Question

Solve

The energy of activation for a first-order reaction is 104 kJ/mol. The rate constant at 25°C is 3.7 × 10^–5 s ^–1. What is the rate constant at 30°C? (R = 8.314 J/K mol)

Answer 1

Given:

Activation energy (E_a) = 104 kJ mol^-1 = 104 × 10³ J mol^-1

Rate constant (k₁) = `3.7 xx 10^-5 "s"^-1`

Temperatures; T₁ = 25 + 273 = 298 K, T₂ = 30 + 273 = 303 K

R = 8.314 J K^-1 mol^-1

To find:

Rate constant (k₂) at 30°C

Formula:

`"log"_10 "k"_2/"k"_1 = "E"_"a"/(2.303"R") (("T"_2 - "T"_1)/("T"_2"T"_1))`

Calculation: 

`"log"_10 "k"_2/(3.7 xx 10^-5 "s"^-1) = (104 xx 10^3 "J"  "mol"^-1)/(2.303 xx 8.314 "J"  "K"^-1 "mol"^-1) ((303 "K" - 298 "K")/(303 "K" xx 298 "K"))`

∴ `"log"_10  "k"_2/(3.7 xx 10^-5 "s"^-1) = 104000/(2.303 xx 8.314) xx 5/(303 xx 298)`

∴ `"log"_10 "k"_2/(3.7 xx 10^-5 "s"^-1) = 0.301`

∴ `"k"_2/(3.7 xx 10^-5 "s"^-1)` = antilog(0.301) = 2.00

k₂ = `2.00 xx 3.7 xx 10^-5 "s"^-1 = 7.4 xx 10^-5 "s"^-1`

The rate constant of the reaction is `7.4 xx 10^-5 "s"^-1`.