A first order reaction,A --->B,requires activation energy of 70 kJ/mol.When a 20 % solution of A was kept at 25oC for 20 minutes,25 % decomposition took place. What will be the percent decomposition in the same time in a 30 % solution maintained at 40oC ? Assume the activation energy remains constant in this range of temperature.
Only method to solve it will do.
1357Find k at 25oC from the given conditions
then find k at 40oC
And then find the % dissociation
1Ea = 2.303 RT1TT1-T2log10k2k1
Using given values, we get k2k1 = 3.872 ,
for first order reaction,
k1 = 2.30320log10100100 - 25 = 0.014386 min-1
thus k2 = 0.014386 x 3.872 = 0.05571 min-1
now at 313 K , k2 = 0.05571 min-1
So, 0.05571 = 2.30320log10 100100 - x
we get x = 67.17 %
1but, how does changing % of A in soln shows no effect
effective no of collisions shud increase as there will be lesser sterric hindrance due to less amt of solvent in the 2nd soln
plz exp??
