Question

_13) The activation energy of a reaction is 100 kJ/mole. The increase in the rate constant when its temperature is raised from 295 to 300 K is approximately: A) 50% B) 100% C) 10% D) 25%

Answer 1

given activation energy Ea =100kJ/mol = 100*1000 J/mol

Ea = 100000 J/mol

given initially T1 =295 K and finally T2 =300ak

by using arrhenius equation by compariong rate constant at 2 different temperatures is given as

ln(K2/K1) = (Ea/R). (1/T1 - 1/T2)

ln(K2/K1) = (100000/8.314).(1/295 - 1/300)

ln(K2/K1) = 0.67954

K2/K1 = e^(0.67954) = 1.97297

K2/K1 = 1.97297. (A)

we have to find increase in K1,

let K1 is increased by x%

then K2 = K1 +K1 * x%

k2 = K1(1+x/100)

put in (A)

1+ x/100 = 1.97297

x/100 = 0.97297

x= 97.297 %, thus K2 is increased by 97.297% whuch is nearly 100%.

so we found increase in k when temperature is increased.