Class 12 Chemistry MCQ – Chemical Kinetics

This set of Class 12 Chemistry Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Chemical Kinetics”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.

1. For a second-order reaction, what is the unit of the rate of the reaction?
a) s-1
b) mol L-1s-1
c) mol-1 L s-1
d) mol-2 L2 s-1
View Answer

Answer: c
Explanation: The unit of the rate of the reaction (k) is (mol L-1) 1-n s-1, where n is the order of the reaction.
For a second-order reaction, n=2
(mol L-1) 1-n s-1 = (mol L-1)1-2 s-1 = mol-1 L s-1.

2. The rate constant of a reaction is k=3.28 × 10-4 s-1. Find the order of the reaction.
a) Zero order
b) First order
c) Second order
d) Third order
View Answer

Answer: b
Explanation: Given,
k= 3.28 × 10-4 s-1
The general formula to find the units for rate constant, k=(mol L-1)1-ns-1 where n is the order of the reaction. The value of n must be 1 for (mol L-1)1-ns-1 to become s-1. Therefore, k=3.28 × 10-4s-1 represents a first order reaction.

3. For a reaction A +B → C, the experimental rate law is found to be R=k[A]1[B]1/2. Find the rate of the reaction when [A] = 0.5 M, [B] = 0.1 M and k=0.03.
a) 4.74 × 10-2 (L/mol)1/2 s-1
b) 5.38 × 10-2 (L/mol)1/2 s-1
c) 5.748 × 10-2 (L/mol)1/2 s-1
d) 4.86 × 10-2 (L/mol)1/2 s-1
View Answer

Answer: a
Explanation: Given, [A] = 0.5 M, [B] = 0.1 M and k= 0.03
From the rate law it is evident that the order of the reaction is 1+ 0.5 = 1.5 = \(\frac{3}{2}\)
Therefore the unit of k= (mol L-1)1-1.5 s-1 = (L/mol)1/2 s-1
R= k[A]1[B]1/2 = 0.03 × 0.5 × 0.11/2 = 4.74 × 10-2(L/mol)1/2 s-1.
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4. The reaction NO2 + CO → NO + CO2 takes place in two steps. Find the rate law.
2NO2 → NO + NO3 (k1) – slow
NO3 + CO → CO2 + NO2 (k2) – fast
a) R = k1 [NO2]3
b) R = k2 [NO3] [CO]
c) R = k1 [NO2]
d) R = k1 [NO2]2
View Answer

Answer: d
Explanation: In any reaction the slowest step is the rate determining step, the rate of the overall reaction depends on this step. So, 2NO2 → NO + NO3(k1) is the rate determining step. Therefore the rate law R= k1[NO2]2.

5. For the reaction A + H2O → products, find the rate of the reaction when [A] = 0.75 M, k= 0.02.
a) 0.077 s-1
b) 0.085 s-1
c) 0.015 s-1
d) 0.026 s-1
View Answer

Answer: c
Explanation: Given,
[A] = 0.75 M, k= 0.02
The reaction belongs to pseudo first order reaction so, the unit is s-1
R= k [A]= 0.02 × 0.75= 0.015 s-1.
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6. What is the rate law for acid hydrolysis of an ester such as CH3COOC2H5 in aqueous solution?
a) k [CH3COOC2H5]
b) k [CH3COOC2H5] [H2O]
c) k [CH3COOC2H5]2
d) k
View Answer

Answer: a
Explanation: Acid hydrolysis of ester, CH3COOC2H5 + H2O → CH3COOH + C2H5OH
The order of the reaction may be altered sometimes by taking reactant in excess compared to the other.
The rate law R= k [CH3COOC2H5] [H2O] however water is present in excess.
So, R= k [CH3COOC2H5].

7. What is the concentration of the reactant in a first order reaction when the rate of the reaction is 0.6 s-1 and the rate constant is 0.035?
a) 26.667 M
b) 17.143 M
c) 26.183 M
d) 17.667 M
View Answer

Answer: b
Explanation: Given, R=0.6 s-1 and k= 0.035
For a first order reaction R= k [A]
[A]=\(\frac{R}{k}\) = \(\frac{0.6}{0.035}\) = 17.143 M.
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8. How many times will the rate of the elementary reaction 3X + Y → X2Y change if the concentration of the substance X is doubled and that of Y is halved?
a) r2= 4.5r1
b) r2= 5r1
c) r2= 2r1
d) r2= 4r1
View Answer

Answer: d
Explanation: Since it is an elementary reaction, its rate law r1= k [A] 3[B]
When the concentrations are changed the new rate will be r2= k (2[A])3([B]/2) = 4k[A]3[B]
So, r2=4r1.

9. What is the rate law for the reaction C2H4 + I2 → C2H4I2?
a) R= [C2H4] [I2]3/2
b) R= [C2H4] [I2]3
c) R= [C2H4] [I2]2
d) R= [C2H4] [I2]
View Answer

Answer: a
Explanation: Fractional order reactions are reaction whose order is a fraction. This reaction is an example of fractional order reaction, where the order of the reaction is \(\frac{5}{2}\).
The rate law for the reaction is known to be R= [C2H4] [I2]3/2.
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10. The rate law for the reaction involved in inversion of cane sugar is R=k [C12H22O11] [H2O].
a) True
b) False
View Answer

Answer: b
Explanation: The reaction for the inversion of cane sugar is C12H22O11 + H2O → glucose + fructose.
In this reaction water is present in excess and belongs to a pseudo first order reaction, even though the molecularity is 2 the order of the reaction is 1 so the rate law R=k[C12H22O11].

More MCQs on Class 12 Chemistry Chapter 4:

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