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Published on: 28/05/2021
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Questions + Answers key
Take MCQ Chemistry Test1.
In the reversible reaction.
Find out the rate of disappearance of NO2
2.
Decomposition of NH3 (g) on surface of catalyst
2NH3 ⟶ N2 (g) + 3H2(g)
Under low pressure follows first order kinetics while at high pressure it is zero order reaction. Why?
3.
The rate for a reaction between the substance A and B is given by Rate = k[A]n [B]m
On doubling the conc. of A and halving the conc. of B, find out the ratio of new rate to that of earlier rate of reaction.
4.
Proposed mechanism for below given reaction :
2NO + Br2 ⟶ 2NOBr is as follows
NO(g) + Br2(g) ⟶ NOBr2(g)
NOBr2(g) + NO(g) ⟶ 2NOBr(g)
Find out the order w.r.t. NO (g)
5.
A systematic plot of ln Keq versus 1/T for a reaction has been shown below: Prove that this reaction is exothermic
1.
Rate of reaction = -1/2 d[NO2]/dt
= K1 [NO2]2 - K2 [N2O4]
therefore rate of disapperance of NO2
= - d[NO2]/dt = 2K1 [NO2]2 –2 K2 [N2O4]
2.
In heterogenous catalysis molecules of NH3 are absorbed on surface. Under lower conc. the surface of catalyst is not completely occupied . When pressure is high the surface is completely occupied and further increase in pressure (conc.) does not affect the rate.
3.
Earlier rate R = K[A]n [B]m
Now rate R1 = K[2A]n [B/2]m
R1/R = 2n-m
4.
Rate = K [NOBr2][NO]
But [NOBr2]/[NO][Br2] = K
Or [NOBr2] = K[NO][Br2]
Substituting values of [NOBr2] in rate law
Rate = K’[NO]2 [Br2]
5.
log K2 /K1 = ∆H/ 2.303R (1/T1 –1/T2)
log 6 /2 = ∆H/ 2.303R (1.5 x 10-3 –2.0 x10-3)
∆H comes negative. Hence exothermic.
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