Now here are some questions worthdoing,Swarna do try it.....

1.SN1,SN2,E1,E2,E1cb
In how many of these cases is entropy change positive?(Give reason for ur answer)

2.A catalyst decreases the energy of activation of the reaction A→B,del.H=37.012Kcal,by 2%.At 27°C 4*10^-8 %
of product molecules is able to cross over the potential barrier in absence of catalyst.How many times the catalyst will increase the reate of reaction at 27°C?

3.Two beakers one containing 20ml of a 0.05M aq. sol. of a non-volatile non-electrolyte and the other containing 20ml of a 0.03M NaCl sol. are placed side by side in a closed enclosure.What are the volumes of the two beakers when equlibrium is obtained?

4 Answers

996
Swarna Kamal Dhyawala ·

(2) ln k = -Ea / (RT) + ln A where k is the rate constant, Ea is activation energy, R is the universal gas constant and A is some constant specific to a reaction
A more useful way to write this is in this case is:
k = Ae^ (-Ea / (RT) (Arrehenius equation)
k1k2 = (A.e^ (-Ea1 / (RT)) (A.e^ (-Ea2 / (RT))
The A cancels out.
Ea1 = (37012 - 2% of 37012)calorie
Ea2 = 37012 calorie
R=1.985 cal K-1 mol-1
T=300 K
u can calculate it .......
here we are getting the ratio of k1 and k2
u can also calculate k2and k1 individually as percentage of product molecules are given....

996
Swarna Kamal Dhyawala ·

volume of first beaker 17 ml
volume of second beaker 23 ml (approx)

996
Swarna Kamal Dhyawala ·

E2 > E1cb > E1 >SN2 >SN1
elimination reaction have higher entropy than substitution becoz elimination occurs at higher temperature ( entropy increases when temperature increase) and second order kinetics have greater entropy than first order kinetics becoz more number of molecules are involved ....

481
Anurag Ghosh ·

Hmmmmm.....gt it..

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