Question

For reaction $A\to B$ and $P\to Q$ Arrhenius constants are ${10}^{10}$ and ${10}^{12}$ respectively. If $E_{A\to B}=800$ cal/ mole and $E_{P\to Q}=1400$ cal/mole, then find the temperature at which their rate constant are same.
(Given : $R=2{\text{cal mol}}^{-1}{\text{K}}^{-1})$

• A. $\frac{300}{4.606}\text{K}$
• B. $\frac{4.606}{600}\text{K}$
• C. $300\times 4.606\text{K}$
• D. $600\text{K}$

Answer 1

The correct option is A $\frac{300}{4.606}\text{K}$

For reaction $A\to B$
$k_1=A_1{e}^{\frac{-E_{a_1}}{RT}}$

For reaction $P\to Q$
$k_2=A_2{e}^{\frac{-E_{a_2}}{RT}}$

According to question, $k_1=k_2$
$\Rightarrow A_1{e}^{\frac{-E{a}_1}{RT}}=A_2{e}^{\frac{-E{a}_2}{RT}}$
$\Rightarrow \frac{A_2}{A_1}=e^{(E{a}_2-E{a}_1)/RT}$
$\Rightarrow \frac{{10}^{12}}{{10}^{10}}=e^{(1400-800)/RT}$
$\Rightarrow {10}^2=e^{\left(600/RT\right)}\{\because R=2{\text{cal mol}}^{-1}{\text{K}}^{-1}\}$
$2\ln 10=\frac{600}{2T}$
$\Rightarrow T=\frac{600}{2\times 2\times \ln 10}$

$T=\frac{300}{2\times 2.303}=\frac{300}{4.606}\text{K}$