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Arrhenius Equation

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Question

According to Arrhenius equation, the rate constant ( $k$ ) is related to temperature ( $T$ ) as:

A

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

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B

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

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C

ln (\frac{k_{2}}{k_{1}}) = \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 A $ln (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$
$k_{1} = {A e}^{- \frac{E_{a}}{R T_{1}}}; k_{2} = {A e}^{- \frac{E_{a}}{R T_{2}}} ∴ \frac{k_{1}}{k_{2}} = e^{- \frac{E_{a}}{R T_{1}} (\frac{1}{T_{1}} - \frac{1}{T_{2}})} ln (\frac{k_{1}}{k_{2}}) = - \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) ln (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

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Similar questions

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

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:

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:

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. Two reactions

R_{1}

R_{2}

have identical pre-exponential factors. Activation energy of

R_{1}

exceeds that of

R_{2}

10 k J m o l^{- 1}

k_{1}

k_{2}

are rate constants for reactions

R_{1}

R_{2}

respectively at 300 K, then

l n (\frac{k_{2}}{k_{1}})

is equal to :
(R = 8.314

J m o l^{- 1} K^{- 1}

)

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