Question

The rate constant for the first-order reaction is given by log₁₀ k = 14.34 – 1.25 × 10⁴ T. Calculate activation energy of the reaction.

Answer 1

The given rate constant equation is

log₁₀k = 14.34 – 1.25 × 10⁴ T ...(1)

Arrhenius equation is

k = `"Ae"^((-"E"_"a")/("RT"))`

∴ In k = InA - `"E"_"a"/("RT")`

∴ log₁₀k = log₁₀A - `"E"_"a"/(2.303 "RT")`  ...(2)

Comparing (1) and (2),

`"E"_"a"/(2.303  "RT") = 1.25 xx 10^4 "T"`

∴ `"E"_"a"/(2.303  "R") = 1.25 xx 10^4`

∴ `"E"_"a"/(2.303 xx 8.314) = 1.25 xx 10^4`

∴ E_a = 1.25 × 10⁴ × 2.303 × R

∴ E_a = 1.25 × 10⁴ × 2.303 × 8.314

= 23.93 × 10⁴ J

= 239.3 kJ mol^-1

The energy of activation of the reaction is 239.3 kJ mol^-1