Question

A certain batch of fireflies were observed to "flash" at the rate of 13.0 times per minute at 25°C, and at the lower rate of 5.0 times a minute at a temperature of 15°C. Assume that the flashing is the result of an overall chemical reaction that has a single rate limiting step with the highest activation energy. Use this data to estimate the activation energy for this slowest step. You can assume that the concentrations of "reactants" in the fireflies do not depend on temperature.

Answer 1

Ea= 1.8 x 10⁵ J

Step-by-step explanation:

Use the Arrehenius equation to obtain the activation energy by comparing them at the two temperatures as follows:

k = A e^-Ea/RT where k = rate constant

A = Arrhenius constant

Ea = activation energy

T = temperature in degrees K

since we know the constants k₁ and k₂

k₁ = A e^ -Ea/RT₁ and K₂ = A e - Ea/RT₂

dividing them side by side:

k₁ = A e^ -Ea/RT₁ / A e - Ea/RT₂

ln k₁/k₂ = Ea/R (1/T₂ - 1/T₁ )

We have to be careful with what we call T₁ an T ₂ and be consistent then with k₁ and k₂. Lets use T₁ = 25 ºC =( 273 + 25 ) K = 298 K

so k₁ = 13.0 times/min

and likewise for k₂ = 5.0 times/min ( T = 273 + 15 ) K = 288 K

ln ( 13.0/ 5.0 ) = Ea/ 8.314 j//k ( 1/288K - 1/298K ) =

ln (2.60) x 8.314 = Ea ( 1.2 x 10 ⁻⁴) Ea = 1.8 x 10⁵ J