NCERT Solutions for Class 12 Chemistry Chapter 3 Chemical Kinetics Intext Questions

Chapter 3 Chemical Kinetics Intext Questions NCERT Solutions Class 12 Chemistry- PDF Download

Page No. 66

3.1. For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Solution

3.2. In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 minutes. Calculate the rate during this interval?

Solution

= 0.005 mol L⁻¹ min⁻¹

= 5 × 10⁻³ M min⁻¹

3.3. For a reaction, A + B → Product; the rate law is given by, r=k[A]1/2[B]2 . What is the order of the reaction?

Solution

The order of the reaction = 2 + ½

= 5/2

= 2.5

3.4. The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Solution

The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k[X]² ...(i)

Let [X] = a mol L⁻¹, then equation (1) can be written as:

Rate₁ = k.(a)²

= ka²

If the concentration of X is increased to three times, then [X] = 3a mol L⁻¹

Now, the rate equation will be:

Rate = k (3a)²

= 9(ka²)

Hence, the rate of formation will increase by 9 times.

Page No. 78

3.5. A first order reaction has a rate constant 1.15 10⁻³s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?

Solution

From the question, we can write down the following information:

Initial amount = 5 g

Final concentration = 3 g

Rate constant = 1.15 10⁻³s⁻¹

We know that for a 1^st order reaction,

= 444.38 s

= 444 s (approx)

3.6. Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.

Solution

We know that for a 1^st order reaction,

It is given that t_1/2 = 60 min

= 0.693/60

= 0.01155 min^-1

= 1.155 min^-1

Or, k = 1.925 × 10^-4 s^-1

Page No. 84

3.7. What will be the effect of temperature on rate constant?

What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?

Solution

The rate constant of a reaction is nearly doubled with a 10° rise in temperature. However, the exact dependence of the rate of a chemical reaction on temperature is given by Arrhenius equation,

K = Ae^-Ea/RT

where,

A is the Arrhenius factor or the frequency factor

T is the temperature

R is the gas constant

E_a is the activation energy

3.8. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_a.

Solution

It is given that T₁ = 298 K

∴ T₂ = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k₁ = k and that of k₂ = 2k

Also, R = 8.314 J K⁻¹ mol⁻¹

Now, substituting these values in the equation:

We get,

= 52897.78 J mol⁻¹

= 52.9 kJ mol⁻¹

3.9. The activation energy for the reaction 2HI_(g) → H₂ + I_2(g) is 209.5 kJ mol⁻¹ at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Solution

In the given case:

E_a = 209.5 kJ mol⁻¹ = 209500 J mol⁻¹

T = 581 K

R = 8.314 JK⁻¹ mol⁻¹

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

x = e^-Ea/RT

Now, x = Antilog (−18.8323)

= 1.471 × 10^-19