The chapter “Chemical Kinetics” from the 12th standard chemistry part 1 NCERT Board textbook explores the rates of chemical reactions and the factors that influence them. Here’s a summary of the key concepts:
1. Rate of Reaction:
- Definition: The change in concentration of reactants or products per unit time.
- Expression: Rate can be expressed in terms of the decrease in reactant concentration or the increase in product concentration.
- Units: Typically mol L⁻¹ s⁻¹ or related units.
- Average vs. Instantaneous Rate: The average rate is measured over a period, while the instantaneous rate is the rate at a specific moment in time.
2. Factors Affecting Reaction Rates:
- Concentration of Reactants: Generally, increasing reactant concentration increases the rate (though the relationship isn’t always linear).
- Temperature: Higher temperature almost always increases the rate.
- Surface Area: For reactions involving solids, greater surface area increases the rate.
- Nature of Reactants: Some reactions are inherently faster than others due to the specific substances involved.
3. Rate Law and Rate Constant:
- Rate Law (Rate Equation): An experimentally determined equation that expresses the rate of reaction in terms of the concentrations of reactants raised to certain powers (orders).
- Order of Reaction: The sum of the powers in the rate law, indicating how the rate changes with concentration. It can be zero, a whole number, or even a fraction.
4. Integrated Rate Equations:
- Mathematical equations that relate reactant concentration to time for specific reaction orders.
- Zero Order: [A]t = [A]₀ – kt
- First Order: ln[A]t – ln[A]₀ = -kt
- Second Order: 1/[A]t – 1/[A]₀ = kt
5. Half-Life:
- It’s related to the rate constant and can be used to determine the order of a reaction.
6. Temperature Dependence of Reaction Rates:
- Arrhenius Equation: k = Ae^(-Ea/RT)
- Activation Energy (Ea): The minimum energy required for reactant molecules to react.
- Effect of Temperature: Increasing temperature increases the fraction of molecules with enough energy to react, thus increasing the rate constant and the reaction rate.
7. Collision Theory:
- Explains reaction rates in terms of collisions between reactant molecules.
- Effective Collisions: Collisions that lead to product formation. They require sufficient energy (activation energy) and proper orientation of the colliding molecules.
8. Transition State Theory:
- A more advanced theory that describes the reaction pathway in terms of an activated complex or transition state – an unstable intermediate configuration of atoms during the reaction.
Exercise
1. From the rate expression for the following reactions determine their order of reaction and the dimensions of the rate constants:
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2. For the reaction ; 2A + B → A2B, the reaction rate = k [A][B]2 with k = 2·0 x 10-6 mol-2 L2 s-1. Calculate the initial rate of the reaction when [A] = 0·1 mol L-1; [B] = 0·2 mol L-1. Also calculate the reaction rate when [A] is reduced to 0·06 mol L-1.
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1. Calculate the initial rate of the reaction:
- Given the rate law: rate = k[A][B]²
- Given the rate constant: k = 2.0 x 10⁻⁶ mol⁻² L² s⁻¹
- Given the initial concentrations: [A] = 0.1 mol L⁻¹ and [B] = 0.2 mol L⁻¹
Substitute the values into the rate law:
rate = (2.0 x 10⁻⁶ mol⁻² L² s⁻¹) * (0.1 mol L⁻¹) * (0.2 mol L⁻¹)²
rate = (2.0 x 10⁻⁶) * (0.1) * (0.04) mol L⁻¹ s⁻¹
rate = 8.0 x 10⁻⁹ mol L⁻¹ s⁻¹
2. Calculate the reaction rate when [A] is reduced to 0.06 mol L⁻¹:
- Assuming the concentration of B remains the same, as no information is provided about its change.
- Given the new concentration: [A] = 0.06 mol L⁻¹
Substitute the values into the rate law:
rate = (2.0 x 10⁻⁶ mol⁻² L² s⁻¹) * (0.06 mol L⁻¹) * (0.2 mol L⁻¹)²
rate = (2.0 x 10⁻⁶) * (0.06) * (0.04) mol L⁻¹ s⁻¹
rate = 4.8 x 10⁻⁹ mol L⁻¹ s⁻¹
Therefore:
- The initial rate of the reaction is 8.0 x 10⁻⁹ mol L⁻¹ s⁻¹.
- The reaction rate when [A] is reduced to 0.06 mol L⁻¹ is 4.8 x 10⁻⁹ mol L⁻¹ s⁻¹.
3. The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if Ar=2.5 x 10-4 mol-1 Ls-1.
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1. Balanced Chemical Equation:
The decomposition of ammonia (NH₃) into nitrogen (N₂) and hydrogen (H₂) is represented by the following balanced equation:
2NH₃ → N₂ + 3H₂
2. Stoichiometry:
- For every 2 moles of NH₃ that react, 3 moles of H₂ are produced.
3. Zero-Order Reaction:
Since the reaction is zero order, the rate of the reaction (Ar) is independent of the concentration of NH₃. It’s given as:
Ar = 2.5 x 10⁻⁴ mol L⁻¹ s⁻¹
4. Rates of Production:
- Rate of production of N₂:
Since 1 mole of N₂ is produced for every 2 moles of NH₃ consumed, the rate of production of N₂ is half the rate of the reaction.
Rate of production of N₂ = (1/2) * Ar
Rate of production of N₂ = (1/2) * 2.5 x 10⁻⁴ mol L⁻¹ s⁻¹
Rate of production of N₂ = 1.25 x 10⁻⁴ mol L⁻¹ s⁻¹
- Rate of production of H₂:
Since 3 moles of H₂ are produced for every 2 moles of NH₃ consumed, the rate of production of H₂ is 1.5 times the rate of the reaction.
Rate of production of H₂ = (3/2) * Ar
Rate of production of H₂ = (3/2) * 2.5 x 10⁻⁴ mol L⁻¹ s⁻¹
Rate of production of H₂ = 3.75 x 10⁻⁴ mol L⁻¹ s⁻¹
Therefore:
- The rate of production of N₂ is 1.25 x 10⁻⁴ mol L⁻¹ s⁻¹.
- The rate of production of H₂ is 3.75 x 10⁻⁴ mol L⁻¹ s⁻¹.
4. The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and die reaction, rate is given by Rate=k [CH3OCH3]3/2 The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also, be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate= k (PCH3OCH3)3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
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5. Mention the factors which affect the rate of a chemical reaction.
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Nature of Reactants: The chemical identity of the reactants plays a crucial role. Some substances are inherently more reactive than others due to their electronic structure, bond strengths, and other chemical properties.
Concentration of Reactants: Generally, increasing the concentration of reactants increases the reaction rate. More reactant molecules in a given space lead to more frequent collisions, making it more likely that effective collisions (those with sufficient energy and proper orientation) will occur.
Temperature: Increasing the temperature almost always increases the reaction rate. Higher temperatures mean the molecules have greater kinetic energy, leading to more frequent and more energetic collisions. This increases the likelihood of collisions overcoming the activation energy barrier.
Surface Area (for reactions involving solids): If the reaction involves a solid reactant, increasing the surface area of the solid increases the reaction rate. A larger surface area provides more contact points for the reaction to occur. Think of how kindling (small pieces of wood) catches fire faster than a large log.
Presence of a Catalyst: A catalyst is a substance that increases the reaction rate without being consumed in the overall reaction. Catalysts provide an alternative reaction pathway with a lower activation energy, making it easier for the reaction to occur.
Pressure (for reactions involving gases): For gaseous reactions, increasing the pressure generally increases the reaction rate. This is essentially equivalent to increasing the concentration of the gaseous reactants.
Radiation (light): Some reactions are initiated or accelerated by exposure to light or other forms of radiation. These are called photochemical reactions. The light provides the activation energy needed for the reaction to start.
pH (for reactions in solution): For reactions in solution, the pH of the solution can affect the reaction rate, especially if acids or bases are involved in the reaction mechanism. pH affects the availability of H⁺ or OH⁻ ions, which can act as catalysts or reactants.
Solvent Effects (for reactions in solution): The choice of solvent can influence reaction rates.
Solvents can affect the stability of reactants or intermediates, the strength of intermolecular forces, and the ability of reactants to come into contact.
6. A reaction is second order with respect to a reactant How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
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7. What is the effect of temperature on the rate constant of reaction? How can this temperature effect on the rate constant be represented quantitatively?
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8. In pseudo first order hydrolysis of ester in water, the following results were obtained:
t/s 0 30 60 90
[Ester] mol L-1 0-55 0-31 0 17 0 085
(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
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9. A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B is doubled?
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(i) Differential Rate Equation:
Since the reaction is first order in A and second order in B, the differential rate equation is:
Rate = -d[A]/dt = -d[B]/2dt = k[A][B]²
Where:
- Rate is the rate of the reaction
- t is time
- k is the rate constant
(ii) Effect of Tripling [B]:
If the concentration of B is tripled (i.e., [B] becomes 3[B]), the new rate (Rate_new) will be:
Rate_new = k[A](3[B])² = k[A] * 9[B]² = 9 * (k[A][B]²)
Since the original rate was k[A][B]², the new rate is 9 times the original rate.
Therefore, tripling the concentration of B will increase the reaction rate by a factor of 9.
(iii) Effect of Doubling [A] and [B]:
If both [A] and [B] are doubled (i.e., [A] becomes 2[A] and [B] becomes 2[B]),
the new rate (Rate_new) will be:
Rate_new = k(2[A])(2[B])² = k * 2[A] * 4[B]² = 8 * (k[A][B]²)
Therefore, doubling the concentrations of both A and B will increase the reaction rate by a factor of 8.
10. In a reaction between A and B, the initial rate of reaction (r0 ) was measured for different initial concentrations of A and B as given below:
What is the order of the reaction with respect to A and B?
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11. The following results have been obtained during the kinetic studies of the reaction.
2A+B ——–> C + D
Determine the rate law and the rate constant for the reaction.
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Rate = k[A]^x[B]^y
where:
- Rate is the rate of the reaction
- k is the rate constant
- [A] are the concentrations of reactants A
- [B] are the concentrations of reactants B
- x and y are the orders of the reaction with respect to A and B (which we need to find)
1. Finding the Order with Respect to A (x)
- Compare Experiments I and IV:
- [A] changes from 0.1 to 0.4 (4 times)
- [B] is constant at 0.1
- Rate changes from 6.0 x 10⁻³ to 2.40 x 10⁻² (4 times)
- Since the rate increased by a factor of 4 when [A] increased by a factor of 4, the reaction is first order with respect to A (x = 1).
2. Finding the Order with Respect to B (y)
- Compare Experiments II and III:
- [A] is constant at 0.3
- [B] changes from 0.2 to 0.4 (2 times)
- Rate changes from 7.2 x 10⁻² to 2.88 x 10⁻¹ (4 times)
- Since the rate increased by a factor of 4 when [B] increased by a factor of 2, the reaction is second order with respect to B (y = 2).
3. The Rate Law
Now that we know x and y, we can write the complete rate law:
Rate = k[A][B]²
4. Finding the Rate Constant (k)
Let’s use Experiment I:
- Rate = 6.0 x 10⁻³ mol L⁻¹ min⁻¹
- [A] = 0.1 M
- [B] = 0.1 M
6.0 x 10⁻³ = k (0.1) (0.1)²
k = (6.0 x 10⁻³) / (0.1 * 0.01) k = 6
5. Complete Rate Law
The complete rate law, including the value of k, is:
Rate = 6[A][B]²
In Summary
- Rate Law: Rate = 6[A][B]²
- Rate Constant (k): 6 mol⁻² L² min⁻¹ (or 6 M⁻² min⁻¹)
12. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
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13. Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 s-1 (ii) 2 min-1
(iii) 4 years-1
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(i) k = 200 s⁻¹
t₁/₂ = 0.693 / 200 s⁻¹
t₁/₂ = 0.003465 s
t₁/₂ ≈ 3.47 x 10⁻³ s
(ii) k = 2 min⁻¹
t₁/₂ = 0.693 / 2 min⁻¹
t₁/₂ ≈ 0.347 min
(iii) k = 4 years⁻¹
t₁/₂ = 0.693 / 4 years⁻¹
t₁/₂ ≈ 0.173 years
14. The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.
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15. The experimental data for decomposition of N2O5
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16. The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16 th value ?
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ln((1/16)[A]₀/[A]₀) = -60t
ln(1/16) = -60t
-2.7726 = -60t
t = -2.7726 / -60
t ≈ 0.0462 seconds
Therefore, it will take approximately 0.0462 seconds to reduce the initial concentration of the reactant to its 1/16th value.
17.During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1 µg of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically ?
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1. Calculate the decay constant (λ):
The decay constant (λ) is related to the half-life (t₁/₂) by the following equation:
λ = ln(2) / t₁/₂
Where ln(2) ≈ 0.693
λ = 0.693 / 28.1 years ≈ 0.0247 per year
2. Calculate the remaining amount after 10 years:
We can use the following formula to calculate the remaining amount (N) after a certain time (t):
N = N₀ * e^(-λt)
Where:
- N₀ is the initial amount (1 µg)
- t is the time (10 years)
N = 1 µg * e^(-0.0247 * 10) N ≈ 0.781 µg
3. Calculate the remaining amount after 60 years:
Using the same formula, but with t = 60 years:
N = 1 µg * e^(-0.0247 * 60) N ≈ 0.228 µg
Answer:
- After 10 years, approximately 0.781 µg of ⁹⁰Sr will remain.
- After 60 years, approximately 0.228 µg of ⁹⁰Sr will remain.
18. Show that for a first order reaction the time required for 99% completion of a reaction is twice the time required to complete 90% of the reaction.
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1. Integrated Rate Law for a First-Order Reaction:
ln([A]t/[A]₀) = -kt
Where:
- [A]₀ is the initial concentration of reactant A
- k is the rate constant
- t is the time
2. Time for 99% Completion (t99):
If the reaction is 99% complete, then 1% of the reactant remains. So, [A]t = 0.01[A]₀.
ln(0.01[A]₀/[A]₀) = -kt99
ln(0.01) = -kt99
t99 = -ln(0.01) / k t99 = 4.605 / k
3. Time for 90% Completion (t90):
If the reaction is 90% complete, then 10% of the reactant remains. So, [A]t = 0.10[A]₀.
ln(0.10[A]₀/[A]₀) = -kt90
ln(0.10) = -kt90
t90 = -ln(0.10) / k t90 = 2.303 / k
4. Comparing t99 and t90:
Now, let’s compare the two times:
t99 / t90 = (4.605 / k) / (2.303 / k)
t99 / t90 = 4.605 / 2.303 t99 / t90 ≈ 2
Therefore, t99 = 2 * t90
19. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.
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1. Integrated Rate Law for a First-Order Reaction:
ln([A]t/[A]₀) = -kt
Where:
- k is the rate constant
- t is the time
2. Apply the given information:
- 30% decomposition means 70% of the reactant remains. So, [A]t = 0.70[A]₀
- The time for this decomposition is t = 40 minutes.
3. Solve for the rate constant (k):
ln(0.70[A]₀/[A]₀) = -k * 40 min
ln(0.70) = -40k
-0.3567 = -40k
k = -0.3567 / -40 k ≈ 0.00892 min⁻¹
4. Calculate the half-life (t₁/₂):
The half-life for a first-order reaction is related to the rate constant by:
t₁/₂ = 0.693 / k
t₁/₂ = 0.693 / 0.00892 min⁻¹
t₁/₂ ≈ 77.7 minutes
Therefore, the half-life of this first-order reaction is approximately 77.7 minutes.
20. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.
Calculate the rate constant.
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21. The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.
Calculate the rate of the reaction when total pressure is 0.65 atm.
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22. The rate constant for the decomposition of N2O5 at various temperatures is given below :
Draw a graph between In k and 1/7 and calculate the value of A and Ea. Predict the rate constant at 30°C and 50°C.
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23. The rate constant for the decomposition of a hydrocarbon is 2·418 x 10-5 s-1 at 546 K. If the energy of activation is 179·9 kJ mol-1, what will be the value of pre-exponential factor?
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24. Consider a certain reaction A → Products with k = 2.0 × 10-2 s-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L-1.
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25. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
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26. The decomposition of a hydrocarbon follows the equation
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27. The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.34 – 1.25 × 104 K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
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28. The decomposition of A into product has value of k as 4.5 × 103 s-1 at 10°C and energy of activation 60 kJ mol-1. At what temperature would k be 1.5 × 104 s-1 ?
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29. The time required for 10% completion of the first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 1010 s-1, calculated at 318 K and Ea.
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30. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
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