Table of Contents

## Chapter 4 Chemical Kinetics Class 12 Chemistry Notes for CBSE Students

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

**Ans:** Average rate=

Average rate = 4 x 10^{-4} mol L^{-1} min^{-1} 4×10^{-4}

= 60 mol L^{-1 }s^{-1} = 6.66 x 10-6 mol L^{-1} s^{-1}

Hence average rate = 4 x 10^{-4} mol L^{-1} min^{-1}

Average rate = 6.66 x 10^{-6} mol L^{-1} s^{-1}

### 4.2 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?

**Ans:**

### 4.3 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?

**Ans:** Rate law(r) = k [A] 1/2[B]2

order of reaction = 12+2 =2 1 / 2 or 2.5

### 4.4 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?

**Ans:** The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k[X]^{2} (1)

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.

### 4.5 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?

**Ans:**

### 4.6 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.

**Ans:** We know that for a 1st order reaction, t_{1/2}=0.693/k

It is given that t_{1/2} = 60 min

### 4.7 What will be the effect of temperature on rate constant?

**Ans:** In general, the rate constant for a reaction nearly becomes double with about 10° rise in temperature because of the fact that the effective collisions become almost double. The exact dependence of the reaction rate on temperature is given by Arrhenius equation; k=Ae−Ea/Rt.

Where A is the Arrhenius factor or the frequency factor. It is also called pre exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole (J mol^{-1}).

### 4.8 The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea .

**Ans:**

### 4.9 The activation energy for the reaction 2 HI(g) → H2 + I2 (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?

**Ans: **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}

**⇒** In x=−Ea/RT **⇒ **log x=−Ea/2.303RT **⇒ **log x=−209500 J mol^{−1}/2.303×8.314 JK^{−1}mol^{−1}×581= −18.8323

==Now, x =Antilog (−18.8323)

= = 1.471×10−19