Calculate the activation energy of a reaction whose rate constant is tripled by $10^{o} C$ rise in the temperature from $22^{o} C$ .
Take
$l n (3) = 1.1$

46.3 k J m o l^{- 1}

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82.2 k J m o l^{- 1}

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168 J m o l^{- 1}

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2403 J m o l^{- 1}

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Solution

The correct option is A $46.3 k J m o l^{- 1}$
The temperature rises from $22^{o} C to 32^{o} C$ .
Thus it rises from $295 K to 305 K$
During this rise in temperature, the rate constant of reaction is tripled.
$∴$
$\frac{k_{305}}{k_{295}} = 3$

According to Arrhenius the relationship between temperature and rate constant is given by,
$k = A e^{- E_{a} / R T}$

$\frac{k_{305}}{k_{295}} = \frac{A e^{- E_{a} / R T_{305}}}{A e^{- E_{a} / R T_{295}}} = 3$

$\frac{e^{- E_{a} / R T_{305}}}{e^{- E_{a} / R T_{295}}} = 3$

Taking ln on both side,
$l n 3 = - \frac{E_{a}}{R T_{305}} + \frac{E_{a}}{R T_{295}}$
$l n 3 = - \frac{E_{a}}{R} [\frac{1}{305} - \frac{1}{295}] l n 3 = \frac{E_{a}}{R} [\frac{10}{305 \times 295}]$

$E_{a} = \frac{l n 3 \times 8.314 \times 305 \times 295}{10} E_{a} = 82285.7 J m o l^{- 1} E_{a} \approx 82.2 k J m o l^{- 1}$

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