For a particular reaction taking place at \(T = 300K\), \(E_a = 300 kJ/mol\) and \(log k = -3.45\); calculate the temperature at which the rate constant becomes 4 times that of \(k\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 304 K
Use the Arrhenius equation for two temperatures:
log (k2/k1) = Ea/(2.303 R) × (1/T1 − 1/T2)
Given: k2/k1 = 4, Ea = 300 kJ mol−1 = 300000 J mol−1, T1 = 300 K, R = 8.314 J K−1 mol−1. (The value log k = −3.45 is not needed, since only the ratio of rate constants enters.)
log 4 = 0.602
Ea/(2.303 R) = 300000 / (2.303 × 8.314) = 300000 / 19.147 = 15668 K
1/T1 − 1/T2 = 0.602 / 15668 = 3.84 × 10−5 K−1
1/T2 = 1/300 − 0.0000384 = 0.0033333 − 0.0000384 = 0.0032949 K−1
T2 = 303.5 K ≈ 304 K
With such a large activation energy, a rise of only about 4 K is enough to make the reaction four times faster.