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Arrhenius Equation

Activation en...

Question

Activation energy $(E_{a})$ and rate constant ( $k_{1}$ and $k_{2}$ ) of a chemical reaction at two different temperatures ( $T_{1}$ and $T_{2}$ ) are related by :

A

ln \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})

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B

ln \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} (\frac{1}{T_{2}} + \frac{1}{T_{1}})

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C

ln \frac{k_{1}}{k_{2}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

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D

ln \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

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Solution

The correct option is B $ln \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$
Activation energy $(E_{a})$ and rate constant ( $k_{1}$ and $k_{2}$ ) of a chemical reaction at two different temperatures ( $T_{1}$ and $T_{2}$ ) are related by Arrhenius equation which can be represented as either $ln \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$

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Similar questions

Q. Activation energy

(E_{a})

and rate constant

(k_{1}

k_{2})

of a chemical reaction at two different temperatures

(T_{1}

T_{2})

are related by:

Q. Activation energy

(E_{a})

and rate constants (

k_{1}

k_{2}

) of a chemical reaction at two different temperature (

T_{1}

T_{2}

) are related by:

k_{1} =

rate constant at temperature

T_{1}

k_{2} =

rate constant at temperature

T_{2}

for a first-order reaction, then which of the following relations is correct?
[

E_{a} =

activation Energy]

Q. Assertion :Th equilibrium constant of the exothermic reaction at high temperature decreases. Reason: Since

ln \frac{K_{2}}{K_{1}} = \frac{{Δ H}^{o}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]

and for exothermic reaction,

{Δ H}^{o} = - v e

\frac{K_{2}}{K_{1}} < 1

K_{1}

K_{2}

are the equilibrium constants at temperature

T_{1}

T_{2}

T_{2} > T_{1}

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