Chemical Kinetics
Welcome to HSLC Guru! This English-medium study guide for ASSEB Class 12 Chemistry Chapter 4 — Chemical Kinetics — has been built to help you understand the rate at which chemical reactions occur, the factors that influence reaction speed, and the mathematical tools used to describe reaction progress. Chemical kinetics is one of the most application-rich chapters of the syllabus, with numerical problems on rate law, order of reaction, integrated rate equations and the Arrhenius equation appearing every year. The notes below cover the entire ASSEB prescribed syllabus with worked examples, MCQs, fill in the blanks, true/false, a glossary and a formula table.
Chapter Summary
Rate of a chemical reaction is the change in concentration of a reactant or product per unit time. For a reaction R to P, the average rate is the change in concentration over a measurable time interval, while the instantaneous rate is the rate at a particular instant of time, obtained by taking the limit as the time interval approaches zero. The average rate is expressed as the negative of change in concentration of reactant divided by change in time, or as the positive change in concentration of product divided by change in time. The instantaneous rate equals the slope of the concentration-time curve at that instant. Rates are usually expressed in mol L^-1 s^-1 for solutions and in atm s^-1 for gases.
The rate of reaction depends on several factors: concentration of reactants (higher concentration usually increases rate), temperature (a 10 K rise in temperature roughly doubles or triples the rate), nature of reactants, surface area of solid reactants, and presence of a catalyst (which lowers activation energy without being consumed). The dependence of rate on concentration is expressed quantitatively by the rate law: Rate = k[A]^x[B]^y, where k is the rate constant and the exponents x and y are determined experimentally. The sum (x + y) is the overall order of reaction, while molecularity is the number of reacting species in an elementary step. Order can be zero, fractional or negative, but molecularity is always a positive whole number.
Integrated rate equations express concentration as a function of time. For a zero order reaction, [R] = [R]_0 – kt, and the half-life t(1/2) = [R]_0 / 2k depends on initial concentration. For a first order reaction, k = (2.303/t) log([A]_0/[A]) and the half-life t(1/2) = 0.693/k is independent of initial concentration. Pseudo-first order reactions are higher order reactions that appear first order because one reactant is in large excess (e.g., hydrolysis of ester in dilute aqueous solution, inversion of cane sugar). The unit of k for a zero order reaction is mol L^-1 s^-1, for a first order reaction is s^-1, and for a second order reaction is L mol^-1 s^-1.
Collision theory states that for a reaction to occur, reactant molecules must collide with sufficient kinetic energy (greater than threshold energy) and proper orientation. Only effective collisions lead to product formation. The Arrhenius equation k = A e^(-Ea/RT) relates the rate constant to temperature, where A is the frequency or pre-exponential factor and Ea is the activation energy. Taking logarithm gives log k = log A – Ea/(2.303RT), which plots as a straight line of slope -Ea/(2.303R) versus 1/T. The two-temperature form log(k2/k1) = (Ea/2.303R)((T2 – T1)/T1 T2) lets us calculate Ea from rate constants at two temperatures, or estimate the rate constant at a new temperature. A catalyst speeds up a reaction by providing an alternative pathway with a lower activation energy.
Key Formulae (KaTeX)
$\text{Rate} = k[A]^x[B]^y$
$[R] = [R]_0 – kt$
$t_{1/2} = \frac{[R]_0}{2k}$
$k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}$
$t_{1/2} = \frac{0.693}{k}$
$k = A e^{-E_a/RT}$
$\log k = \log A – \frac{E_a}{2.303RT}$
$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 – T_1}{T_1 T_2}\right)$
$\text{Avg rate} = -\frac{\Delta[R]}{\Delta t}$
1 Mark Questions and Answers
Q1. Define rate of a chemical reaction.
Answer: The change in concentration of a reactant or a product per unit time is called the rate of reaction. Its SI unit is mol L^-1 s^-1.
Q2. What is meant by instantaneous rate of reaction?
Answer: The rate of reaction at a particular instant of time, obtained by making the time interval approach zero, is called the instantaneous rate. It is equal to the slope of the concentration-time graph at that instant.
Q3. Write the SI unit of rate constant of a first order reaction.
Answer: The unit of rate constant of a first order reaction is s^-1 (per second).
Q4. Define order of reaction.
Answer: The sum of the powers of the concentration terms of reactants in the experimentally determined rate law expression is called the order of reaction. It can be zero, integral or fractional.
Q5. Define molecularity of a reaction.
Answer: The total number of reacting species (atoms, ions or molecules) that collide simultaneously in an elementary reaction step is called molecularity. It is always a positive whole number.
Q6. What is a pseudo-first order reaction? Give one example.
Answer: A reaction whose actual order is more than one but appears first order experimentally because one of the reactants is taken in large excess is called a pseudo-first order reaction. Example: acid-catalysed hydrolysis of ester in dilute aqueous solution.
Q7. What is activation energy?
Answer: The minimum extra energy that the reactant molecules must possess over and above their average energy in order to react and form products is called activation energy (Ea).
Q8. State the effect of a catalyst on activation energy.
Answer: A catalyst provides an alternative reaction pathway with a lower activation energy, thereby increasing the rate of reaction. It does not change the equilibrium constant or the enthalpy change of the reaction.
Q9. What is the half-life of a first order reaction?
Answer: The time required for the concentration of reactant to fall to half of its initial value in a first order reaction is given by t(1/2) = 0.693/k. It is independent of initial concentration.
Q10. Write the Arrhenius equation in exponential form.
Answer: The Arrhenius equation is k = A e^(-Ea/RT), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant and T is the absolute temperature.
2-3 Marks Questions and Answers
Q11. Distinguish between average rate and instantaneous rate of a reaction.
Answer: Average rate is the change in concentration of reactant or product over a definite measurable time interval, expressed as -Delta[R]/Delta t or +Delta[P]/Delta t. Instantaneous rate is the rate at a particular instant of time, obtained as the time interval approaches zero, and is equal to the slope of the concentration-time curve at that point. Average rate gives an overall picture; instantaneous rate gives the actual rate at any moment.
Q12. Differentiate between order and molecularity of a reaction.
Answer: Order is the sum of powers of concentration terms in the experimentally determined rate law and may be zero, fractional, integral or negative. Molecularity is the number of species colliding in an elementary step and is always a small positive whole number. Order is an experimental quantity applicable to overall reactions; molecularity is a theoretical concept applied only to elementary reactions.
Q13. List four factors that affect the rate of a chemical reaction.
Answer: (i) Concentration of reactants — higher concentration generally increases rate. (ii) Temperature — for most reactions, rate roughly doubles for every 10 K rise. (iii) Catalyst — lowers activation energy and speeds up the reaction. (iv) Surface area — finely divided solids react faster than lumps. Nature of reactants and exposure to light also influence rate.
Q14. Derive the integrated rate equation for a zero order reaction.
Answer: For a zero order reaction R to P, rate = -d[R]/dt = k[R]^0 = k. Rearranging, d[R] = -k dt. Integrating between [R]_0 at t = 0 and [R] at time t gives [R] – [R]_0 = -kt, i.e., [R] = [R]_0 – kt. A plot of [R] versus t is a straight line with slope -k and intercept [R]_0.
Q15. Show that the half-life of a first order reaction is independent of initial concentration.
Answer: For a first order reaction, k = (2.303/t) log([A]_0/[A]). At t = t(1/2), [A] = [A]_0/2, so k = (2.303/t(1/2)) log 2 = 0.693/t(1/2). Thus t(1/2) = 0.693/k. Since this expression contains only k (a constant at given T), it does not depend on initial concentration.
Q16. The rate constant of a first order reaction is 2.303 x 10^-3 s^-1. Calculate the half-life.
Answer: t(1/2) = 0.693/k = 0.693/(2.303 x 10^-3) = 300.9 s. Hence the half-life is approximately 301 seconds.
Q17. Briefly explain the postulates of collision theory.
Answer: Collision theory assumes that (i) reactant molecules must collide for a reaction to occur, (ii) only those collisions that involve energy equal to or greater than the threshold energy result in product formation, and (iii) molecules must be properly oriented at the moment of collision. The fraction of effective collisions is given by the Arrhenius factor e^(-Ea/RT) multiplied by a steric factor P.
5-7 Marks Questions and Answers
Q18. Derive the integrated rate equation for a first order reaction and define its half-life.
Answer: Consider a first order reaction R to P with initial concentration [R]_0. The differential rate law is -d[R]/dt = k[R]. Rearranging gives d[R]/[R] = -k dt. Integrating between limits [R]_0 at t = 0 and [R] at time t, we get ln([R]/[R]_0) = -kt, or ln([R]_0/[R]) = kt. Converting to common logarithm, k = (2.303/t) log([R]_0/[R]). Setting [R] = [R]_0/2 at t = t(1/2) gives k = (2.303/t(1/2)) log 2 = 0.693/t(1/2), so t(1/2) = 0.693/k. The half-life of a first order reaction is independent of initial concentration. A plot of log [R] versus t is a straight line of slope -k/2.303.
Q19. 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?
Answer: Using k = (2.303/t) log([A]_0/[A]) with [A]_0 = 5 g and [A] = 3 g, t = (2.303/k) log(5/3) = (2.303/(1.15 x 10^-3)) log(1.6667) = 2002.6 x 0.2218 = 444.2 s. Hence approximately 444 seconds (about 7 minutes 24 seconds) are required.
Q20. The rate constants of a reaction at 500 K and 700 K are 0.02 s^-1 and 0.07 s^-1 respectively. Calculate the activation energy of the reaction. (R = 8.314 J K^-1 mol^-1)
Answer: Using log(k2/k1) = (Ea/2.303R)((T2 – T1)/T1 T2), log(0.07/0.02) = log 3.5 = 0.5441. (T2 – T1)/(T1 T2) = (700 – 500)/(500 x 700) = 200/350000 = 5.714 x 10^-4. So 0.5441 = (Ea/(2.303 x 8.314)) x 5.714 x 10^-4, giving Ea = (0.5441 x 2.303 x 8.314)/(5.714 x 10^-4) = 10.418/5.714 x 10^-4 = 18234 J/mol = 18.23 kJ/mol.
Q21. The decomposition of N2O5 is a first order reaction with rate constant 5.0 x 10^-4 s^-1 at 318 K. How much time is needed for 80% decomposition?
Answer: If 80% has decomposed, [A]/[A]_0 = 0.20, so [A]_0/[A] = 5. Using k = (2.303/t) log([A]_0/[A]), t = (2.303/k) log 5 = (2.303/(5.0 x 10^-4)) x 0.6990 = 4606 x 0.6990 = 3219.6 s, i.e., approximately 3220 seconds (about 53.7 minutes).
Q22. Explain the Arrhenius equation. How can activation energy be determined graphically from this equation?
Answer: The Arrhenius equation is k = A e^(-Ea/RT), where k is the rate constant, A is the pre-exponential (frequency) factor, Ea is the activation energy, R is the gas constant and T is absolute temperature. Taking the logarithm gives log k = log A – Ea/(2.303RT). A plot of log k versus 1/T is a straight line of slope -Ea/(2.303R) and intercept log A. By measuring the slope of this Arrhenius plot, Ea can be calculated as Ea = -2.303R x slope. The exponential term e^(-Ea/RT) represents the fraction of molecules with energy equal to or greater than Ea. As T increases, this fraction increases, so k and the rate increase. A small increase in T can produce a large increase in rate, especially for reactions with high Ea.
Multiple Choice Questions (MCQs)
Q1. The unit of rate constant for a zero order reaction is:
(a) s^-1 (b) mol L^-1 s^-1 (c) L mol^-1 s^-1 (d) L^2 mol^-2 s^-1
Answer: (b) mol L^-1 s^-1.
Q2. The half-life of a first order reaction with k = 6.93 x 10^-2 s^-1 is:
(a) 10 s (b) 100 s (c) 1 s (d) 50 s
Answer: (a) 10 s, since t(1/2) = 0.693/k = 0.693/(6.93 x 10^-2) = 10 s.
Q3. Molecularity of an elementary reaction is always:
(a) zero (b) fractional (c) negative (d) a positive whole number
Answer: (d) a positive whole number.
Q4. The rate of a chemical reaction generally:
(a) increases with temperature (b) decreases with temperature (c) is independent of temperature (d) becomes zero at high temperature
Answer: (a) increases with temperature.
Q5. Inversion of cane sugar in dilute acid is an example of:
(a) zero order (b) first order (c) pseudo-first order (d) second order
Answer: (c) pseudo-first order.
Q6. The Arrhenius factor A is also known as:
(a) frequency factor (b) activation factor (c) collision factor (d) rate factor
Answer: (a) frequency factor.
Q7. A catalyst increases the rate of a reaction by:
(a) increasing activation energy (b) decreasing activation energy (c) increasing temperature (d) shifting equilibrium
Answer: (b) decreasing activation energy.
Q8. For a zero order reaction, a plot of [R] versus time is:
(a) curved (b) straight line with positive slope (c) straight line with negative slope (d) parabola
Answer: (c) straight line with negative slope.
Q9. The slope of a plot of log k against 1/T equals:
(a) -Ea/R (b) -Ea/2.303R (c) Ea/2.303R (d) Ea/R
Answer: (b) -Ea/2.303R.
Q10. Order of reaction can be:
(a) only zero (b) only positive integer (c) zero, fractional, integral or negative (d) only fractional
Answer: (c) zero, fractional, integral or negative.
Fill in the Blanks
Q1. The rate of reaction is the change in __________ per unit time.
Answer: concentration.
Q2. The unit of rate constant of a first order reaction is __________.
Answer: s^-1.
Q3. A reaction that appears first order due to large excess of one reactant is called __________ reaction.
Answer: pseudo-first order.
Q4. The half-life of a zero order reaction depends on __________ concentration.
Answer: initial.
Q5. In the Arrhenius equation, A stands for __________ factor.
Answer: pre-exponential (frequency).
True or False
Q1. The order of a reaction can be fractional.
Answer: True.
Q2. Molecularity of a reaction can be zero.
Answer: False. Molecularity is always a positive whole number.
Q3. A catalyst changes the equilibrium constant of a reaction.
Answer: False. A catalyst does not change the equilibrium constant; it only speeds up attainment of equilibrium.
Q4. The half-life of a first order reaction is independent of initial concentration.
Answer: True.
Q5. A plot of log k against 1/T gives a straight line.
Answer: True, with slope -Ea/(2.303R).
Glossary
| Term | Meaning |
|---|---|
| Rate of reaction | Change in concentration of reactant or product per unit time. |
| Average rate | Change in concentration over a measurable time interval. |
| Instantaneous rate | Rate at a particular instant of time; slope of the concentration-time curve. |
| Rate law | Experimentally determined expression Rate = k[A]^x[B]^y relating rate to concentrations. |
| Rate constant (k) | Proportionality constant in the rate law; depends on temperature and catalyst, not on concentration. |
| Order of reaction | Sum of powers of concentrations in the rate law; can be zero, fractional or integral. |
| Molecularity | Number of species colliding in an elementary step; always a positive whole number. |
| Pseudo-first order | Reaction that appears first order because one reactant is present in large excess. |
| Half-life (t1/2) | Time taken for the concentration of reactant to fall to half its initial value. |
| Activation energy (Ea) | Minimum energy required by reactant molecules above their average energy to react. |
| Threshold energy | Minimum total energy that colliding molecules must possess for an effective collision. |
| Arrhenius equation | k = A e^(-Ea/RT); relates rate constant to temperature and activation energy. |
| Frequency factor (A) | Pre-exponential factor in the Arrhenius equation; related to collision frequency and orientation. |
| Catalyst | Substance that increases reaction rate by lowering activation energy without being consumed. |
| Effective collision | Collision with energy greater than threshold and proper orientation, leading to product. |
Formula Table
| Formula | Meaning / Use |
|---|---|
| Rate = k[A]^x[B]^y | General rate law; x + y is overall order. |
| Avg rate = -Delta[R]/Delta t | Average rate of reaction in terms of reactant. |
| Instantaneous rate = -d[R]/dt | Rate at a specific instant. |
| [R] = [R]_0 – kt | Integrated rate equation for zero order reaction. |
| t(1/2) = [R]_0 / 2k | Half-life of a zero order reaction. |
| k = (2.303/t) log([A]_0/[A]) | Rate constant for a first order reaction. |
| t(1/2) = 0.693/k | Half-life of a first order reaction. |
| k = A e^(-Ea/RT) | Arrhenius equation in exponential form. |
| log k = log A – Ea/(2.303RT) | Logarithmic form of the Arrhenius equation. |
| log(k2/k1) = (Ea/2.303R)((T2 – T1)/T1 T2) | Two-temperature form for finding Ea. |
| Slope of log k vs 1/T = -Ea/2.303R | Graphical determination of activation energy. |
| Unit of k (zero order) | mol L^-1 s^-1. |
| Unit of k (first order) | s^-1. |
| Unit of k (second order) | L mol^-1 s^-1. |
This completes the HSLC Guru English-medium notes for Chapter 4 — Chemical Kinetics. Practise the numerical problems on first order reactions and the Arrhenius equation thoroughly, since these carry high weightage in the ASSEB Class 12 Chemistry examination.