Question

In: Chemistry

The activation energy for the isomerization of methyl isonitrile is 160KJ/Mol. 1) Calculate the fraction of...

The activation energy for the isomerization of methyl isonitrile is 160KJ/Mol.

1) Calculate the fraction of methyl isonitrile molecules that have an energy of 160.0 kJ or greater at 508 K .

2) Calculate this fraction for a temperature of 525 K .

3) What is the ratio of the fraction at 525 K to that at 508 K ?

please help!

Solutions

Expert Solution

Answer : The effect of temperature on rate of reaction and hence on rate constant k is given by Arrhenius equation:
k = Ae^-(Ea/RT)

where A is called pre-exponential factor..

Ea is activation enrgy ..

T is absolute temperature

R is gas constant = 8.314 j/K/mole

and k is rate constant..

the factor e^(-Ea/RT) gives the fraction of molecules having energy equal to or greater than the activation enrgy Ea..
suppose the fraction is represented as x then :

x = e^(-Ea/RT)

taking log both sides..

ln x = -Ea/RT

converting ln to log ...multiply by 2.303

2.303 X log x = -Ea/RT

log x = -Ea/(2.303RT)

Now , On the basis of this we can calculate the all requiered things

1] LogX = - 160000 / 2.303 * 8.314 * 508

log X = - 160000/ 9726.74814 = -16.4494

X = 3.55 * 10^-17

Hence, the fraction of methyl isonitrile molecules is 3.55 * 10^-17

2] Similarly this one is solve at 525 K

Log X = - 160000 / 2.303 * 8.314 * 525

= - 160000 / 10052.2496 = - 15.9168

X = 1.211 * 10^-16

Hecne the fraction is comes out to be 1.211 * 10^-16

3 ] Now the ratio between 525K / 508 K = 1.211 * 10^-16 / 3.55 * 10^-17

= 3.41

Hecne the ratio of fraction is 3.4

Thank you :)

queen_honey_blossom answered 2 years ago

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