# 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:

Rate_{1} = k.(a)^{2}

= ka^{2}

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(ka^{2})

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 ^{−3}s^{−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^{−3}s^{−1}

We know that for a 1^{st} order reaction,

= 444.38 s

= 444 s (approx)

**3.6. Time required to decompose SO _{2}Cl_{2} 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 *T*_{1} = 298 K

∴ *T*_{2} = (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 *k*_{1} = *k* and that of *k*_{2} = 2*k*

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)}→ H_{2}+ 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}