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 |
Are you looking for Ncert Chemistry Class 12 intext questions solutions chapter 3 Chemical Kinetics? Now you can download Chemical Kinetics intext questions solutions from here.
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
- Calculate the ratio:
\(\frac{[A]_0}{[A]} = \frac{5 \, \text{g}}{3 \, \text{g}} = \frac{5}{3} \approx 1.6667\)
- 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 – 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
Question 3.8: The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298 K. Calculate Ea.
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:
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
\(\ln x = \frac{-E_a}{RT}\)
\(\log x = -\frac{E_a}{2.303RT}\)
\(\log x = -\frac{209500 \, \text{J mol}^{-1}}{2.303 \times 8.314 \, \text{J K}^{-1} \text{mol}^{-1} \times 581}\)
=−18.8323
Now,x= Antilog (−18.8323)
=1.471×10−19