Question

The second-order rate constant for the decomposition of nitrous oxide to nitrogen molecules and oxygen atoms has been determined at various temperatures:
k(1/M·s) t(°C)
1.87 x10^-3 600
0.0113 650
0.0569 700
0.244 750

Determining the activation energy graphically.

Answer 1

Answer:

The value of the activation energy is 240.96 kJ/mol.

Explanation:

According to the Arrhenius equation,

$k=A\times e^{\frac{-Ea}{RT}}$

$\ln k=-\frac{E_a}{RT}+\ln A$

$\log k=-\frac{E_a}{2.303RT}+\log A$

The graph between log(k) and (1/T) will give straight line with negative slope along with the intercept corresponding to the value of A.

Slope f the line = $-\frac{E_a}{2.303R}$

The slope of the line :

$=\frac{\log (k_2)-\log(k_1)}{\frac{1}{T_2}-\frac{1}{T_1}}$

$T_1=600^oC=600+273K=873 K$

$k_1=1.87\times 10^{-3} (Ms)^{-1}$

$T_2=650^oC=650+273K=923K$

$k_2=0.0113 (Ms)^{-1}$

$-\frac{E_a}{2.303R}=\frac{\log (k_2)-\log(k_1)}{\frac{1}{T_2}-\frac{1}{T_1}}$

$-\frac{E_a}{2.303\times 8.314J/K mol}=\frac{\log (0.0113 (Ms)^{-1})-\log(1.87\times 10^{-3} (Ms)^{-1})}{\frac{1}{923 K}-\frac{1}{873 K}}$

$-E_a=-240,959.466J/mol$

$E_a=240,959.466 J/mol=240.96 kJ/mol$

The value of the activation energy is 240.96 kJ/mol.