Ncert Chemistry Class 12 intext questions solutions chapter 3 Chemical Kinetics

Chemical Kinetics intext questions solutions: Class 12 Chemistry Chapter 3 intext questions solutions

Textbook	Ncert
Class	Class 12
Subject	Chemistry
Chapter	Chapter 3
Chapter Name	Chemical Kinetics Class 12 intext questions solutions
Category	Intext questions
Medium	English

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Question 3.1: 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.

Solution 3.1: To calculate the average rate of reaction, we use the following formula:

Average rate of reaction = \(- \frac{\Delta [\text{R}]}{\Delta t}\)

where:

• \(\Delta [\text{R}]\) is the change in concentration of the reactant ([R]) over time.
• \(\Delta t\) is the change in time.

Calculate \(\Delta [\text{R}]\)

The change in concentration of the reactant is:

\(\Delta [\text{R}]\) = [R]_final – [R]_initial

\(= 0.02\ \text{M} – 0.03\ \text{M} = -0.01\ \text{M}\)

Time conversion

The time interval \(\Delta t\) is 25 minutes.

In seconds: 25 minutes = 25 × 60 = 1500 seconds

Average rate of reaction

In minutes:

Average rate = \(- \frac{-0.01\ \text{M}}{25\ \text{min}} = 0.0004\ \text{M/min}\)

In seconds:

Average rate = \(- \frac{-0.01\ \text{M}}{1500\ \text{seconds}} = 6.67 \times 10^{-6}\ \text{M/s}\)

• The average rate of reaction is \(0.0004\ \text{M/min}\).
• The average rate of reaction is \(6.67 \times 10^{-6}\ \text{M/s}\).

Question 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 3.2: To calculate the rate of the reaction for the process 2A ⟶ Products, we use the formula for the rate of disappearance of reactant (A):

Rate of reaction = \(-\frac{\Delta [A]}{\Delta t}\)

where:

• \(\Delta [A]\) is the change in concentration of (A) over the time interval,
• \(\Delta t\) is the time interval.

Given:

• Initial concentration of (A), \([A]_0 = 0.5 \, \text{mol L}^{-1}\),
• Final concentration of (A), \([A]_f = 0.4 \, \text{mol L}^{-1}\),
• Time interval, \(\Delta t = 10 \, \text{min}\).

Calculate \(\Delta [A]\)

\(\Delta [A] = [A]_f – [A]_0 \)

\(= 0.4 – 0.5 = -0.1 \, \text{mol L}^{-1}\)

Since the concentration of (A) is decreasing, we take the absolute value of \(\Delta [A]), i.e., (0.1 \, \text{mol L}^{-1}\).

Calculate the rate of reaction

Now, divide the change in concentration by the time interval:

Rate of reaction = \(\frac{0.1 \, \text{mol L}^{-1}}{10 \, \text{min}} \)

\(= 0.01 \, \text{mol L}^{-1} \, \text{min}^{-1}\)

However, because the reaction is 2A ⟶ Products, the rate of disappearance of (A) is twice the rate of the reaction.

Thus, the actual rate of the reaction is:

Rate of reaction = \(\frac{0.01}{2} = 0.005 \, \text{mol L}^{-1} \, \text{min}^{-1}\)

The rate of reaction during this interval is: Rate of reaction = \(0.005 \, \text{mol L}^{-1} \, \text{min}^{-1}\)

Question 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 3.3: The order of the reaction r  = k [A]^½ [B]²

order of reaction = \(\frac { 1 }{ 2 } +2=2\frac { 1 }{ 2 } \quad or\quad 2.5\)

Question 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 3.4: 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:

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.

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

Solution 3.5: To determine how long it takes for 5 g of a reactant to reduce to 3 g in a first-order reaction, we can use the first-order rate equation:

\(\ln \left( \frac{[A]_0}{[A]} \right) = kt\)

Where:

• \([A]_0\) = initial concentration (or amount) of the reactant
• [A] = concentration (or amount) of the reactant at time (t)
• k = rate constant
• t = time

Given:

• \([A]_0 = 5 \, \text{g}\)
• \([A] = 3 \, \text{g}\)
• \(k = 1.15 \times 10^{-3} \, \text{s}^{-1}\)

Plug the values into the equation

1. Calculate the ratio:

\(\frac{[A]_0}{[A]} = \frac{5 \, \text{g}}{3 \, \text{g}} = \frac{5}{3} \approx 1.6667\)

2. Take the natural logarithm:

\(\ln \left( \frac{5}{3} \right) \approx \ln(1.6667) \approx 0.511\)

Use the first-order rate equation to find (t)

\(\ln \left( \frac{[A]_0}{[A]} \right) = kt\)

Substituting the values:

\(0.511 = (1.15 \times 10^{-3}) t\)

Solve for (t)

\(t = \frac{0.511}{1.15 \times 10^{-3}} \approx 444.35 \, \text{s}\)

Thus, the time (t) is approximately 444 seconds.

Question 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 3.6: To calculate the rate constant ( k ) for a first-order reaction, we can use the half-life equation specific to first-order kinetics:

\(t_{1/2} = \frac{0.693}{k}\)

Given:

• Half-life \(t_{1/2} = 60 \, \text{minutes} \)

Convert time from minutes to seconds

Since the rate constant is typically expressed in seconds, we need to convert minutes to seconds:

60 minutes = 60 × 60 seconds = 3600 seconds

Rearrange the half-life equation to solve for ( k )

\(k = \frac{0.693}{t_{1/2}}\)

Substitute the value of \( t_{1/2} \)

\(k = \frac{0.693}{3600 \, \text{s}}\)

Calculate ( k )

\(k = \frac{0.693}{3600} \)

\(\approx 1.925 \times 10^{-4} \, \text{s}^{-1}\)

Thus, the rate constant ( k ) for the decomposition of \( \text{SO}_2\text{Cl}_2 \) is approximately \( 1.925 \times 10^{-4} \, \text{s}^{-1}\).

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

Solution 3.7: 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 – E_a / 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

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

Solution 3.8: 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:

Here’s the given equation in MathJax format:

\(\log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left[ \frac{T_2 – T_1}{T_1 T_2} \right]\)

We get:

\(\log \left(\frac{2k}{k}\right) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{10}{298 \times 308} \right]\)

\(\Rightarrow \log 2 = \frac{E_a}{2.303 \times 8.314} \left[ \frac{10}{298 \times 308} \right]\)

\(\Rightarrow E_a = \frac{2.303 \times 8.314 \times 298 \times 308 \times \log 2}{10}\)

= 52897.78 J mol⁻¹

= 52.9 kJ mol⁻¹

Question 3.9: 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?

Solution 3.9: In the given case:

E_a = 209.5 kJ mol⁻¹ = 209500 J mo^l−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}

\(\ln x = \frac{-E_a}{RT}\)

\(\log x = -\frac{E_a}{2.303RT}\)

\(\log x = -\frac{209500 \, \text{J mol}^{-1}}{2.303 \times 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \times 581}\)

=−18.8323

Now,x= Antilog (−18.8323)

=1.471×10⁻¹⁹