1. Tardigrade
  2. Question
  3. Chemistry
  4. For a certain reaction the variation of the rate constant with temperature is given by the equation ln Kt= ln K0+(( ln 3/10)) t(t ≥ 0° C ) The value of the temperature coefficient of the reaction rate is therefore.

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Q. For a certain reaction the variation of the rate constant with temperature is given by the equation
$\ln K_{t}=\ln K_{0}+\left(\frac{\ln 3}{10}\right) t\left(t \geq 0^{\circ} C \right)$
The value of the temperature coefficient of the reaction rate is therefore.

Chemical Kinetics

Solution:

$\ln \left(\frac{K_{t}}{K_{0}}\right)=\frac{t}{10}(\ln 3) ;$ when $t=10 ;$
$\frac{K_{t}}{K_{0}}=$ temperature coefficient
$\therefore \ln ($ temperature coefficient $)=(\ln 3) \times \frac{10}{10}=\ln 3$
or temperature coefficient $=3$.