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3 years ago

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Be sure to answer all parts. By what factor does the fraction of collisions with energy equal to or greater than activation ener

gy of 100. kJ/mol change if the temperature increases from 34.0°C to 52.0°C? Give your answer in scientific notation.

1 answer:

qwelly [4]3 years ago

8 0

Answer:

Factor = 8.77

Explanation:

Fraction of collision with energy greater than or equal to the activation energy can be given by the formula:

$f = \exp(\frac{-E_{a} }{RT} )$

$E_{a} = Activation Energy\\E_{a} = 100 kJ /mol\\E_{a} = 10^{5} J /mol\\$

R = 8.314 J/mol.K

When Temperature = 34⁰C = 34 + 273 = 307 K

$f_{1} = \exp(\frac{-10^{5} }{8.314 * 307} )\\f_{1} = \exp(\frac{-10^{5} }{2552.398} )\\f_{1} = \exp(-39.18)\\f_{1} = 964.59 * 10^{-20}$

When Temperature = 52⁰C = 52 + 273 = 325 K

$f_{2} = \exp(\frac{-10^{5} }{8.314 * 325} )\\f_{2}= \exp(\frac{-10^{5} }{2702.05} )\\f_{2}= \exp(-37.009)\\f_{2} = 8456.6 * 10^{-20}$

$\frac{f_{2}}{f_{1}}= \frac{8456.6 * 10^{-20} }{964.59 * 10^{-20} }$

Factor = 8.77

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