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bazaltina [42]
3 years ago
12

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.
Physics
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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