(A) For every $10 \ K$ rise in temperature,the rate constant of a reaction approximately doubles. Given that the temperature increases from $290 \ K$

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(A) For every $10 \ K$ rise in temperature,the rate constant of a reaction approximately doubles. Given that the temperature increases from $290 \ K$ to $310 \ K$,which is a rise of $20 \ K$. This corresponds to two intervals of $10 \ K$ each. Therefore,the rate constant will increase by a factor of $2^2 = 4$. New rate constant $= 3.2 \times 10^{-3} \times 4 = 12.8 \times 10^{-3} = 1.28 \times 10^{-2}$.

Class 12 Chemistry Chemical Kinetics Collision theory, Energy of activation and Arrhenius equation

Medium

The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $310 \ K$,it will be about:

A
$1.28 \times 10^{-2}$
B
$9.6 \times 10^{-3}$
C
$6.4 \times 10^{-3}$
D
$3.2 \times 10^{-4}$

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $310 \ K$,it will be about:

Similar Questions

Explain: How is the value of activation energy determined based on the Arrhenius equation?

For an endothermic reaction $X \rightarrow Y$,the activation energies for the forward and backward reactions are $E_f$ and $E_b$ respectively. Then,in general:

The rate constant of a reaction depends on ................

The activation energy of a chemical reaction can be determined by:

$A$ first order reaction is $50\%$ completed in $20\, \text{minutes}$ at $27\, ^oC$ and in $10\, \text{minutes}$ at $47\, ^oC$. The energy of activation of the reaction is ....... $Kcal$.

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