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Rate Constant

Activation en...

Question

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

A

l n \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]

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B

l n \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

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C

l n \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{R} [\frac{1}{T_{2}} + \frac{1}{T_{1}}]

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D

l n \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 D $l n \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
We know that $k = A e^{- E_{a} / R T}$
$\Rightarrow l n k = l n A - \frac{E_{a}}{R T}$
At temperatures $t_{1}$ and $t_{2}$ , let rate constants be $k_{1}$ and $k_{2}$
$l n k_{1} = l n A - \frac{E_{a}}{R T_{1}}$ .....(i)
$l n k_{2} = l n A - \frac{E_{a}}{R T_{2}}$ .....(ii)
Subtracting (i) and (ii)
$l n k_{2} - l n k_{1} = \frac{E_{a}}{R T_{1}} - \frac{E_{a}}{R T_{2}}$
$l n \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

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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. 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} =

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]

K_{1}

K_{2}

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T_{1}

T_{2}

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Q. The effect of temperature on equilibrium constant is expressed as

(T_{2} > T_{1})

l o g \frac{K_{2}}{K_{1}} = \frac{- Δ H}{2.303} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

. For endothermic reaction false statement is:

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