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CBSE Class 12 Chemistry 2013 Solved Paper

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Question : 28
Total: 30
(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−
Ea
2.303R
(
1
T
)
Where Ea is the activation energy. When a graph is plotted for log‌k Vs.
1
T
, 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:
dx
dt
=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 2
Rate 1
=
K×a×(3b)2
K×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 2
Rate 1
=
K×(2a)×(2b)2
K×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=
2.303
K
‌log‌
a
a−x
t99%=
2.303
K
‌log‌
100
1
=
2.303
K
‌log‌100
=
2.303×2
K
=
4.606
K
and t90%=
2.303
K
‌log‌
100
10
=
2.303
K
‌log‌10
=
2.303
K
∵
t99%
t90%
=2
t99%=2×t90%
(b) log‌K=−
Ea
2.303R
(
1
T
)
−
Ea
2.303R
=−4250
Ea=4250×2.303×8.314
=81375Jmol−1
=81.375kJmol−1
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