Question

A reaction has an activation energy of 60 kJ/mol at 25 °C. The reaction proceeds slowly
What might happen to the reaction rate if the temperature is increased to 35 celsius explain using the concept of activation energy

Answer 1

The rate will increase — raising the temperature increases the fraction of molecules with energy ≥ the activation energy, so more collisions are effective.

Quantitatively, use the Arrhenius equation:
k = A exp(−Ea/RT), so
k2/k1 = exp[−Ea/R (1/T2 − 1/T1)].

Take Ea = 60 kJ/mol = 60000 J/mol, T1 = 25 °C = 298.15 K, T2 = 35 °C = 308.15 K, R = 8.314 J·mol−1·K−1:

Exponent = −(60000/8.314) (1/308.15 − 1/298.15) ≈ +0.785

k2/k1 = exp(0.785) ≈ 2.2.

So the rate at 35 °C will be roughly 2.2 times the rate at 25 °C. (This assumes the pre‑exponential factor A and the mechanism don’t change with temperature.)