Difficult
The rate constant of a reaction is $2 \times 10^{-3} \ min^{-1}$ at $300 \ K$. When the temperature is increased by $20 \ K$,the rate constant becomes three times its initial value. Calculate the activation energy $(E_a)$ of the reaction. Also,determine the rate constant at $310 \ K$.
(N/A) Given: $k_1 = 2 \times 10^{-3} \ min^{-1}$ at $T_1 = 300 \ K$,$k_2 = 3 \times k_1 = 6 \times 10^{-3} \ min^{-1}$ at $T_2 = 320 \ K$.
Using the Arrhenius equation: $\ln(k_2/k_1) = (E_a/R) \times (1/T_1 - 1/T_2)$.
$\ln(3) = (E_a / 1.987) \times (1/300 - 1/320)$.
$1.0986 = (E_a / 1.987) \times (20 / 96000)$.
$E_a = 1.0986 \times 1.987 \times 4800 \approx 10480 \ cal/mol$.
To find $k$ at $310 \ K$: $\ln(k_3/k_1) = (E_a/R) \times (1/T_1 - 1/T_3)$.
$\ln(k_3 / 2 \times 10^{-3}) = (10480 / 1.987) \times (1/300 - 1/310) = 5274.3 \times (10 / 93000) = 0.5671$.
$k_3 / 2 \times 10^{-3} = e^{0.5671} = 1.763$.
$k_3 = 3.526 \times 10^{-3} \ min^{-1}$.
Using the Arrhenius equation: $\ln(k_2/k_1) = (E_a/R) \times (1/T_1 - 1/T_2)$.
$\ln(3) = (E_a / 1.987) \times (1/300 - 1/320)$.
$1.0986 = (E_a / 1.987) \times (20 / 96000)$.
$E_a = 1.0986 \times 1.987 \times 4800 \approx 10480 \ cal/mol$.
To find $k$ at $310 \ K$: $\ln(k_3/k_1) = (E_a/R) \times (1/T_1 - 1/T_3)$.
$\ln(k_3 / 2 \times 10^{-3}) = (10480 / 1.987) \times (1/300 - 1/310) = 5274.3 \times (10 / 93000) = 0.5671$.
$k_3 / 2 \times 10^{-3} = e^{0.5671} = 1.763$.
$k_3 = 3.526 \times 10^{-3} \ min^{-1}$.
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