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oee [108]
3 years ago
6

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).
Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
4 0

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

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