Question

The elementary reaction $A+B\to$ products has $k=2\times {10}^{-5}{M}^{-1}{s}^{-1}$ at a temperature of ${27}^{o}C$. Several experimental runs are carried out using stoichiometric proportion. The reaction has a temperature coefficient value of 2.0. At what temperature should the reaction be carried out if inspite of halving the concentrations, the rate of reaction is desired to be 50% higher than a previous run. (Given: $\frac{ln6}{ln2}=2.585)$.

• A. ${46}^{o}C$
• B. ${53}^{o}C$
• C. ${57}^{o}C$
• D. ${37}^{o}C$

Answer 1

The correct option is B ${53}^{o}C$

For a certain run $r_2=k_2{[A{]}_2}^1{[B{]}_2}^1$......(1)

For a previous run $r_1=k_1{[A{]}_1}^1{[B{]}_1}^1$......(2)

Divide equation (2) with equation (1)

$\frac{r_2}{r_1}=\frac{k_2}{k_1}\frac{[A{]}_2}{[B{]}_1}\frac{[B{]}_2}{[B{]}_1}$

Substitute the values in the above equation.

$1.5=2^{\left(\frac{t_2-27}{10}\right)}\times \frac{1}{2}\times \frac{1}{2}$

$6=2^{\frac{t_2-27}{10}}$

$\frac{t_2-27}{10}ln2=ln6$

$\frac{t_2-27}{10}=\frac{ln6}{ln2}$

Hence, $\frac{t_2-27}{10}=2.585$

$t_2={52.85}^{o}C\simeq {53}^{o}C$