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Q.

The variation of rate constant with temperature follows the equation In =3040T+20T230 ln T

based on this mformation, identify the most appropriate option.

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a

EA will remain constant at all temperatures.

b

There is some temperature greater than 50 K where EA will become negative.

c

EA at 300 K is approximately equal to 1.08×109R.

d

Rate constant will follows Arrhenius equation.

answer is A.

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Detailed Solution

ln k=3040T+20T230ln T

d(ln k)dT=+40T2+40T30T

=EaRT2

Ea=R40+40T330T

At T=300K, Ea1.08×109R

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