The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (r. atkinson and j. n. pitts, j. phys. chem.79, 295 ( in the reaction with benzene the rate constantsare 1.44 ×107dm3mol−1s−1at 300.3 k, 3.03 ×107dm3mol−1s−1at 341.2 k, and 6.9 ×107dm3mol−1s−1at 392.2 k. find the pre-exponential factor andactivation energy of the reaction.

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹

Ea = 16,7 kJ/mol

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

A = 1,13x10¹⁰

I hope it helps!

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