Question
A first order reaction is $$50\% $$ complete in 30 minutes at $${27^ \circ }C$$ and in 10 minutes at $${47^ \circ }C.$$ The reaction rate constant at $${27^ \circ }C$$ and the energy of activation of the reaction are respectively
A.
\[k=0.0231\,{{\min }^{-1}},\] $${E_a} = 43.848\,kJ\,mo{l^{ - 1}}$$
B.
\[k=0.017\,{{\min }^{-1}},\] $${E_a} = 52.54\,kJ\,mo{l^{ - 1}}$$
C.
\[k=0.0693\,{{\min }^{-1}},\] $${E_a} = 43.848\,kJ\,mo{l^{ - 1}}$$
D.
\[k=0.0231\,{{\min }^{-1}},\] $${E_a} = 28.92\,kJ\,mo{l^{ - 1}}$$
Answer :
\[k=0.0231\,{{\min }^{-1}},\] $${E_a} = 43.848\,kJ\,mo{l^{ - 1}}$$
Solution :
$${\text{Since}}$$ $${t_{\frac{1}{2}}} = \frac{{0.693}}{k},\,\,\therefore \,\,k = \frac{{0.693}}{{{t_{\frac{1}{2}}}}}$$
$${\text{Given}}\,{\text{:}}\,{t_{\frac{1}{2}}} = 30\,\min \,\,{\text{at}}\,\,{27^ \circ }C$$ $${\text{and}}\,\,{t_{\frac{1}{2}}} = 10\,\min \,\,{\text{at}}\,\,{47^ \circ }C$$
\[\begin{align}
& \therefore \,{{k}_{{{27}^{\circ }}C}}=\frac{0.693}{30}{{\min }^{-1}} \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=0.0231\,{{\min }^{-1}} \\
\end{align}\]
$${\text{and}}$$ \[{{k}_{{{47}^{\circ }}C}}=\frac{0.693}{10}=0.0693\,{{\min }^{-1}}\]
$$\eqalign{
& {\text{We know that}} \cr
& \log \frac{{{k_{{{47}^ \circ }C}}}}{{{k_{{{27}^ \circ }C}}}} = \frac{{{E_a}}}{{2.303R}} \times \frac{{\left( {{T_2} - {T_1}} \right)}}{{{T_2} \times {T_1}}} \cr
& \therefore \,{E_a} = \frac{{2.303\,R \times {T_1} \times {T_2}}}{{\left( {{T_2} - {T_1}} \right)}}\log \frac{{{k_{{{47}^ \circ }C}}}}{{{k_{{{27}^ \circ }C}}}} \cr} $$
$${\text{or}}$$ $${E_a} = \frac{{2.303 \times 8.314 \times {{10}^{ - 3}} \times 300 \times 320}}{{\left( {320 - 300} \right)}}$$ $$\log \frac{{0.0693}}{{0.0231}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,= 43.848\,kJ\,mo{l^{ - 1}}$$