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. $k=0.0231 min^{-1}$, $E_{a}$ $=43.848 KJ$ $mol^{-1}$
• B. $k=0.017 min^{-1}$, $E_{a}$ $=52.54 KJ$ $mol^{-1}$
• C. $k=0.00693 min^{-1}$, $E_{a}$ $=43.848 KJ$ $mol^{-1}$
• D. $k=0.0231 min^{-1}$, $E_{a}$ $=28.92 KJ$ $mol^{-1}$

Answer 1

Answer: (A)

Since ${t_{1 /2}}$ $=\frac{0.693}{k}$, $\therefore $ $k=\frac{0.693}{t_{1 /2}}$
Given : $t_{1 / 2}$ $=30 \,min$ at $27^{\circ}C$ and $t_{1 /2}=10$ min at $47^{\circ}C$
$\therefore $ $\quad$ $k_{27^{\circ}C}$ $=\frac{0.693}{30}min^{-1}$ $=0.0231 min^{-1}$
and $k_{47^{\circ}c}$ $=\frac{0.693}{10}=0.0693 min^{-1}$
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 $ $\quad$ $E_{a}$ $=\frac{2.303 R\times T_{1}\times T_{2}}{\left(T_{2}-T_{1}\right)} log$ $\frac{k_{47^{\circ}c}}{k_{27^{\circ}c}}$
or$\quad$ $E_{a}$ $=\frac{2.303\times8.314\times10^{-3} \times300\times320}{\left(320-300\right)}log\frac{0.0693}{0.0231}$
$=43.848 kJ$ $mol^{-1}$