Question

The specific rate for a reaction increases by a factor of 4, if the temperature is increased by ${20}^{\circ }C$ from T. The activation energy for the reaction is $55.776kJmo{l}^{-1}$. The value of T is
(Take $ln2=0.7$, $R=8.3Jmo{l}^{-1}{K}^{-1})$

• A. $25{}^{\circ }C$
• B. $27{}^{\circ }C$
• C. $300{}^{\circ }C$
• D. $47{}^{\circ }C$

Answer 1

The correct option is B $27{}^{\circ }C$

We know,
$ln\frac{k_2}{k_1}=\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})$
$\Rightarrow ln4=\frac{55776}{8.3}(\frac{1}{T}-\frac{1}{T+20})$
$\Rightarrow 2\times 0.7=6720\left(\frac{20}{T\times (T+20)}\right)$
$\Rightarrow T\times (T+20)=96000=300\left(320\right)$
$\Rightarrow T=300K=27{}^{\circ }C$