The activation energy Ea for a particular reaction is 50.0 kJ/mol. How much faster is the reaction at 320 K than at 310.0 K? (R = 8.314 J/mol • K)

Answer 1

Hi Cori J.!

The solution to this question derives from the Arrhenius equation. The equation necessary is as follows:

lnk₂ - lnk₁ = (E_a/R)*[(1/T₁) - (1/T₂)]

From here, there are multiple ways to find a solution:

lnk₂ - lnk₁ = (E_a/R)*[(1/T₁) - (1/T₂)]

ln(k₂/k₁) = (E_a/R)*[(1/T₁) - (1/T₂)]

k₂/k₁ = e^{(Ea/R)*[(1/T1) - (1/T2)]}

<=> e^{[(50000 J/mol)/8.314 J/mol*K)*[(1/310K) - (1/320K)]}

k₂/k₁ = 1.83 => k₂ = 1.83k₁

Alternatively, you could keep the ratio of the rate constants within the natural log until you have an actual number. At which point, you could then liberate it and find the same answer as above:

lnk₂ - lnk₁ = (E_a/R)*[(1/T₁) - (1/T₂)]

ln(k₂/k₁) = (E_a/R)*[(1/T₁) - (1/T₂)]

<=> (50000 J/mol)/8.314 J/mol*K)[(1/310K) - (1/320K)]

ln(k₂/k₁) = 0.606

k₂/k₁ = e^0.606

k₂/k₁ = 1.83 => k₂ = 1.83k₁

A third solution involves assuming the first rate constant is equal to 1, in which case it drops out. You are still finding the second rate constant relative to the first.

lnk₂ - lnk₁ = (E_a/R)*[(1/T₁) - (1/T₂)]

lnk₂ = (E_a/R)*[(1/T₁) - (1/T₂)] + lnk₁

=> k₁ = 1

lnk₂ = (E_a/R)*[(1/T₁) - (1/T₂)]

From this point, the solution is the same as above.

Notice that in all of these cases, we find that the rate of the reaction at the higher temperature is 1.83 times greater than at the lower temperature.

Please let me know if you have any follow up questions!

Cheers