Sucrose is converted to a mixture of glucose and fructose in a pseudo first order process under alkaline conditions. The reaction has a half-life of 28.4 min. The time required for the reduction of a 8.0 mM sample of sucrose to 1.0 mM is
A. 56.8 min
B. 170.4 min
C. 85.2 min
D. 227.2 min
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At T = 300 K, the thermal energy [kB T] in cm-1 is approximately
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The half-life of a first order reaction varies with temperature according to
A. In t1/2 ∝ 1/T
B. In t1/2 ∝ T
C. t1/2 ∝ 1/T2
D. t1/2 ∝ T2
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For a given first order reaction, the reactant reduces to 4 1 th
A. 0.1386 min-1
B. 0.0693 min-1
C. 0.1386 mol L-1 min-1
D. 0.0693 mol L-1 min-1
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For the reaction, 2X3 ⇌ 2 , the rate of formation of X2 is
A. 3 ( d t − d [ X 3 ] )
B. 2 1 ( d t − d [ X 3 ] )
C. 3 1 ( d t − d [ X 3 ] )
D. 2 3 ( d t − d [ X 3 ] )
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The half-life time for a reaction at initial concentrations of 0.1 and 0.4 mol-1 are 200 s and 50 s respectively. The order of the reaction is
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For the reaction shown below,
the value of k
1 is 1 × 10
-4 s
-1 . If the reaction starts from X, the ratio of the concentrations of Y and Z at any given time during the course of the reaction is found to be
[ Z ] [ Y ] = 4 1 . The value of k
2 is
A. 1 × 10-4 s-1
B. 2.5 × 10-5 s-1
C. 4 × 10-4 s-1
D. 3.5 × 10-4 s-1
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Consider an exothermic reaction, A k − 1 ⇌ k 1 I
A. k1 , k-1 and k1 /k-1 increase
B. k1 increases, k-1 decreases and k1 /k-1 increases
C. k1 , k-1 and k1 /k-1 decrease
D. k1 and k-1 decrease and k1 /k-1 increases
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1 g of 90 Sr gets converted to 0.953 g after 2 yr. The half-life of 90 Sr, and the amount of 90 Sr remaining after 5 yr are
A. 1.44 yr and 0.916 g
B. 57.6 yr and 0.75 g
C. 28.8 yr and 0.887 g
D. 100 yr and 0.982 g
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The reaction, 2NO(g) + O
2 (g) → 2NO
2 (g) proceeds via the following steps:
NO + NO
K a N
2 O
2
N
2 O
2 K a’ NO + NO
N
2 O
2 + O
2 k b ( slow ) NO
2 + NO
2
The rate of this reaction is equal to
A. 2kb [NO][O2 ]
B. (2Ka kb [NO]2 [O2 ])/(Ka + kb [O2 ])
C. 2kb [NO]2 [O2 ]
D. Ka [NO]2 [O2 ]
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A reaction proceeds through the formation of an intermediate B in a unimolecular reaction
A k a B k b C
The integrated rate law for this reaction is
A. [ A ] = [ A ] 0 e − k a t
B. [ A ] = [ A ] 0 ( e − k a t − e − k b t )
C. [ A ] = 2 [ A ] 0 ( 1 + k a − k b k a e − k b t − k b e − k a t )
D. [ A ] = [ A ] 0 ( 1 + e − k a t − e − k b t )
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If ka ≫ kb , then concentration versus time plot for the reaction is
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In a homogeneous catalytic reaction, 1.0 M of a substrate band 1.0 μM of a catalyst yields 1.0 mM of a product in 10 s. The turnover frequency (TOF) of the reaction (S-1 ) is
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For the reaction, P + Q + R → S, experimental data for the measured initial rates is given below:
Expt. Initial conc. P Initial conc. Q Initial conc. R Initial rate
1
0.2
0.5
0.4
8.0 × 10-5
2
0.4
0.5
0.4
3.2 × 10-4
3
0.4
2.0
0.4
1.28 × 10-3
4
0.1
0.25
1.6
4.0 × 10-5
The order of the reaction with respect to P, Q and R respectively, is
A. 2, 2, 1
B. 2, 1, 2
C. 2, 1, 1
D. 1, 1, 2
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A reaction follows second order rate law, d t − d [ A ] = k [ A ] 2 ,
A. a plot of [A] versus t is a straight line
B. a plot of 1/[A] versus t is a straight line
C. a plot of ln[A] versus t is a straight line
D. a plot of e[A] versus t is a straight line
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The half-life (t
1/2 ) for the hydrolysis of an ester varies with the initial concentration of the reactant ([E]
0 ) as follows:
[E]0 /10 mol/L
5.0
4.0
3.0
t1/2 /s
240
300
400
The order of the reaction is
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The specific rate constant of decomposition of a compound is represented by ln k = 5.0 − T 12000
A. 24 kcal/mol
B. 12 kcal/mol
C. 24 cal/mol
D. 12 cal/mol
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In a consecutive first order reaction,1 and k2 are the respective rate constants) species B has transient existence. Therefore,
A. k1 ≈ k2
B. k1 = 2k2
C. k1 ≫ k2
D. k1 ≪ k2
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The fluorescence life time of a molecule in solution is 10 ns. If the fluorescence quantum yield is 0.1, the rate constant of fluorescence decay is
A. 1 × 109 s-1
B. 1 × 108 s-1
C. 1 × 107 s-1
D. 9 × 107 s-1
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The value of the rate constant for the gas phase reaction, 2NO2 + F2 → 2NO2 F is 38 dm3 mol-1 s-1 at 300K. The order of the reaction is
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