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Thermal Energy

If k 1 = ra...

Question

If $k_{1} =$ rate constant at temperature $T_{1}$ and $k_{2} =$ rate constant at temperature $T_{2}$ for a first-order reaction, then which of the following relations is correct?
[ $E_{a} =$ activation Energy]

A

log \frac{k_{1}}{k_{2}} = \frac{{2.303 E}_{a}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})

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B

log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})

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C

log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{1} T_{2}}{T_{2} - T_{1}})

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D

log \frac{k_{1}}{k_{2}} = \frac{E_{a}}{2.303 R} (\frac{T_{1} T_{2}}{T_{2} - T_{1}})

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Solution

The correct option is B $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
Arrhenius gave mathematical expression to deduce the relationship between rate constant and temperature

$k = A e^{- E_{a} R T}$

where,
$A$ is frequency factor and it is a constant
$E_{a}$ is activation energy
$R$ is gas constant, $T$ is temperature

On taking $l n$ on both sides

$ln k = ln A - \frac{E_{a}}{R T}$

If $k_{1}$ and $k_{2}$ are rate constants at temperatures $T_{1}$ and $T_{2}$ respectively, then

$log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{2} T_{1}})$
Hence, the correct option is $B$

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