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The activation energy of a reaction is 58.3 kJ/mole. The ratio

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 $0.1667 = 1.468$ ]

1.25

1.75

1.5

2.0

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Evelyn Discussed

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

]

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

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}} = an t i l o g (0.1666)

= 1.468 \approx 1.5

Hence, 1.5 is the answer.

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Question Text	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 $0.1667 = 1.468$ ]
Answer Type	Text solution:1
Upvotes	161