**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)

#### Solution

**Given:**

Activation energy (E_{a}) = 104 kJ mol^{-1} = 104 × 10^{3} J mol^{-1}

Rate constant (k_{1}) = `3.7 xx 10^-5 "s"^-1`

Temperatures; T_{1} = 25 + 273 = 298 K, T_{2 }= 30 + 273 = 303 K

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

**To find: **

Rate constant (k_{2}) 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

k_{2 }= `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`.