Answer to Question #147941 in General Chemistry for Joven

Question

The rate constant of a first-order reaction is 3.75
x 10-4
s
-1 at 300°C. If the activation energy is 101
kJ/mol, calculate the temperature at which its rate
constant is 7.50 x 10-4
s
-1
.

Answer 1

Arrhenius equation:

K = Ae^{-\frac{E_a}{RT}}

K – rate constant

A – Arrhenius constant

Ea – activation energy

R – gas constant

T – temperature

$logk_1 = logA - \frac{E_a}{2.303RT_1} \\ logk_2 = logA - \frac{E_a}{2.303RT_2} \\ logk_2 – logk_1 = \frac{E_a}{2.303 RT_1} - \frac{E_a}{2.303 R T_2} \\ log\frac{k_2}{k_1} = \frac{E_a}{2.303 R}(\frac{1}{T_1} - \frac{1}{T_2}) \\ ln\frac{k_2}{k_1} = \frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2}) \\ k_1 = 3.75 \times 10^{-4} \;s^{-1} \\ k_2 = 7.50 \times 10^{-4} \;s^{-1} \\ T_1 = 300 + 273 = 573 \;K \\ E_a = 101 \times 10^3 \;J \\ ln\frac{7.50 \times 10^{-4}}{3.75 \times 10^{-4}} = \frac{101 \times 10^3}{8.314}(\frac{1}{573} - \frac{1}{T_2}) \\ 5.6798 \times 10^{-5} = 174.52 \times 10^{-5} - \frac{1}{T_2} \\ \frac{1}{T_2} = 168.8402 \times 10^{-5} \\ T_2 = 592.27 \; K \\ T_2 = 592 - 273 = 319 \;C$

Answer: 319 ºC