Question

A first order reaction is $50$% complete in $30$ minutes at $27°C$ and in $10$ minutes at $47°C$. The reaction rate constant at $27°C$ and the energy of activation of the reaction are respectively

• A. $k=0.0231mi{n}^{-1}$, $E_a$ $=43.848KJ$ $mo{l}^{-1}$
• B. $k=0.017mi{n}^{-1}$, $E_a$ $=52.54KJ$ $mo{l}^{-1}$
• C. $k=0.00693mi{n}^{-1}$, $E_a$ $=43.848KJ$ $mo{l}^{-1}$
• D. $k=0.0231mi{n}^{-1}$, $E_a$ $=28.92KJ$ $mo{l}^{-1}$

Answer 1

Answer: (A)

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 $t_{1/2}=10$ min 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 $\frac{k_{47^{\circ }C}}{k_{27^{\circ }C}}$ $=\frac{E_a}{2.303R}\times \frac{\left(T_2-T_1\right)}{T_2\times T_1}$
$\therefore$ $E_a$ $=\frac{2.303R\times T_1\times T_2}{\left(T_2-T_1\right)}log$ $\frac{k_{47^{\circ }c}}{k_{27^{\circ }c}}$
or $E_a$ $=\frac{2.303\times 8.314\times 10^{-3}\times 300\times 320}{\left(320-300\right)}log\frac{0.0693}{0.0231}$
$=43.848kJ$ $mo{l}^{-1}$