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: $\frac{l n 6}{l n 2} = 2.585)$ .

46^{o} C

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53^{o} C

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57^{o} C

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37^{o} C

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Solution

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.
$1.5 = 2^{⎛ ⎝ \frac{t_{2} - 27}{10} ⎞ ⎠} \times \frac{1}{2} \times \frac{1}{2}$
$6 = 2^{\frac{t_{2} - 27}{10}}$
$\frac{t_{2} - 27}{10} l n 2 = l n 6$
$\frac{t_{2} - 27}{10} = \frac{l n 6}{l n 2}$

Hence, $\frac{t_{2} - 27}{10} = 2. 585$
$t_{2} = {52.85}^{o} C ≃ 53^{o} C$

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