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The elementary reaction A+B\rightarrow \; products\; has\; k=2\times 10^{-5}M^{-1}S^{-1}   at a temperature of 27^{\circ}C. Several experimental runs are carried out using stoichiometric proportion. The reaction has a temperature coefficient value of 2.0. At what temperature (in ^{\circ}C ) should the reaction be carried out if inspite of halving the concentrations, the rate of reaction is desired to be  50^{o}/_{o}  higher than a previous run. \left ( Given\; \; \frac{\: ln\; 6}{l\: n\; 2}=2.585 \right ). 

(response should be upto only one decimal digit like 78.7).

Option: 1

47


Option: 2

52.8


Option: 3

57


Option: 4

37


Answers (1)

best_answer

r_{2}=k_{2}\left [ A \right ]_{2}^1\left [ B \right ]_{2}^1           for a certain run

r_{1}=k_{1}\left [ A \right ]_{1}^1\left [ B \right ]_{1}^1           for a previous run

dividing, we get

\frac{r_{2}}{r_{1}}=\frac{k_{2}}{k_{1}}\frac{\left [ A \right ]_{2}}{\left [ A \right ]_{1}}\frac{\left [ B \right ]_{2}}{\left [ B \right ]_{1}}

Substituting the given information

1.5=_{2}\left [ \frac{t_{2}-27}{10} \right ]\times \frac{1}{2}\times \frac{1}{2}

\Rightarrow 6=_{2}\left [ \frac{t_{2}-27}{10} \right ]\; \; \Rightarrow \frac{t_{2}-27}{10}\; ln2=\; l\; n\; 6\; \; \Rightarrow \frac{t_{2}-27}{10}=\frac{l\; n\; 6}{l\; n\; 2}\Rightarrow t_{2}    =52.85^{\circ}C\approx 53^{\circ}C

Posted by

Shailly goel

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