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A reaction is found to have an activation energy of 38.0 kJ/mol. If the rate constant...

A reaction is found to have an activation energy of 38.0 kJ/mol. If the rate constant for this reaction is 1.60

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Answer #1

1Joule------------ 0,00987Latm

-38000Jolue-----X= -375,06Latm

Arrhenius Equation its

K= Ae(-Ea/RT)

where: K: rate constant, A: colision factor, Ea: activation energy, R: ideal gases constant, T: temperature

1,60*10^2 1/Ms= Ae(-375,06Latm/mole/0,082Latm/Kmole* 249K)

A= 1,60*10^2 1/Ms/ e(-375,06Latm/mole/0,082latm/Kmole* 249K)

A= 1,60*10^2 1/Ms/ e (-18.37)

A= 1,60*10^2 1/Ms /1,05*10-8

A= 1,52*10^10 1/Ms

To the second temperature the Ea and A will be the same so we can calculate K

K= 1,52*10^10 1/Ms* e(-375,06Latm/0,082Latm/Kmole*436K)

K= 1,52*10^10 1/Ms* 2,78*10-5

K=4,23*10^5 1/Ms

The closest answer is the 3th 4,20*10^5M-1s-1 the difference its probably because to the significant number taken into account.

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