# 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 1st 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 1st order reaction,

It is given that t1/2 = 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

Ea 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 Ea.

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 + I2(gis 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:

Ea = 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