The activation energy for a certain reaction is 113 kJ/mol. By what factor (how many times) will the rate constant increase when the temperature is raised from 310 K to 325 K

Question

Quizplus · Accepted Answer

The rate constant can be expressed as:

k = A * exp(-Ea/RT)

where A is the frequency factor, Ea is the activation energy, R is the gas constant (8.314 J/mol*K), and T is the temperature in Kelvin.

If we take the ratio of the rate constants at two different temperatures, say T1 and T2, we get:

k2/k1 = (A*exp(-Ea/RT2))/(A*exp(-Ea/RT1))

Simplifying this expression, we get:

k2/k1 = exp[(Ea/R)*((1/T1) - (1/T2))]

Plugging in the given values, we get:

k2/k1 = exp[(113*10^3 J/mol)/(8.314 J/mol*K)*((1/310 K) - (1/325 K))]

k2/k1 = exp(13.11)

k2/k1 = 7.6

Therefore, the rate constant will increase by a factor of 7.6 when the temperature is raised from 310 K to 325 K. The answer is choice C.

The Activation Energy for a Certain Reaction Is 113 KJ/mol