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For the gas phase decomposition of t-butyl alcohol , the rate constant has been

ID: 1020580 • Letter: F

Question


For the gas phase decomposition of t-butyl alcohol,


the rate constant has been determined at several temperatures. When ln k in s-1 is plotted against the reciprocal of the Kelvin temperature, the resulting linear plot has a slope of -3.30×104 K and a y-intercept of 33.8.

The activation energy for the gas phase decomposition of t-butyl alcohol is kJ/mol?

Now lets say, For the gas phase decomposition of acetic anhydride,


the rate constant in s-1 has been determined at several temperatures. When ln k is plotted against the reciprocal of the Kelvin temperature, the resulting linear plot has a slope of -1.73×104 K and a y-intercept of 27.6.

The value of the rate constant for the gas phase decomposition of acetic anhydride at 496 K is ?

Explanation / Answer

Gas phase decomposition of t-butyl alcohol:

(CH3)3C-OH -----------> (CH3)2C=CH2 + H2O

A plot on ln k (in s-1) is plotted against 1/T where T is the Kelvin temperature, the plot has a slope of (-3.30*104)K and the y-intercept is 33.8.

We know that

k = Aexp(-Ea/RT) where A is the Arrhenius constant and R is the gas constant; this is the Arrhenius equation. Taking logarithm on both sides,

ln k = ln A – Ea/RT = ln A – (Ea/R).1/T

Re-arranging,

ln k = (-Ea/R).1/T + ln A ……..(1)

This equation is of the form of a straight line y = mx + c where m is the slope and c is the y-intercept.

Our present problem follows the same y = mx + c equation. We are given m = -3.30*104 K and ln A = 33.8

Now, m = (-Ea/R)

or, (-3.30*104 K) = (-Ea/R)

===> Ea = (3.30*104 K)*R = (3.30*104 K)*(8.314 J/mol.K) = 274,362 J/mol = 274.362 kJ/mol (ans)

Gas phase decomposition of acetic anhydride:

As per equation (1) above, (-Ea/R) = (-1.73*104 K); therefore,

Ea = (1.73*104 K)*R = (1.73*104 K)*(8.314 J/mol.K) = 143,832.2 J/mol = 143.8322 kJ/mol.

Also, ln A = 27.6

====> A = e27.6 = 9.6945*1011

Now, T = 496 K; therefore, (-Ea/RT) (s-1) = (-1.73*104 K/496 K) = -34.879 and exp(-Ea/RT) = e-34.789 = 7.786*10-16

Therefore, at 496 K, k =Aexp(-Ea/RT) = (9.6945*1011)*(7.786*10-16) s-1 = 7.548*10-4 s-1 7.5*10-4 s-1 (ans)