NEET and AIPMT NEET Chemistry Chemical Kinetics MCQ questions with solutions. Get access to MCQs on Chemical Kinetics NEET Questions With Answers.

These questions are aligned with the NEET syllabus and help you better preparation. Free Question Bank for NEET Chemistry Chemical Kinetics / रासायनिक बलगतिकी. NEET Previous Year Question Paper Chemistry – Chapter-Wise Class 12 Chapter 4 Chemical Kinetics. That is why the NEET Question Papers are very prestigious. Chemical Kinetics NEET Chemistry MCQ. NEET Chemistry students should refer to the following multiple-choice questions with answers for Chemical Kinetics in NEET. Chemical Kinetics NEET Questions With Answers

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## Class 12 Chapter 4 Chemical Kinetics NEET Questions with Answers

**NEET 2020:** The rate constant for a first order reaction is 4.606×10^{−3}s^{−1}. The time required to reduce 2.0g of the reactant to 0.2g is :

- a) 200 s
- b) 500 s
- c) 1000 s
- d) 100 s

Answer:

**NEET 2020:** An increase in the concentration of the reactants of a reaction leads to change in:

(1) heat of reaction

(2) threshold energy

(3) collision frequency

(4) activation energy

Answer: *Collision frequency ∝ number of reactant molecules per unit volume*

As the concentration of reactants of a reaction increase the number of reactant molecules per unit volume increase which increases the collision frequency

**NEET 2020:** The half-life for a zero order reaction having 0.02 M initial concentration of reactant is 100 s. The rate constant (in mol L^{-1} s^{-1}) for the reaction is

(1) 1.0×10^{−4}

(2) 2.0×10^{−4}

(3) 2.0×10^{−3}

(4) 1.0×10^{−2}

Answer: For zero order reaction

t_{½}=a/2k

⟹k= a/2t_{½}

k= 0.02/2×100=1.0×10^{−4} molL^{−1}s^{−1}

**NEET 2020:** In collision theory of chemical reaction, ZAB represents

(1) the fraction of molecules with energies greater than Ea

(2) the collision frequency of reactants, A and B

(3) steric factor

(4) the fraction of molecules with energies equal to Ea

Answer: (2) the collision frequency of reactants, A and B

Sol. The number of collisions per second per unit volume of the reaction mixture (A and B) is known as collision frequency Z_{AB}.

**NEET 2019:** If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:

- t=2.303/k
- t=0.693/k
- t=6.909/k
- t=4.606/k

Answer: Solution: Correct option is D

t=4.606/k

First order rate constant is given as ,

t= (2.303/k)log(a/a-x)

For 99% completed reaction,

t= (2.303/k)log(100/100-99)

t= (2.303/k)log(100)

t= (2.303/k)log10^{2}

t= (2.303/k)x2

t=4.606/k

**NEET 2019:** For the chemical reaction N_{2}(g)+3H_{2}(g) ⇌ 2NH_{3}(g) the correct option is:

Answer: Solutions 4

**NEET 2019:** A first order reaction has a rate constant of 2.303×10−3 s−1. The time required for 40 g of this reactant to reduce to 10 g will be [Given that log2= 0.3010] **NCERT Problem**

(1) 230.3 s

(2) 301 s

(3) 2000 s

(4) 602 s

Answer: Solution: Correct option is D

t=602 s

First order rate constant is given as ,

t= (2.303/k)log(a/a-x)

For 40 g of this reactant to reduce to 10 g ,

t= (2.303/k)log(40/10)

t= (2.303/2.303×10^{-3})log(4)

t= (10^{3})log2^{2}

t= (10^{3})x2 log2 = (10^{3})x2 x 0.3010

t=602.0

**NEET 2019:** For a reaction, activation energy E_{a}=0 and the rate constant at 200 K is 1.6 X 10^{6}s^{−1}. The rate constant at 400K will be [Given that gas constant, R=8.314 J K^{−1} mol^{−1}] **NCERT Problem**

(1) 3.2 x 10^{4} s^{−1}

(2) 1.6 x 10^{6} s^{−1}

(3) 1.6 x 10^{3} s^{−1}

(4) 3.2 x 10^{6} s^{−1}

Answer: Solution: (2) 1.6 x 10^{6} s^{−1}

log(K_{2}/K_{1})=(E/2.303R)(T_{2}−T_{1}/T_{1}T_{2})

E_{a}=0

log(K_{2}/K_{1})=0

K_{2}/K_{1}=10^{0} =1

⇒ K_{2}=K_{1}

K_{2}= 1.6 x 10^{6} s^{−1}

at 400K

**NEET 2018:** The correct difference between first- and second-order reactions is that **NCERT Problem**

- the rate of a first-order reaction does not depend on reactant concentration; the rate of a second-order reaction does depend on reactant concentrations.
- the half-life of a first-order reaction does not depend on [A]
_{o}; the half-life of a second-order reaction does depend on [A]_{0} - a first-order reaction can be catalyzed; a second-order reaction cannot be catalyzed.
- the rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations

**Answer:** Solutions- Correct option is – (2) The half-life of a first-order reaction does not depend on [A]_{o} ; the half-life of a second-order reaction does depend on [A]_{0}

For first order reaction, t_{½}= 0.693/k

For second order reaction, t_{½}=1/k[A]_{0}

**NEET 2018:** When the initial concentration of the reactant is doubled, the half-life period of a zero order reaction **NCERT Problem**

- is halved
- is doubled
- is tripled
- remains unchanged

**Answer:** Solutions- Correct option is – (2) For zero order reaction, t_{½}= [A]_{0}/2k

**NEET 2017:** Mechanism of a hypothetical reaction

X2 + Y2 ——-> 2XY

is given below

(A) X2 ⇌ X + X (fast)

(B) X + Y2 —–→ XY + Y (slowest)

(C) X + Y —-−→XY (fast)

The overall order of the reaction will be

- 1
- 2
- 0
- 1.5

**Answer:** Solutions- Correct option is – (4) For complex reaction,

**NEET 2017:** A first order reaction has a specific reaction rate of 10-2s-1. How much time will it take for 20 g of the reactant to reduce to 5 g? **NCERT Problem**

(a) 238.6 s

(b) 138.6 s

(c) 346.5 s

(d) 693.0 s

**Answer:** Solutions- Correct option is – (2) For fast order reaction,

t= (2.303/k)log(a/a-x)

t= (2.303/10^{-2})log(20/5)

t= (2.303/10^{-2})log(4)

t= (2.303/10-^{2})log(2^{2})

t= (2.303×10^{2})x2xlog(2)

t= (2.303×10^{2})x2x0.3010

t= 138.6 s

**NEET 2017:** The equilibrium constant of the following are: The equilibrium constants of the following are:

N_{2}+3H_{2} ⇌ 2NH_{3}; K_{1}

N_{2}+O_{2} ⇌ 2NO; K_{2}

H_{2}+ ½ O_{2} → H_{2}O; K_{3}

The equilibrium constant (K) of the reaction:

2NH_{3}+5/2 O_{2} ⇌ 2NO+3H_{2}O ; K will be:

- K
_{2}K_{3}/K_{1} - K
_{2}K_{3}^{3}/K_{1} - K
_{2}^{3}K_{3}/K_{1} - K
_{1}K_{3}^{3}/K_{2}

**Answer:** Solutions- Correct option is – (2)

**NEET 2017:** Which one of the following statements is not correct?

- The value of equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium.
- Enzymes catalyse mainly bio-chemical reactions
- Coenzymes increase the catalytic activity of enzyme
- Catalyst does not initiate any reaction

Answer: Solutions- Correct option is – (1) The value of equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium

**NEET 2016:** The decomposition of phosphine (PH3) on tungsten at low pressure is a first-order reaction. It is because the

(1) rate is proportional to the surface coverage

(2) rate is inversely proportional to the surface coverage

(3) rate is independent of the surface coverage

(4) rate of decomposition is very slow

**Answer:** Solutions- Correct option is (1) rate is proportional to the surface coverage

In case of gases (like PH_{3}) pressure of gas on metal/catalyst surface acts as activity/concentration hence the rate of reaction depends upon the pressure of gas. Rate=k[P_{PH3}]

According to Freundlich adsorption isotherm:

x/m=kP^{1/n}

Here x/m is mass of gas absorbed per unit mass of metal/catalyst. At low pressures n = 1 and it keeps increasing on increasing the pressure. ** Additional Information:** As the pressure increases or at high pressure the surface of metal/catalyst becomes saturated and thus pressure no longer affects the rate of reaction.

**NEET 2016:** The rate of a first-order reaction is 0.04 mol L^{-1} s^{-1} at 10 sec and 0.03 mol L^{-1} s^{-1} at 20 sec after initiation of the reaction. The half-life period of the reaction is

- 34.1 s
- 44.1 s
- 54.1 s
- 24.1 s

**Answer:** Solutions- Correct option is (B) 24.1 sec

The rate law expression for the first order reaction is =k×[A].

Substitute values in the above expression.

0.04=K[A]_{10}……(1)

0.03=K[A]_{20}……(2)

Divide equation (1) by equation (2).

[A]_{10}/[A]_{20}= 0.04/0.03= 4/3

At 10 sec, the expression for the rate constant will be t=(2.303/k)log[A]_{10}/[A]_{20}.

10=(2.303/k)log(4/3)

Hence, the rate constant k=(2.303/10)log(4/3)=0.0288min^{−1}

The half life period is t _{½} =0.693/k= 0.693/0.0288= 24.06min^{−1}

**NEET 2015:** The activation energy of a reaction can be determined from the slope of which of the following graphs?

- ln K vs T
- ln K/T vs T
- ln K vs 1/T
- T/lnK vs 1/T

Answer: Solutions- Correct option is (3) ln K vs 1/T

The Arrhenius equation is lnk= lnA− E_{a}/RT.

Thus, the activation energy of a reaction can be determined from the slope of which of the graph ln K vs. 1/T.

Slope= −E_{a}/R

Hence, Ea= −slope×R

**NEET 2015:** Find the order of the reaction if the half-life is independent of its initial concentration.

- Zero
- First
- Second
- More than zero but less than first

**Answer:** Solutions- Correct option is – (2) The half-life of a first-order reaction does not depend on [A]_{o}

For first order reaction, t_{½}= 0.693/k

NEET 2015: The rate constant of the reaction A → B is 0.6 x 10^{-3} molar per second. If the

concentration of A is 5 M then concentration of B after 20 min is

- 1.08 M
- 3.60 M
- 0.36 M
- 0.72 M

**Answer:** Solutions- Correct option is (4): The rate constant has unit mole per second which indicates zero order reaction.

x = kt

The concentration of B after 20 minutes is 20×60×0.6×10^{−3} = 0.72M

**NEET 2013:** What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (R= 8.314 J mol^{-1} K^{-1})

- 342 kJ mol
^{-1} - 269 kJ mol
^{-1} - 34.7 kJ mol
^{-1} - 15.1 kJ mol
^{-1}

Answer: Solutions- Correct option is (3): 34.7 kJ mol^{-1}

Given, initial temperature,

T_{1} = 20+273= 293 K

Final temperature

T_{2} = 35+273 = 308 K

R: 8.314 J mol^{-1} K^{-1}

Since, rate becomes double on raising temperature,

therefore,

r_{2} =2r_{1} or r_{2}/r_{1} = 2

As rate constant, k a r

From Arrhenius equation, we know that

Ea= 34.7 kJ mol−1

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