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Molecularity of Reaction

The activatio...

Question

The activation energy of a reaction is $58.3 k J / m o l e$ . The ratio of the rate constants at $305 K$ and $300 K$ is about:
[ $R = 8.3 J k^{- 1} {m o l}^{- 1}$ and Antilog $0.1667 = 1.468$ ]

A

1.25

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B

1.75

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C

1.5

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D

2.0

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Solution

The correct option is C $1.5$
$l o g \frac{K_{2}}{K_{1}} = \frac{E_{o}}{2.303 X R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$= \frac{58.3}{2.303 X 8.3} [\frac{305 - 300}{305 X 300}]$

$= 3.0499 X 5.4644 X 1 0^{- 5}$

$= 1.666 X 1 0^{- 4}$

= $l o g \frac{K_{2}}{K_{1}} = 0.1666 X 1 0^{- 3}$

= $\frac{K_{2}}{K_{1}} = a n t i l o g (0.1666)$

$= 1.468 \approx 1.5$

Hence, 1.5 is the answer.

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