Question

The reaction A(aq) → 2 B(aq) is a second order reaction with respect to A(aq). Its activation energy is 41.3 kJ/mol. When the concentration of A(aq) is 0.100 M and the temperature is 25.0^oC, the rate of reaction is 0.333 M/s. What is the rate of reaction when we increase the concentration of A(aq) to 0.272 M and we raise the temperature to 54.9 ^oC?

Answer 1

Given that, given reaction is a second order reaction.

Therefore, rate law expression of this reaction at 25^oC will be:

rate = K₁.[A]²

Hence, rate constant at, T₁ = 25^oC = 298.15 K will be:

K₁ = (rate) / ([A]²)

K₁ = (0.333 M/s) / ([0.100]²)

K₁ = 33.3 M^-1s^-1

Also we have, activation energy of reaction, Ea = 41.3 kJ/mol = 41300 J/mol

Hence, rate constant (K₂) at, T₂ = 54.9^oC = 328.05 K will be:

Therefore, Rate of reaction at second condition will be:

----------- (**Answer**)

_{=============================================================}

(in case of anything wrong/have any doubts, please reach out to me via comments. I will help you)