Question

The activation energy, Ea, for a particular reaction is 13.6 kJ/mol. If the rate constant at 754 °C is 24.5/min, at what temperature (in °C) will the rate constant be 10.7/min? (R = 8.314 J/mol • K)

Answer 1

Answer:

402.2 degrees Celsius

Explanation:

We are given that

Activation energy=$E_a=13.6 KJ/mol=13.6\times 10^3J/mol$

$1 KJ=10^3 J$

Rate constant=$k_2=24.5/min$

$T_2=754^{\circ}C=754+273=1027 K$

$k_1=10.7/min$

R=8.314 J/mol k

Formula:$log\frac{k_2}{k_1}=\frac{E_a}{2.303R}(\frac{T_2-T_1}{T_2T_1}$

Using the formula

$log\frac{24.5}{10.7}=\frac{13.6\times 10^3}{2.303\times 8.314}\times (\frac{1027-T_1}{1027T_1}$$)$

$\frac{1027-T_1}{1027T_1}=log\frac{24.5}{10.7}\times \frac{8.314\times 2.303}{13.6\times 10^3}$

$\frac{1027-T_1}{1027T_1}=(log 24.5-log10.7)\times 1.408\times 10^{-3}$

$\frac{1027-T_1}{1027T_1}=0.3598\times 1.408\times 10^{-3}=0.507\times 10^{-3}$

$\frac{1027-T_1}{T_1}=1027\times 0.507\times 10^{-3}=0.521$

$1027-T_1=0.521T_1$

$1027=0.521T_1+T_1=1.521T_1$

$T_1=\frac{1027}{1.521}=675.2$

$T_1=675.2-273=402.2^{\circ}C$

Hence, the rate constant will be 10.7/min at temperature 402.2 degrees Celsius