CBSE Class 12 Chemistry 2013 Solved Paper

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Question : 28 of 30
 
Marks: +1, -0
(a) A reaction is second order in A and first order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
(b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t1∕2 this reaction.
( Given log1.428=0.1548)
OR
(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
(b) Rate constant ' k ' of a reaction varies with temperature ' T ' according to the equation:
logk=logA−Ea2.303R(1T)
Where Ea is the activation energy. When a graph is plotted for logk Vs. 1T, a straight line with a slope of −4250K is obtained. Calculate ' Ea ' for the reaction. (R=8.314JK−1mol−1)
Solution:
(a) (i) Differential Rate equation:
dxdt=K[A][B]2

(ii) Let [A]=a,[B]=b
if [B] increases three times
[B]=3b
∴ Rate =K[A][B]2
Rate1=K×a×b2.......(i)
Rate2=K×a×(3b)2.......(ii)
From eqn (i) and eqn (ii)
Rate 2Rate 1=K×a×(3b)2K×a×b
rate 2=9× Rate 1
∵ The rate becomes 9 times when the concentration of B is tripled.
(iii) If [A] and [B] is doubled then [A]=2a,[B]=2b
Rate1=K×a×b2.......(i)
Rate2=K×(2a)×(2b)2.......(ii)
From eqn (i) and eqn (ii)
Rate 2Rate 1=K×(2a)×(2b)2K×a×b2=8
Rate 2=8× Rate 1
∵ The rate becomes eight times when the concentration of both A and B is doubled.
OR
(i) For the first order reaction:
t=2.303Klogaa−x
t99%=2.303Klog1001
=2.303Klog100
=2.303×2K
=4.606K
and t90%=2.303Klog10010
=2.303Klog10
=2.303K
∵t99%t90%=2
t99%=2×t90%
(b) logK=−Ea2.303R(1T)
−Ea2.303R=−4250
Ea=4250×2.303×8.314
=81375Jmol−1
=81.375kJmol−1
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