If a reaction has a rate constant of 3.7 x 10-3 s-1 at 25oC, and an activation energy of 43.6 kJ/mol, what will be the rate constant for this reaction at 75oC?

The rate constant for the reaction at 75 °C = 4.6 × 10^(-2) s^(-1)

Explanation:

By Arrhenius equitation

k = Ae^(-Ea/(RT))

Therefore

3.7 x 10-3 s-1 = A×e^(43600J/mol/(8.3145×298))

Therefore A = 160900.320

Therefore at 75 °C T = 348.15 K and k = 160900.320×e^(-43600/(8.314×348.15)

k = 4.6 × 10^(-2) s^(-1)

the rate constant for this reaction at 75 °C is k = 4.6 × 10^(-2) s^(-1)

Which shows that the rate of reaction increases as temperature increases that is the proportions of effective collisions with energy equal to or greater than the activation energy increases.

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )