ — Finally compute $k_{2} = 970 \times 3.0 \times 1 0^{- 3} s^{- 1} \approx 2.9 s^{- 1}$ .

Step-by-step solution

Use the ratio of rate constants to eliminate the pre‑exponential factor: $\frac{k _{2}}{k _{1}} = e^{- E_{a} / (R T_{2}) + E_{a} / (R T_{1})}$ .
Compute the exponent difference: $Δ = - \frac{E _{a}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ .
Insert numbers: $E_{a} = 120000 J mol^{- 1}$ , $R = 8.314 J mol^{- 1} K^{- 1}$ , $T_{1} = 300 K$ , $T_{2} = 350 K$ .\n $Δ = - \frac{120000}{8.314} (\frac{1}{350} - \frac{1}{300})$ .
Evaluate the fraction: $\frac{1}{350} - \frac{1}{300} = - 4.7619 \times 1 0^{- 4}$ .\nMultiply: $- \frac{120000}{8.314} \times (- 4.7619 \times 1 0^{- 4}) \approx 6.87$ .
Thus $\frac{k _{2}}{k _{1}} = e^{6.87} \approx 970$ .
Finally compute $k_{2} = 970 \times 3.0 \times 1 0^{- 3} s^{- 1} \approx 2.9 s^{- 1}$ .

Common mistake: Using the wrong sign in the exponent or forgetting to convert units of activation energy to joules.

Concept tested: Arrhenius rate law