Question

The decomposition of A into product has value of k as $4.5 \times 10^3 s^{-1}$ at $10°C$ and energy of activation $60\ kJ mol^{-1 }$. At what temperature would k be $1.5 \times 10^4 s^{-1}$?

Answer 1

$From \ Arrhenius \ equation,\ we \ obtain$

$log \ \frac {k_2}{k_1} = \frac {E_a}{2.303\ R}(\frac {T_2-T_1}{T_1T_2})$

$Also,$ $k_1 = 4.5 × 10^3 s^{-1}$
$T_1 = 273 + 10 = 283 \ K$
$k_2 = 1.5 \times 10^4 s^{-1}$
$E_a = 60 \ kJ mol^{-1} = 6.0 \times 10^4 J mol^{-1}$
$Then,$

$log \ \frac {1.5 \times 10^4}{4. \times 10^3}$ = $\frac {6.0 \times 10^4 J mol^{-1}}{2.303 \times 8.314 \ j K^{-1} mol^{-1}}$ $(\frac {T_2-283}{283\ T_2})$

⇒ $0.5229 = 3133.627$ $(\frac {T_2-283}{283\ T_2})$

⇒ $\frac {0.5229 \times 283 \ T_2}{3133.627}$ = $T_2 - 283$

⇒ $0.0472 \ T_2 = T_2-283$
⇒ $0.9528 \ T_2 = 283$
⇒ $T_2 = 297.019 \ K$ $(approximately)$
⇒ $T_2 = 297\ K$
⇒ $T_2 = 24°C$
$Hence,\ k \ would\ be$ $1.5 \times 10^4 s^{-1}$ at $24°C$.