Question

For a zero order reaction at $200K$ reaction complete in 5 minutes while at $300K$, same reaction, completes in $2.5$ minutes. What will be the activation energy in calorie? $(R=2Cal/(mol.K);ln2=0.7)$

Answer 1

For zero order reaction, $A=A_o-Kt$

When reaction completes, then A becomes zero.

So, $A_o=Kt$ or $K=\frac{A_o}{t}$

$K_1=\frac{A_o}{t_1}=\frac{A_o}{5}$, $T_1$=200 K

$K_2=\frac{A_o}{t_2}=\frac{A_o}{2.5}$, $T_1$=300 K

By using equation,

$\log \left(\frac{K_2}{K_1}\right)=\frac{E_q}{2.303R}(\frac{1}{T_1}-\frac{1}{T_2})⟹\ln \left(\frac{\frac{A_o}{2.5}}{\frac{A_o}{5}}\right)=\frac{E_q}{R}(\frac{1}{200}-\frac{1}{300})⟹\ln 2=\frac{E_q}{2cal}(\frac{1}{200}-\frac{1}{300})⟹E_q=\frac{0.7\times 2cal{mol}^{-1}{K}^{-1}[300\times 200]K^2}{300-200}=0.7\times 2\times 600=840cal{mol}^{-1}$