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A first order reaction has a rate constant of 0.924 at 25 oC. Given that the activation energy is 89.8 kJ/mol, calculate the rate constant at 33.3 oC.

Answer :

sebassandin

Answer:

[tex]k_2=2.46[/tex]

Explanation:

Hello!

In this case, according to the Arrhenius equation for variable temperature:

[tex]ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]

Given the temperatures to be computed in kelvins and the activation energy, we obtain:

[tex]ln(\frac{k_2}{0.924} )=-\frac{89800J/mol}{8.314J/(mol*K)}(\frac{1}{306.45K} -\frac{1}{298.15K} )\\\\ln(\frac{k_2}{0.924} )=0.981\\\\k_2=0.924*exp(0.981)\\\\k_2=2.46[/tex]

Best regards!

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