Chemical Kinetics MCQs

Ans. c) mol^-1 L s^-1

Explanation: The rate of reaction unit is - (mol L^-1) ^1-n s^-1

Here, the order of the reaction is denoted as n. Hence, for the secondary reaction we know that n= 2

Now, if we put this in the equation, it will be - (mol L^-1)^1-2 s^-1 = mol^-1 L s^-1

Ans. a) 4.74 × 10^-2 (L/mol)1/2 s^-1

Explanation: It is mentioned in the question that, [A] = 0.5 M, [B] = 0.1 M and k= 0.03

And we know that according to the rate law- Rate = k[A] x [B] y

Therefore, from this law it is clear that the order of the reaction will be- 1+ 0.5 = 1.5 = 32

Hence, the unit k= (mol L^-1)^1-1.5 s^-1 = (L/mol)^1/2 s^-1

Now, if we put all this in the equation it will be-

R= k[A]1[B]1/2 = 0.03 × 0.5 × 0.11/2

= 4.74 × 10^-2 (L/mol) 1/2 s^-1.

Ans. a) k [CH₃COOC₂H₅]

Explanation: We know that the acid hydrolysis of ester can be expressed as CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH

Sometimes, if the reactant is used excessively compared to the other. The order of the reaction can be altered or even controlled.

According to the rate law, Rate = k [A] x [B] y

= k [CH₃COOC₂H₅] [H₂O]

But as the water is excessive, it will be R= k [CH₃COOC₂H₅].

Ans. (c) 1.5

Explanation: As we know, the rate law is- Rate = k[A] x [B]y

Hence, according to this-

Ans. (b) catalyst used

Explanation: We know that in chemistry, the substance, that interferes with the rate of any chemical reaction, yet remains unchanged itself, is called a catalyst. A catalyst has the power to increase the rate of a chemical reaction.

Hence, the equilibrium constant Kp mentioned in the question, is dependant on the usage of the catalyst.

Ans. b) First order

Explanation: From the question, we get to know that, k= 3.28 × 10^-4 s^-1

The formula for rate constant, k = (mol L^-1)1-ns^-1

Here, n = the order of the reaction. The value of n needs to be 1 for the equation (mol L^-1)1-ns^-1 to convert into s^-1.

Hence, k = 3.28 × 10^-4 s^-1 expresses a first order reaction.

Ans. d) R = k1 [NO₂ ]²

Explanation: We know that, the rate determining step is the slowest step in every reaction, and the whole reaction time depends on this basic step.

Hence, 2NO₂ → NO + NO₃(k1) must be the determinant step for this reaction.

So, we can conclude that the rate law R= k1[NO₂]².

Ans. c) 0.015 s^-1

Explanation: From the question we know that,

[A] = 0.75 M, and k= 0.02

As this reaction represents a pseudo-first-order reaction, s-1 will be counted as the unit.

Hence, R= k [A]= 0.02 × 0.75

= 0.015 s^-1

Ans. (b) 1

Explanation: We know that, rate = k[A]x

From the question we get to know that we need to calculate 100 x rate = k [10]x

Now if we solve this equation, we will see that ⇒ x = 2.

Ans. b) False

Explanation: It is mentioned that the reaction related to the inversion of cane sugar is C₁₂H₂₂O₁₁ + H₂O → glucose + fructose.

In this particular reaction, there is excessive water. Also, this reaction belongs to the pseudo-first order.

Hence, the molecularity remains 2 the order of the reaction remains 1

Therefore, the rate law R=k[C₁₂H₂₂O₁₁]

Ans. d) Molecularity

Explanation: The concept- Molecularity of a reaction can be described as the number of molecules, atoms and ions, that need to be collided parallelly and continuously with each other, so that the chemical reaction can take place.

This process of Molecularity does not affect the process of any chemical reaction in a direct way.

Ans. (d) infinite

Explanation: It is mentioned in the question that, the first order reaction was not fully completed.

Hence, t 100% = infinity

Ans. a) Rate of reaction increases

Explanation: The average kinetic energy of the elements taking part in any reaction, can be increased by increasing the temperature of the reaction. The movement of the particle speeds up with the kinetic energy. For this reason, the particle collides with each other faster and often. Hence it makes easier for chemical reaction to take place so we can conclude that increasing the temperature can increase the rate of the reaction.

Ans. c) It quadruples

Explanation: Suppose the initial concentration of reactant is [A]

Hence, as mentioned in the question, the concentration of reactant = 2 x [A]

As per the second order reaction rate, it will be = k x [A]²

According to this equation the rate for 2 x [A] = k x 4 x [A]²

Finally from this equation, we can conclude that, when the concentration of the reactant is doubled in a second order chemical kinetics reaction, the reaction rate will be quadruples.

Ans. (d) E_a

Explanation: In chemistry, the substance, that interferes with the rate of any chemical reaction, yet remain unchanged itself, is called a catalyst. A catalyst has the power to increase the rate of a chemical reaction. From the characteristics of a catalyst, we know that it affects the activation energy.

In this question, there is only one such option, which is E_a.

Ans. (a) R= [C₂H₄] [I₂]^3/2

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 5/2. The rate law for the reaction is known to be R= [C₂H₄] [I₂]^3/2.

Ans. (d) r₂= 4r₁

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.

Ans. b

Explanation: Given, R=0.6 s^-1 and k= 0.035

For a first order reaction R= k [A]

[A]=R/k = 0.6/0.035 = 17.143 M.

Ans. 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.

Ans. 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 = 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 × 0.1^1/2 = 4.74 × 10^-2(L/mol)^1/2 s^-1.