Blog Posts
Chemical Kinetics
By
bash
cat > /mnt/user-data/outputs/chemical_kinetics_class12_chemistry.html << 'HTMLEOF'
Chemical Kinetics Class 12 Notes | Chapter 4 Chemistry NCERT
Class 12ChemistryChapter 4NCERTCBSE
Chemical Kinetics - Class 12 Chemistry Chapter 4
📌 Quick Overview: Chapter 4 deals with Chemical Kinetics — the study of the speed (rate) of chemical reactions and the factors that affect it. You will learn how to express rate of reaction, write rate laws, determine order of reaction, derive and use integrated rate equations for zero, first and second order reactions, calculate half-life, and understand the effect of temperature using the Arrhenius equation. This chapter has important numericals and is consistently tested in CBSE board exams.
Topics Covered:
- > Rate of Reaction
- > Factors Affecting Rate
- > Rate Law and Rate Constant
- > Order and Molecularity
- > Zero Order Reaction
- > First Order Reaction
- > Second Order Reaction
- > Half Life
- > Pseudo First Order
- > Temperature and Rate
- > Arrhenius Equation
- > Activation Energy
- > Collision Theory
1. Rate of Reaction
Rate of Reaction: The change in concentration of a reactant or product per unit time is called the rate of reaction.
For reaction: aA + bB → cC + dD
Rate = -(1/a)(d[A]/dt) = -(1/b)(d[B]/dt) = +(1/c)(d[C]/dt) = +(1/d)(d[D]/dt)
Unit of rate: mol L^(-1) s^(-1) or mol L^(-1) min^(-1)
For reaction: aA + bB → cC + dD
Rate = -(1/a)(d[A]/dt) = -(1/b)(d[B]/dt) = +(1/c)(d[C]/dt) = +(1/d)(d[D]/dt)
Unit of rate: mol L^(-1) s^(-1) or mol L^(-1) min^(-1)
⭐ Average Rate vs Instantaneous Rate:
> Average Rate = change in concentration / time interval = Delta[X] / Delta t
> Instantaneous Rate = rate at a particular instant = d[X]/dt (slope of concentration-time graph at that point)
> Rate decreases as reaction proceeds (reactant concentration decreases)
> Average Rate = change in concentration / time interval = Delta[X] / Delta t
> Instantaneous Rate = rate at a particular instant = d[X]/dt (slope of concentration-time graph at that point)
> Rate decreases as reaction proceeds (reactant concentration decreases)
2. Factors Affecting Rate of Reaction
| Factor | Effect on Rate | Reason |
|---|---|---|
| Concentration of reactants | Rate increases with concentration | More collisions per unit time |
| Temperature | Rate increases (~doubles per 10°C rise) | More energetic collisions, more molecules exceed Ea |
| Catalyst | Increases rate (positive) or decreases (negative) | Lowers activation energy |
| Surface area | Increases rate (for solid reactants) | More surface available for reaction |
| Light/Radiation | Increases rate of photochemical reactions | Provides energy to activate molecules |
| Pressure | Increases rate for gaseous reactions | Equivalent to increasing concentration |
3. Rate Law and Rate Constant ⭐
Rate Law (Rate Expression): The mathematical expression that relates the rate of a reaction to the concentration of reactants is called the rate law.
For aA + bB → products:
Rate = k [A]^x [B]^y
where k = rate constant, x and y are orders with respect to A and B (determined experimentally, NOT from stoichiometry).
For aA + bB → products:
Rate = k [A]^x [B]^y
where k = rate constant, x and y are orders with respect to A and B (determined experimentally, NOT from stoichiometry).
Rate Constant (k): It is the proportionality constant in the rate law. When concentration of each reactant = 1 mol/L, Rate = k.
Unit of k depends on order of reaction:
Unit of k = (mol L^(-1))^(1-n) s^(-1)
where n = overall order of reaction.
Unit of k depends on order of reaction:
Unit of k = (mol L^(-1))^(1-n) s^(-1)
where n = overall order of reaction.
UNITS OF RATE CONSTANT:
Zero order: mol L^(-1) s^(-1)
First order: s^(-1)
Second order: L mol^(-1) s^(-1)
nth order: (mol L^(-1))^(1-n) s^(-1)
4. Order and Molecularity ⭐
| Property | Order of Reaction | Molecularity |
|---|---|---|
| Definition | Sum of powers of concentration terms in rate law | Number of molecules/atoms/ions that collide simultaneously to give products |
| Determined by | Experimentally | Theoretically (from mechanism) |
| Value | Can be 0, 1, 2, 3, fraction, negative | Always a positive integer (1, 2, 3) |
| Applies to | Overall reaction | Elementary step only |
| Can be fraction? | Yes | No |
⭐ Key Points:
> Order of reaction is determined EXPERIMENTALLY — never from the balanced equation (unless it is an elementary reaction).
> Molecularity is always a whole number and is never zero or negative.
> Unimolecular: 1 molecule (e.g. radioactive decay)
> Bimolecular: 2 molecules (most common)
> Termolecular: 3 molecules (rare, very few known)
> Reactions of higher molecularity are not known because simultaneous collision of more than 3 molecules is extremely unlikely.
> Order of reaction is determined EXPERIMENTALLY — never from the balanced equation (unless it is an elementary reaction).
> Molecularity is always a whole number and is never zero or negative.
> Unimolecular: 1 molecule (e.g. radioactive decay)
> Bimolecular: 2 molecules (most common)
> Termolecular: 3 molecules (rare, very few known)
> Reactions of higher molecularity are not known because simultaneous collision of more than 3 molecules is extremely unlikely.
5. Integrated Rate Equations ⭐ (Most Important)
5.1 Zero Order Reaction
Rate does not depend on concentration of reactant.
Rate = k[A]^0 = k (constant)
Rate = k[A]^0 = k (constant)
ZERO ORDER:
Rate = k
Integrated: [A] = [A]0 - kt
k = ([A]0 - [A]) / t
Half life: t(1/2) = [A]0 / 2k
Unit of k: mol L^(-1) s^(-1)
Graph: [A] vs t is a straight line with slope = -k
Examples: Decomposition of NH3 on Pt surface,
Photochemical reactions (H2 + Cl2)
5.2 First Order Reaction ⭐ (Most Important)
Rate depends on first power of concentration of one reactant.
Rate = k[A]
Rate = k[A]
FIRST ORDER:
Rate = k[A]
Integrated: ln[A] = ln[A]0 - kt
OR: [A] = [A]0 . e^(-kt)
OR: k = (2.303/t) log([A]0/[A])
Half life: t(1/2) = 0.693/k = ln2/k
(INDEPENDENT of initial concentration)
Unit of k: s^(-1) or min^(-1) or h^(-1)
Graphs:
[A] vs t : exponential decay curve
log[A] vs t : straight line, slope = -k/2.303
ln[A] vs t : straight line, slope = -k
Examples: Radioactive decay, decomposition of N2O5,
decomposition of H2O2, inversion of sugar
Example 1: A first order reaction has rate constant 1.15 x 10^(-3) s^(-1). How long will 5g of this reactant take to reduce to 3g?
k = (2.303/t) log([A]0/[A])
1.15 x 10^(-3) = (2.303/t) log(5/3)
1.15 x 10^(-3) = (2.303/t) x 0.2219
t = (2.303 x 0.2219) / (1.15 x 10^(-3))
t = 0.5112 / (1.15 x 10^(-3))
t = 444.5 s
k = (2.303/t) log([A]0/[A])
1.15 x 10^(-3) = (2.303/t) log(5/3)
1.15 x 10^(-3) = (2.303/t) x 0.2219
t = (2.303 x 0.2219) / (1.15 x 10^(-3))
t = 0.5112 / (1.15 x 10^(-3))
t = 444.5 s
Example 2: The half life of a first order reaction is 60 minutes. Find the rate constant and the time for 90% completion.
t(1/2) = 0.693/k → k = 0.693/60 = 0.01155 min^(-1)
For 90% completion: [A] = 10% of [A]0
t = (2.303/k) log([A]0/[A]) = (2.303/0.01155) log(100/10)
= (2.303/0.01155) x 1
= 199.4 minutes
Note: t(90%) = 2 x t(1/2) is only approximate. Exact: t(90%) ≈ 3.32 x t(1/2)
t(1/2) = 0.693/k → k = 0.693/60 = 0.01155 min^(-1)
For 90% completion: [A] = 10% of [A]0
t = (2.303/k) log([A]0/[A]) = (2.303/0.01155) log(100/10)
= (2.303/0.01155) x 1
= 199.4 minutes
Note: t(90%) = 2 x t(1/2) is only approximate. Exact: t(90%) ≈ 3.32 x t(1/2)
5.3 Second Order Reaction
SECOND ORDER:
Rate = k[A]^2
Integrated: 1/[A] = 1/[A]0 + kt
k = (1/t)(1/[A] - 1/[A]0)
Half life: t(1/2) = 1/(k[A]0)
(DEPENDS on initial concentration)
Unit of k: L mol^(-1) s^(-1)
Graph: 1/[A] vs t is a straight line with slope = k
Examples: 2HI → H2 + I2 (in gas phase)
NO2 + NO2 → N2O4
6. Comparison of Zero, First and Second Order ⭐
| Property | Zero Order | First Order | Second Order |
|---|---|---|---|
| Rate law | Rate = k | Rate = k[A] | Rate = k[A]^2 |
| Integrated law | [A] = [A]0 - kt | ln[A] = ln[A]0 - kt | 1/[A] = 1/[A]0 + kt |
| Half life | [A]0/2k | 0.693/k | 1/k[A]0 |
| Unit of k | mol L^-1 s^-1 | s^-1 | L mol^-1 s^-1 |
| Linear graph | [A] vs t | log[A] vs t | 1/[A] vs t |
| t(1/2) depends on [A]0? | Yes | No | Yes |
7. Pseudo First Order Reactions
Pseudo First Order Reaction: A reaction that is actually second or higher order but appears to be first order because the concentration of one reactant is in large excess and remains essentially constant throughout the reaction.
Example: Hydrolysis of ethyl acetate in excess water:
CH3COOC2H5 + H2O → CH3COOH + C2H5OH
Rate = k[CH3COOC2H5][H2O] ≈ k'[CH3COOC2H5]
where k' = k[H2O] = pseudo first order rate constant
Other examples: Inversion of sucrose in excess water, hydrolysis of methyl acetate
Example: Hydrolysis of ethyl acetate in excess water:
CH3COOC2H5 + H2O → CH3COOH + C2H5OH
Rate = k[CH3COOC2H5][H2O] ≈ k'[CH3COOC2H5]
where k' = k[H2O] = pseudo first order rate constant
Other examples: Inversion of sucrose in excess water, hydrolysis of methyl acetate
8. Effect of Temperature on Rate ⭐
Temperature Coefficient: The ratio of rate constants at two temperatures differing by 10°C.
Temperature coefficient = k(T+10) / k(T) ≈ 2 to 3
This means the rate of reaction approximately doubles for every 10°C rise in temperature.
Temperature coefficient = k(T+10) / k(T) ≈ 2 to 3
This means the rate of reaction approximately doubles for every 10°C rise in temperature.
9. Arrhenius Equation ⭐ (Very Important)
ARRHENIUS EQUATION:
k = A . e^(-Ea/RT)
Taking log:
ln k = ln A - Ea/RT
log k = log A - Ea/(2.303RT)
where:
k = rate constant
A = frequency factor (Arrhenius factor / pre-exponential factor)
Ea = activation energy (J/mol)
R = 8.314 J/mol/K
T = temperature in Kelvin
TWO TEMPERATURE FORM (most used in numericals):
log(k2/k1) = (Ea/2.303R) x (T2-T1)/(T1 x T2)
ln(k2/k1) = (Ea/R) x (1/T1 - 1/T2)
Activation Energy (Ea): The minimum extra energy that reactant molecules must possess to overcome the energy barrier and form products.
Frequency Factor (A): It represents the total number of collisions per second (both effective and ineffective). It accounts for the frequency of collisions and their correct orientation.
e^(-Ea/RT): The fraction of molecules having energy equal to or greater than Ea.
Frequency Factor (A): It represents the total number of collisions per second (both effective and ineffective). It accounts for the frequency of collisions and their correct orientation.
e^(-Ea/RT): The fraction of molecules having energy equal to or greater than Ea.
Example 1: The rate constant of a reaction at 500K is 1.6 x 10^(-3) s^(-1) and at 700K is 2.3 x 10^(-2) s^(-1). Find activation energy.
log(k2/k1) = (Ea/2.303R) x (T2-T1)/(T1 x T2)
log(2.3 x 10^(-2) / 1.6 x 10^(-3)) = (Ea/(2.303 x 8.314)) x (200/(500 x 700))
log(14.375) = (Ea/19.147) x (200/350000)
1.1578 = (Ea/19.147) x 5.714 x 10^(-4)
Ea = (1.1578 x 19.147) / (5.714 x 10^(-4))
Ea = 22.18 / (5.714 x 10^(-4))
Ea = 38,813 J/mol ≈ 38.8 kJ/mol
log(k2/k1) = (Ea/2.303R) x (T2-T1)/(T1 x T2)
log(2.3 x 10^(-2) / 1.6 x 10^(-3)) = (Ea/(2.303 x 8.314)) x (200/(500 x 700))
log(14.375) = (Ea/19.147) x (200/350000)
1.1578 = (Ea/19.147) x 5.714 x 10^(-4)
Ea = (1.1578 x 19.147) / (5.714 x 10^(-4))
Ea = 22.18 / (5.714 x 10^(-4))
Ea = 38,813 J/mol ≈ 38.8 kJ/mol
Example 2: The activation energy of a reaction is 75 kJ/mol. Find how many times the rate constant increases when temperature is raised from 25°C to 35°C.
T1 = 298K, T2 = 308K, Ea = 75000 J/mol
log(k2/k1) = (75000/(2.303 x 8.314)) x (308-298)/(298 x 308)
= (75000/19.147) x (10/91784)
= 3917.3 x 1.089 x 10^(-4)
= 0.4266
k2/k1 = 10^(0.4266) = 2.67
The rate constant increases about 2.67 times.
T1 = 298K, T2 = 308K, Ea = 75000 J/mol
log(k2/k1) = (75000/(2.303 x 8.314)) x (308-298)/(298 x 308)
= (75000/19.147) x (10/91784)
= 3917.3 x 1.089 x 10^(-4)
= 0.4266
k2/k1 = 10^(0.4266) = 2.67
The rate constant increases about 2.67 times.
10. Important Graphs in Chemical Kinetics ⭐
| Graph | Order | What it shows |
|---|---|---|
| [A] vs t → straight line | Zero order | Slope = -k |
| log[A] vs t → straight line | First order | Slope = -k/2.303 |
| 1/[A] vs t → straight line | Second order | Slope = k |
| log k vs 1/T → straight line | Any order | Slope = -Ea/2.303R (Arrhenius plot) |
| Rate vs [A] → straight line through origin | First order | Slope = k |
| Rate vs [A] → horizontal line | Zero order | Rate = k (constant) |
| Rate vs [A] → parabola | Second order | Rate = k[A]^2 |
11. Collision Theory of Chemical Reactions
Collision Theory: A chemical reaction occurs when reactant molecules collide with:
(i) Sufficient energy (at least equal to activation energy Ea)
(ii) Proper orientation
Rate = Z x f x p
where Z = collision frequency, f = fraction of effective collisions (f = e^(-Ea/RT)), p = probability (steric) factor (accounts for orientation)
(i) Sufficient energy (at least equal to activation energy Ea)
(ii) Proper orientation
Rate = Z x f x p
where Z = collision frequency, f = fraction of effective collisions (f = e^(-Ea/RT)), p = probability (steric) factor (accounts for orientation)
⭐ Limitations of Collision Theory:
> It considers molecules as hard spheres — ignores the internal structure
> The steric factor p is difficult to calculate theoretically
> Does not explain the effect of bond breaking/forming on rate
Transition State Theory (Activated Complex Theory):
Reactants → Activated Complex (transition state) → Products
The activated complex is an unstable, high-energy intermediate at the peak of the energy profile.
> It considers molecules as hard spheres — ignores the internal structure
> The steric factor p is difficult to calculate theoretically
> Does not explain the effect of bond breaking/forming on rate
Transition State Theory (Activated Complex Theory):
Reactants → Activated Complex (transition state) → Products
The activated complex is an unstable, high-energy intermediate at the peak of the energy profile.
12. Important Board Exam Questions
Q1. What is the difference between order and molecularity of a reaction? Give one example of each.
Order is the sum of powers of concentration terms in the experimentally determined rate law. It can be zero, fractional, or a whole number. Example: H2 + Br2 → 2HBr, Rate = k[H2][Br2]^(1/2), order = 3/2.
Molecularity is the number of molecules/ions that simultaneously collide in an elementary reaction. It is always a positive integer. Example: NO2 → NO + O (unimolecular, molecularity = 1), 2HI → H2 + I2 (bimolecular, molecularity = 2).
Molecularity is the number of molecules/ions that simultaneously collide in an elementary reaction. It is always a positive integer. Example: NO2 → NO + O (unimolecular, molecularity = 1), 2HI → H2 + I2 (bimolecular, molecularity = 2).
Q2. For a first order reaction, show that the half life is independent of initial concentration.
For first order: k = (2.303/t) log([A]0/[A])
At t = t(1/2): [A] = [A]0/2
k = (2.303/t(1/2)) log([A]0/([A]0/2))
k = (2.303/t(1/2)) log(2)
k = (2.303/t(1/2)) x 0.3010
k = 0.693/t(1/2)
t(1/2) = 0.693/k
Since t(1/2) = 0.693/k depends only on k (which is constant at a given temperature) and NOT on [A]0, the half life of a first order reaction is independent of initial concentration. ✓
At t = t(1/2): [A] = [A]0/2
k = (2.303/t(1/2)) log([A]0/([A]0/2))
k = (2.303/t(1/2)) log(2)
k = (2.303/t(1/2)) x 0.3010
k = 0.693/t(1/2)
t(1/2) = 0.693/k
Since t(1/2) = 0.693/k depends only on k (which is constant at a given temperature) and NOT on [A]0, the half life of a first order reaction is independent of initial concentration. ✓
Q3. The decomposition of NH3 on platinum surface is zero order. What are the rates of production of N2 and H2 if k = 2.5 x 10^(-4) mol L^(-1) s^(-1)?
2NH3 → N2 + 3H2
Rate of decomposition of NH3 = k = 2.5 x 10^(-4) mol L^(-1) s^(-1)
Rate = -(1/2)d[NH3]/dt = (1/1)d[N2]/dt = (1/3)d[H2]/dt
Rate of production of N2 = k/2... wait:
-d[NH3]/dt = 2k = 2 x 2.5 x 10^(-4) = 5 x 10^(-4)
Rate of production of N2 = k = 2.5 x 10^(-4) mol L^(-1) s^(-1)
Rate of production of H2 = 3k/2... Let us use the rate expression:
Rate = 2.5 x 10^(-4) mol L^(-1) s^(-1)
d[N2]/dt = 2.5 x 10^(-4) mol L^(-1) s^(-1)
d[H2]/dt = 3 x 2.5 x 10^(-4) = 7.5 x 10^(-4) mol L^(-1) s^(-1)
Rate of decomposition of NH3 = k = 2.5 x 10^(-4) mol L^(-1) s^(-1)
Rate = -(1/2)d[NH3]/dt = (1/1)d[N2]/dt = (1/3)d[H2]/dt
Rate of production of N2 = k/2... wait:
-d[NH3]/dt = 2k = 2 x 2.5 x 10^(-4) = 5 x 10^(-4)
Rate of production of N2 = k = 2.5 x 10^(-4) mol L^(-1) s^(-1)
Rate of production of H2 = 3k/2... Let us use the rate expression:
Rate = 2.5 x 10^(-4) mol L^(-1) s^(-1)
d[N2]/dt = 2.5 x 10^(-4) mol L^(-1) s^(-1)
d[H2]/dt = 3 x 2.5 x 10^(-4) = 7.5 x 10^(-4) mol L^(-1) s^(-1)
Q4. A first order reaction is 40% complete in 50 minutes. Calculate the rate constant and the time for 90% completion.
40% complete means [A] = 60% of [A]0 = 0.6[A]0
k = (2.303/t) log([A]0/[A])
k = (2.303/50) log(1/0.6)
k = (2.303/50) log(1.667)
k = (2.303/50) x 0.2219
k = 0.01022 min^(-1)
For 90% completion: [A] = 0.1[A]0
t = (2.303/0.01022) log(1/0.1)
t = (2.303/0.01022) x 1
t = 225.3 minutes
k = (2.303/t) log([A]0/[A])
k = (2.303/50) log(1/0.6)
k = (2.303/50) log(1.667)
k = (2.303/50) x 0.2219
k = 0.01022 min^(-1)
For 90% completion: [A] = 0.1[A]0
t = (2.303/0.01022) log(1/0.1)
t = (2.303/0.01022) x 1
t = 225.3 minutes
Q5. The rate of a reaction doubles when temperature is raised from 300K to 310K. Find the activation energy. (R = 8.314 J/mol/K)
k2/k1 = 2, T1 = 300K, T2 = 310K
log(k2/k1) = (Ea/2.303R) x (T2-T1)/(T1 x T2)
log 2 = (Ea/(2.303 x 8.314)) x (10/(300 x 310))
0.3010 = (Ea/19.147) x (10/93000)
0.3010 = (Ea/19.147) x 1.075 x 10^(-4)
Ea = (0.3010 x 19.147) / (1.075 x 10^(-4))
Ea = 5.763 / (1.075 x 10^(-4))
Ea = 53,610 J/mol ≈ 53.6 kJ/mol
log(k2/k1) = (Ea/2.303R) x (T2-T1)/(T1 x T2)
log 2 = (Ea/(2.303 x 8.314)) x (10/(300 x 310))
0.3010 = (Ea/19.147) x (10/93000)
0.3010 = (Ea/19.147) x 1.075 x 10^(-4)
Ea = (0.3010 x 19.147) / (1.075 x 10^(-4))
Ea = 5.763 / (1.075 x 10^(-4))
Ea = 53,610 J/mol ≈ 53.6 kJ/mol
Q6. Explain how the rate constant of a reaction varies with temperature using the Arrhenius equation. What is the effect of a catalyst?
According to Arrhenius equation: k = A e^(-Ea/RT)
As temperature (T) increases, Ea/RT decreases, so e^(-Ea/RT) increases, and hence k increases. A 10°C rise approximately doubles the rate.
A catalyst provides an alternate reaction pathway with lower activation energy (Ea'). Since Ea' is less than Ea, more molecules can overcome the energy barrier. This increases k and hence the rate. The catalyst does not change Delta H or the equilibrium constant — it only speeds up attainment of equilibrium.
As temperature (T) increases, Ea/RT decreases, so e^(-Ea/RT) increases, and hence k increases. A 10°C rise approximately doubles the rate.
A catalyst provides an alternate reaction pathway with lower activation energy (Ea'). Since Ea' is less than Ea, more molecules can overcome the energy barrier. This increases k and hence the rate. The catalyst does not change Delta H or the equilibrium constant — it only speeds up attainment of equilibrium.
13. Key Formulas at a Glance
RATE:
Rate = -(1/a)d[A]/dt = +(1/c)d[C]/dt
RATE LAW: Rate = k[A]^x[B]^y
UNITS OF k:
Zero order: mol L^-1 s^-1
First order: s^-1
Second order: L mol^-1 s^-1
INTEGRATED EQUATIONS:
Zero: [A] = [A]0 - kt
First: k = (2.303/t) log([A]0/[A])
Second: 1/[A] - 1/[A]0 = kt
HALF LIFE:
Zero: t(1/2) = [A]0/2k
First: t(1/2) = 0.693/k (independent of [A]0)
Second: t(1/2) = 1/(k[A]0)
ARRHENIUS:
k = A.e^(-Ea/RT)
log(k2/k1) = (Ea/2.303R) x (T2-T1)/(T1.T2)
TIME FOR n% COMPLETION (first order):
t = (2.303/k) log(100/(100-n))
t(50%) = 0.693/k
t(75%) = 2 x t(1/2)
t(87.5%) = 3 x t(1/2)
t(99%) = 6.645 x t(1/2)
14. MCQ Practice (1 Mark)
1. For a first order reaction, the unit of rate constant is:
(a) mol L^(-1) s^(-1) (b) s^(-1) (c) L mol^(-1) s^(-1) (d) mol^2 L^(-2) s^(-1)
Answer: (b) s^(-1)
2. The half life of a first order reaction is 60 s. The rate constant is:
(a) 0.693 s^(-1) (b) 1.155 x 10^(-2) s^(-1) (c) 60 s^(-1) (d) 0.0693 s^(-1)
Answer: (b) k = 0.693/60 = 1.155 x 10^(-2) s^(-1)
3. Which plot gives a straight line for a first order reaction?
(a) [A] vs t (b) 1/[A] vs t (c) log[A] vs t (d) [A]^2 vs t
Answer: (c) log[A] vs t (slope = -k/2.303)
4. The molecularity of a reaction can NEVER be:
(a) 1 (b) 3 (c) 0 (d) 2
Answer: (c) 0 — molecularity is always a positive integer
5. In the Arrhenius equation, the term e^(-Ea/RT) represents:
(a) Frequency factor (b) Fraction of effective collisions (c) Rate constant (d) Activation energy
Answer: (b) Fraction of molecules with energy greater than or equal to Ea
6. If the concentration of a reactant is doubled and the rate becomes 4 times, the order with respect to that reactant is:
(a) 0 (b) 1 (c) 2 (d) 4
Answer: (c) 2 — Rate = k[A]^n; 4 = 2^n → n = 2
15. Exam Tips
- Never determine order from stoichiometric coefficients — order is always determined experimentally from rate law.
- For first order: memorise k = (2.303/t) log([A]0/[A]) and t(1/2) = 0.693/k — these two formulas solve 80% of kinetics numericals.
- When temperature is given in °C, always convert to Kelvin (add 273) before using Arrhenius equation.
- For Arrhenius equation numericals: use log(k2/k1) form when two temperatures are given — it avoids calculating ln.
- Half life of first order is INDEPENDENT of initial concentration — this is a very commonly asked theory point.
- Pseudo first order: water is in large excess so its concentration is constant — only the other reactant's concentration appears in effective rate law.
- For graphical questions: check which variable gives a straight line — [A] vs t (zero), log[A] vs t (first), 1/[A] vs t (second).
- Catalyst lowers Ea but does NOT change Delta H, equilibrium constant, or the thermodynamic feasibility of the reaction.
✅ Summary: Chemical Kinetics is a must-master chapter with both theory and numericals. The most frequently asked topics are: first order integrated rate equation, half-life derivation and calculation, Arrhenius equation numericals (finding Ea), order determination from rate data, and the distinction between order and molecularity. Practice at least 5 Arrhenius equation problems and 5 first-order problems before the board exam.
Study Got - Making Studies Simple | studygot.in
HTMLEOF echo "Done" Output
Post a Comment