Be sure to answer all parts At 50.00 degree C, what is the fraction of collisions...

Question

Question

Be sure to answer all parts At 50.00 degree C, wh

science chemistry

Answer 1

Answer #1

We know the relation for fraction of collisons

ƒ=e^(-Ea/RT)

From given data, activation energy Ea= 84.60 kj/mole = 84600 J/mole

Temperature T = 50 +273 = 323 K

R = 8.314 J/mole.K

Let us plugin all values in the equation, we will get

f =e^(-84600/(8.314*323))
=2.08 x10^-14

Answer 2

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