Question

A certain reaction has an activation energy of 39.5 kJ/mol. As the temperature is increased from 25.0°C to a higher temperature, the rate constant increases by a factor of 5.90. Calculate the higher temperature (in °C).

Answer 1

Answer:

Explanation:

We shall apply Arrhenius equation which is given below .

$ln\frac{k_2}{k_1} = \frac{E_a}{R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

K₂ and K₁ are rate constant at temperature T₂ and T₁ , Ea is activation energy .

Putting the given values

$ln\frac{5.9}{1} = \frac{39500}{8.3} [\frac{1}{298} -\frac{1}{T_2} ]$

$.000373= [\frac{1}{298} -\frac{1}{T_2} ]$

T₂ = 335.27 K

= 62.27 °C