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Arrhenius Equation

Identify the ...

Question

Identify the correct relation between rate constant (k), activation energy $(E_{a})$ , temperature T (K)?

A

log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]

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B

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

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C

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

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D

log \frac{k_{2}}{k_{1}} = \frac{- E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]

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Solution

The correct option is A $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
The arrhenius expression is $k = A e^{- E a / R T} .$
At two different temperatures, this expression takes the form $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}] .$
It can also be written as $ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}] .$

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Similar questions

Q. For a reversible reaction

a A + b B ⇌ c C + d D

; the variation of

K

with temperature is given by

log \frac{K_{2}}{K_{1}} = \frac{- Δ H^{o}}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

Q. Match the column I with column II and mark the appropriate choice.

Column I	Column II
(A) Zero order	(i) $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
(B) First order	(ii) $Δ H > 0$
(C) Endothermic reaction	(iii) $k = \frac{2.303}{t} log \frac{[A]_{0}}{[A]}$
(D) Activation energy	(iv) $k = \frac{1}{t} ([A]_{0} - [A])$

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:

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:

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