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Q1. A reaction was observed for 15 days and the percentage of the reactant remaining after the days indicated was recorded in the following table.
Time (days) % Reactant remaining
0                  100
2                  50
4                  39
6                  25
8                   21
10                18
12                15
14              12.5
15              10
Which one of following best describes the order and the half-life of the reaction? Reaction order Half-life (days)
•  Zero 6
•  First 6
•  First 2
•  Second 2
Solution

Q2.………… of a reaction cannot be determined experimentally
•  Order
•  Molecularity
•  Rate of constant
•  Rate
Solution
Molecularity of reaction is simply the number of molecules reacting in balanced chemical equation. It can be simply determined by examining balanced equation

Q3.  The accompanying figure depicts the change in concentration of species X and Y for the reaction X⟶Y, as a function of time. The point of intersection of the two curves represents

•   Data is insufficient to predict
•  t_(3/4)
•  t_(2/3)
•  t_(1/2)
Solution
The intersection point indicates that half of the reactant X is converted into Y

Q4. A drop of a solution (volume =0.05 mL) contains 6×〖10〗^(-7) mol of H^+. If the rate of disappearance of H^+ is 6.0×〖10〗^5 mol/L×s, how long will it take for H^+ to disappear from the drop
•  8.0×〖10〗^(-8) s
•  2.0×〖10〗^(-2) s
•  6.0×〖10〗^(-6) s
•  2.0×〖10〗^(-8) s
Solution

Q5.CH_3 COOC_2 H_5+H_2 O□(→┴( H+ ) ) CH_3 COOH+C_2 H_5 OH is an example of …… order.
•  Pseudo first order
•  Third
•  Second
•  Zero
Solution
CH_3 COOC_2 H_5+H_2 O□(→┴(H+) ) CH_3 COOH+C_2 H_5 OH Since, in this reaction, water is excess, it is an example of psedo first order reaction (as rate depends only on the concentration of CH_3 COOC_2 H_5).

Q6. Which one of the following is a second order reaction?
•  NH_4 NO_3→N_2+3H_2 O
•  CH_3 COOCH_3+NaOH→CH_3 COONa+H_2 O
• H_2+Br_2→2HBr
•  H_2+Cl_2 □(→┴Sunlight 2HCl)
Solution
The reaction is said to be of second order if its reaction rate is determined by the variation of two concentration terms of reactants. CH_3 COOCH_3+NaOH→CH_3 COONa+H_2 O Is an example of second order reaction

Q7.Hydrogenation of vegetable ghee at 25℃ reduces pressure of H_2 from 2 atm to 1.2 atm in 50 minute. The rate of reaction in terms of molarity per second is:
•  1.09×〖10〗^(-6)
•  1.09×〖10〗^(-8)
•  1.09×〖10〗^(-5)
•  1.09×〖10〗^(-7)
Solution
The change in molarity =n/V=∆P/RT=0.8/(0.0821×273)=0.0327 ∴ rate of reaction = change in molarity per sec =0.0327/(50×60)=1.09×〖10〗^(-5) mol litre^(-1) sec^(-1)

Q8.The chemical reaction 2O_3→3O_2 proceeds as follows O_3⇌O_2+O (fast) O+O_3→2O_2 (slow) The rate law expression should be
•  r=k[O_3 ]^2
•  r=k[O_3 ]^2 [O_2 ]^(-1)
•  r=k[O_3 ][O_2 ]
•  Unpredictable
Solution

Q9.The rate constant of a first order reaction is 4×〖10〗^(-3) sec^(-1). At a reactant concentration of 0.02 M, the rate of reaction would be:
•  4×〖10〗^(-1) M sec^(-1)
•  2×〖10〗^(-1) M sec^(-1)
•  4×〖10〗^(-3) M sec^(-1)
•  8×〖10〗^(-5) M sec^(-1)
Solution
r=K[A]=4×〖10〗^(-3)×0.02=8×〖10〗^(-5) M sec^(-1)

Q10. For A+B⟶C+D,∆H=-20 kJ mol^(-1) the activation energy of the forward reaction is 85kJ mol^(-1). The activation energy for backward reaction is …kJ mol^(-1).
•  85
•  105
•  40
• 65
Solution
For a reaction E_a for forward reaction =E_a for backward reaction +∆H, ∴ 85=A-20 or A=105 kJ mol^(-1)

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