Chemical Kinetics and Nuclear ChemistryHard
Question
In a study of effect of temperature on reaction rate, the value of $\frac{1}{K}.\frac{dK}{dT}$ is found to be $\frac{1.25 \times 10^{6}}{T^{3}}K^{- 1}$. Identify the correct statement(s).
Options
A.The activation energy for the reaction at 250 K is 10 kcal/mol.
B.The activation energy for the reaction at 2000 K is 1.25 kcal/mol.
C.The rate of increase of rate constant with the increase in temperature is higher at lower temperature than at higher temperature.
D.The value of $\frac{d\left( \ln K \right)}{dT}$ is 0.625 at 1000 K.
Solution
$\frac{1}{K}.\frac{dK}{dT} = \frac{d\left( \ln K \right)}{dT} = \frac{1.25 \times 10^{6}}{T^{3}} = \frac{E_{a}}{RT^{2}}$
$\therefore E_{a} = \frac{1.25 \times 10^{6}R}{T} = \frac{1.25 \times 10^{6} \times 2}{250} = 10^{4}\text{ cal/mol}$
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