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. $k=0.0231mi{n}^{-1},E_a=43.848kJmo{l}^{-1}$
• B. $k=0.017mi{n}^{-1},E_a=52.54kJmo{l}^{-1}$
• C. $k=0.0693mi{n}^{-1},E_a=43.848kJmo{l}^{-1}$
• D. $k=0.0231mi{n}^{-1},E_a=28.92kJmo{l}^{-1}$

Answer 1

The correct option is A $k=0.0231mi{n}^{-1},E_a=43.848kJmo{l}^{-1}$

Since $t_{1/2}=\frac{0.693}{k},$

$\therefore k=\frac{0.693}{t_{1/2}}$

Given: $t_{1/2}=30min$ at ${27}^{\circ }C$ and $t_{1/2}=10min$ at ${47}^{\circ }C$

$\therefore k_{{27}^{\circ }C}=\frac{0.693}{30}mi{n}^{-1}=0.0231mi{n}^{-1}$

and $k_{{47}^{\circ }C}=\frac{0.693}{10}=0.0693mi{n}^{-1}$

We know that $\log \frac{k_{{47}^{\circ }C}}{k_{{27}^{\circ }C}}=\frac{E_a}{2.303R}\times \frac{(T_2-T_1)}{T_2\times T_1}$

$\therefore E_a=\frac{2.303R\times T_1\times T_2}{(T_2-T_1)}\log \frac{k_{{47}^{\circ }C}}{k_{{27}^{\circ }C}}$

or $E_a=\frac{2.303\times 8.314\times {10}^{-3}\times 300\times 320}{(320-300)}\log \frac{0.0693}{0.0231}$

$=43.848kJmo{l}^{-1}$.

Hence, the correct answer is option $\text{A}$.