Chemical Kinetics

Chemical Kinetics is the branch of Chemistry that deals with the rate of chemical reactions, the factors affecting reaction rate, and the way a reaction proceeds with time.

In Class 12, this chapter is especially important because it combines concepts + mathematical relationships + numericals. The most important areas are rate of reaction, rate law, order, molecularity, integrated rate equations, half-life, Arrhenius equation and activation energy.

---

1. What is Chemical Kinetics?

Chemical Kinetics is the study of:

• the rate at which a chemical reaction occurs,
• the factors affecting the rate,
• the relationship between concentration and reaction rate,
• the dependence of rate on temperature,
• activation energy and the effect of catalysts.

For example:$$2H_2 + O_2 \rightarrow 2H_2O$$

Thermodynamics can tell us whether a reaction is energetically possible, but Chemical Kinetics tells us how fast the reaction occurs.

A simple example

Consider the decomposition of hydrogen peroxide:$$2H_2O_2 \rightarrow 2H_2O + O_2$$

The concentration of $H_2O_2$ decreases with time, while the concentration of $O_2$ increases.

Chemical Kinetics studies this change with time.

---

2. Rate of a Chemical Reaction

The rate of a reaction is the change in concentration of a reactant or product per unit time.

For a reactant

Since reactant concentration decreases:$$\text{Rate}=-\frac{\Delta[\text{Reactant}]}{\Delta t}$$

The negative sign is used because the change in concentration of the reactant is negative.

For a product

Since product concentration increases:$$\text{Rate}=+\frac{\Delta[\text{Product}]}{\Delta t}$$

Therefore:

Rate = Change in concentration / Change in time

The usual unit of reaction rate is:$$mol\,L^{-1}s^{-1}$$

or$$M\,s^{-1}$$

---

3. Average Rate

The average rate is the change in concentration over a particular time interval.

For a reaction:$$R \rightarrow P$$

Average rate based on reactant:$$\text{Average rate} = -\frac{[R]_2-[R]_1}{t_2-t_1}$$

Average rate based on product:$$\text{Average rate} = \frac{[P]_2-[P]_1}{t_2-t_1}$$

Example

Suppose the concentration of a reactant decreases from 0.80 M to 0.50 M in 10 s.$$\text{Rate} = -\frac{0.50-0.80}{10}$$$$=\frac{0.30}{10}$$$$=0.03\,M\,s^{-1}$$

So, the average rate is:$$\boxed{0.03\,M\,s^{-1}}$$

---

4. Instantaneous Rate

The instantaneous rate is the rate of reaction at a particular instant of time.

It is obtained by considering a very small time interval.

Mathematically:$$\text{Instantaneous rate} = -\frac{d[R]}{dt}$$

for a reactant, and$$\text{Instantaneous rate} = \frac{d[P]}{dt}$$

for a product.

Simple understanding

Average rate tells us:

“How fast did the reaction occur during this interval?”

Instantaneous rate tells us:

“How fast is the reaction occurring at this exact moment?”

---

5. Rate of Reaction and Stoichiometric Coefficients

Consider:$$aA+bB\rightarrow cC+dD$$

The rate of reaction is written as:$$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$$

The stoichiometric coefficients are important because different substances may be consumed or formed at different rates.

Example

For:$$2HI\rightarrow H_2+I_2$$$$\text{Rate} = -\frac{1}{2}\frac{d[HI]}{dt} = \frac{d[H_2]}{dt} = \frac{d[I_2]}{dt}$$

Common mistake

Do not simply write:$$-\frac{d[HI]}{dt}=\frac{d[H_2]}{dt}$$

because two moles of HI are consumed for every one mole of $H_2$ formed.

---

6. Factors Affecting the Rate of a Reaction

The rate of a chemical reaction depends on several factors.

Important factors include:

• concentration of reactants,
• temperature,
• catalyst,
• nature of reactants,
• surface area in heterogeneous reactions,
• physical state of reactants.

For Class 12, the most important factors are concentration, temperature and catalyst.

---

7. Effect of Concentration

Increasing the concentration of reactants generally increases the reaction rate.

Why?

Higher concentration means more particles are present in the same volume.

Therefore, the probability of collisions increases.

लेकिन केवल collision होना पर्याप्त नहीं है। Collision effective भी होना चाहिए।

The rate dependence is expressed through the rate law.

---

8. Rate Law

The experimentally determined relationship between reaction rate and concentration of reactants is called the rate law or rate equation.

For:$$A+B\rightarrow Products$$

a possible rate law is:$$\text{Rate}=k[A]^m[B]^n$$

where:

• $k$ = rate constant
• $[A]$ = concentration of A
• $[B]$ = concentration of B
• $m$ = order with respect to A
• $n$ = order with respect to B

Overall order:$$\boxed{\text{Order}=m+n}$$

Very important

The powers $m$ and $n$ cannot generally be obtained simply from the balanced chemical equation.

They are determined experimentally.

---

9. Rate Constant $k$

In the rate law:$$\text{Rate}=k[A]^m[B]^n$$

$k$ is called the rate constant or specific rate constant.

It is the proportionality constant between reaction rate and the concentration terms.

Important point

For a given reaction at a fixed temperature:

• $k$ has a definite value.
• $k$ does not depend on reactant concentration.
• $k$ changes with temperature.
• $k$ also changes when a catalyst changes the reaction pathway.

Meaning of $k$

If the concentrations of all reactants are unity, numerically:$$\text{Rate}=k$$

Therefore, the rate constant gives an indication of the reaction’s rate under specified conditions.

---

10. Order of a Reaction

The order of a reaction is the sum of the powers of concentration terms appearing in its experimentally determined rate law.

If:$$\text{Rate}=k[A]^2[B]$$

then:

• order with respect to A = 2
• order with respect to B = 1
• overall order = 3

$$\boxed{2+1=3}$$

Another example

$$\text{Rate}=k[A]^0[B]^1$$

Overall order:$$0+1=1$$

So it is a first-order reaction.

---

11. Order Can Be Zero

A reaction can have zero order.

For example:$$\text{Rate}=k[A]^0$$

Since:$$[A]^0=1$$

therefore:$$\text{Rate}=k$$

The rate does not depend on the concentration of A.

---

12. Units of Rate Constant

The units of $k$ depend on the order of the reaction.

Zero-order reaction

$$\text{Rate}=k$$

Therefore:$$[k]=mol\,L^{-1}s^{-1}$$

or$$M\,s^{-1}$$

First-order reaction

$$\text{Rate}=k[A]$$

Therefore:$$[k]=s^{-1}$$

General relationship

For an overall reaction order $n$:$$[k]=(concentration)^{1-n}(time)^{-1}$$

Using $M$:$$[k]=M^{1-n}s^{-1}$$

Quick table

Order	Unit of $k$
Zero	$mol\,L^{-1}s^{-1}$
First	$s^{-1}$
Second	$L\,mol^{-1}s^{-1}$

---

13. Molecularity of a Reaction

Molecularity is the number of reacting species that collide simultaneously in an elementary reaction step to bring about the reaction.

Examples:

Unimolecular reaction

One reacting species:$$A\rightarrow Products$$

Molecularity = 1

Bimolecular reaction

Two reacting species:$$A+B\rightarrow Products$$

Molecularity = 2

Termolecular reaction

Three reacting species:$$A+B+C\rightarrow Products$$

Molecularity = 3

---

14. Order vs Molecularity

This is one of the most frequently tested comparisons.

Order	Molecularity
Obtained from rate law	Defined for an elementary reaction
Determined experimentally	Based on reaction mechanism/elementary step
Can be zero	Cannot be zero
Can be fractional	Always a positive whole number
Can be greater than 3 in overall reaction	Generally 1, 2 or 3 for elementary steps
Applies to overall reaction	Applies to an elementary step

Important relationship

For an elementary reaction, order and molecularity are numerically the same.

But for a complex reaction, the overall order need not be equal to the molecularity.

Remember

Order → experimentally determined
Molecularity → elementary reaction mechanism

---

15. Elementary and Complex Reactions

Elementary reaction

A reaction occurring in a single elementary step.

For an elementary reaction:$$A+B\rightarrow Products$$

the rate law may directly correspond to the molecularity:$$\text{Rate}=k[A][B]$$

Complex reaction

A complex reaction occurs through two or more elementary steps.

The overall balanced equation does not necessarily tell us the rate law.

Therefore:

Never assume the powers in a rate law directly from the overall balanced equation unless the reaction is known to be elementary.

---

16. Integrated Rate Equation

A rate law gives the relationship between rate and concentration.

An integrated rate equation gives the relationship between concentration and time.

For Class 12, the most important integrated equations are:

• zero-order reaction
• first-order reaction

---

17. Zero-Order Reaction

For a zero-order reaction:$$A\rightarrow Products$$

Rate law:$$\text{Rate}=k[A]^0$$

Therefore:$$-\frac{d[A]}{dt}=k$$

Rearranging:$$d[A]=-k\,dt$$

On integration:$$[A]_t=[A]_0-kt$$

Therefore:$$\boxed{[A]_t=[A]_0-kt}$$

where:

• $[A]_0$ = initial concentration
• $[A]_t$ = concentration after time $t$
• $k$ = rate constant
• $t$ = time

---

18. Graph for Zero-Order Reaction

From:$$[A]_t=[A]_0-kt$$

compare with:$$y=c+mx$$

A plot of $[A]$ versus $t$ gives a straight line.

• slope = $-k$
• intercept = $[A]_0$

So:$$\boxed{\text{Slope}=-k}$$

This is an important graph-based question.

---

19. Half-Life of Zero-Order Reaction

Half-life $t_{1/2}$ is the time required for the concentration of a reactant to become half of its initial value.

At half-life:$$[A]_t=\frac{[A]_0}{2}$$

Using:$$[A]_t=[A]_0-kt$$

we get:$$\frac{[A]_0}{2} = [A]_0-kt_{1/2}$$

Therefore:$$kt_{1/2}=\frac{[A]_0}{2}$$

Hence:$$\boxed{t_{1/2}=\frac{[A]_0}{2k}}$$

Important observation

For a zero-order reaction:$$t_{1/2}\propto[A]_0$$

So half-life depends on the initial concentration.

---

20. Solved Numerical – Zero Order

A zero-order reaction has an initial concentration of 0.50 M and rate constant $0.01\,M\,s^{-1}$. Find the time required for the concentration to become 0.20 M.

Given

$$[A]_0=0.50M$$$$[A]_t=0.20M$$$$k=0.01M\,s^{-1}$$

Using:$$[A]_t=[A]_0-kt$$$$0.20=0.50-(0.01)t$$$$0.01t=0.30$$$$t=30s$$

Answer

$$\boxed{t=30s}$$

---

21. First-Order Reaction

For a first-order reaction:$$A\rightarrow Products$$

Rate law:$$\text{Rate}=k[A]$$

Therefore:$$-\frac{d[A]}{dt}=k[A]$$

After integration:$$\ln\frac{[A]_0}{[A]_t}=kt$$

Therefore:$$\boxed{k=\frac{1}{t}\ln\frac{[A]_0}{[A]_t}}$$

Using common logarithm:$$\boxed{k=\frac{2.303}{t}\log\frac{[A]_0}{[A]_t}}$$

This is one of the most important formulas in Chemical Kinetics.

---

22. Exponential Form of First-Order Equation

The first-order equation can also be written as:$$[A]_t=[A]_0e^{-kt}$$

This shows that concentration decreases exponentially with time.

---

23. Graph for First-Order Reaction

For a first-order reaction:$$\ln[A]_t=\ln[A]_0-kt$$

Therefore, a plot of:$$\ln[A]_t \text{ vs } t$$

gives a straight line.

Its:

• slope = $-k$
• intercept = $\ln[A]_0$

Using common logarithm:$$\log[A]_t = \log[A]_0-\frac{k}{2.303}t$$

Therefore, the slope of the plot of $\log[A]$ against $t$ is:$$\boxed{-\frac{k}{2.303}}$$

---

24. Half-Life of a First-Order Reaction

For a first-order reaction:$$\boxed{t_{1/2}=\frac{0.693}{k}}$$

This is an extremely important formula.

Most important feature

The half-life of a first-order reaction is independent of the initial concentration.

That means whether the initial concentration is 1 M, 0.5 M or 0.1 M, the half-life remains the same at a fixed temperature for the same reaction.

---

25. Why is First-Order Half-Life Independent of Initial Concentration?

From:$$t_{1/2}=\frac{0.693}{k}$$

there is no $[A]_0$ term.

Therefore:$$\boxed{t_{1/2}\text{ is independent of }[A]_0}$$

Repeated half-lives

For a first-order reaction:

• after 1 half-life → 50% remains
• after 2 half-lives → 25% remains
• after 3 half-lives → 12.5% remains
• after 4 half-lives → 6.25% remains

So after $n$ half-lives:$$\text{fraction remaining}= \left(\frac12\right)^n$$

---

26. Solved Numerical – First Order

A first-order reaction has a rate constant:$$k=0.693\,min^{-1}$$

Find its half-life.

Using:$$t_{1/2}=\frac{0.693}{k}$$$$t_{1/2} = \frac{0.693}{0.693}$$$$\boxed{t_{1/2}=1\,min}$$

---

27. Solved Numerical – First-Order Concentration

A first-order reaction has $k=0.23\,min^{-1}$. How much time is required for the concentration to decrease from 0.80 M to 0.20 M?

Using:$$k=\frac{2.303}{t}\log\frac{[A]_0}{[A]_t}$$

Therefore:$$t= \frac{2.303}{k} \log\frac{0.80}{0.20}$$$$t= \frac{2.303}{0.23}\log4$$

Since:$$\log4=0.602$$$$t\approx \frac{2.303\times0.602}{0.23}$$$$t\approx6.03\,min$$

So:$$\boxed{t\approx6.0\,min}$$

---

28. Comparison of Zero-Order and First-Order Reactions

Property	Zero Order	First Order
Rate law	$Rate=k$	$Rate=k[A]$
Integrated equation	$[A]_t=[A]_0-kt$	$\ln([A]_0/[A]_t)=kt$
Unit of $k$	$M\,s^{-1}$	$s^{-1}$
Half-life	$[A]_0/2k$	$0.693/k$
Half-life depends on initial concentration?	Yes	No
Straight-line graph	$[A]$ vs $t$	$\ln[A]$ vs $t$
Slope	$-k$	$-k$

---

29. Determination of Order from Experimental Data

The order of a reaction is generally determined experimentally.

Suppose:$$\text{Rate}=k[A]^m[B]^n$$

By changing the concentration of one reactant while keeping the other constant, we can determine the corresponding order.

Example

Suppose experimental data show:

Experiment	[A]	[B]	Rate
1	0.1	0.1	0.02
2	0.2	0.1	0.04
3	0.1	0.2	0.08

Compare Experiment 1 and 2.

[B] remains constant.

[A] doubles:$$0.1\rightarrow0.2$$

Rate also doubles:$$0.02\rightarrow0.04$$

Therefore:$$\text{Rate}\propto[A]^1$$

Order with respect to A = 1.

Now compare Experiment 1 and 3.

[A] remains constant.

[B] doubles:$$0.1\rightarrow0.2$$

Rate becomes four times:$$0.02\rightarrow0.08$$

Therefore:$$\text{Rate}\propto[B]^2$$

Order with respect to B = 2.

Thus:$$\boxed{\text{Overall order}=1+2=3}$$

---

30. Method of Initial Rates

The method of initial rates is commonly used to determine the order of a reaction experimentally.

Suppose:$$Rate=k[A]^m$$

For two experiments:$$Rate_1=k[A_1]^m$$$$Rate_2=k[A_2]^m$$

Dividing:$$\frac{Rate_2}{Rate_1} = \left(\frac{[A_2]}{[A_1]}\right)^m$$

This equation can be used to calculate $m$.

---

31. Pseudo First-Order Reaction

Sometimes a reaction actually involves two or more reactants but behaves like a first-order reaction because one reactant is present in very large excess.

Consider:$$CH_3COOC_2H_5+H_2O \rightarrow CH_3COOH+C_2H_5OH$$

The actual rate law may involve both reactants:$$Rate=k[CH_3COOC_2H_5][H_2O]$$

If water is present in very large excess, its concentration remains almost constant.

Therefore:$$k[H_2O]=k’$$

and:$$Rate=k'[CH_3COOC_2H_5]$$

Thus, the reaction behaves as a first-order reaction.

This is called a pseudo first-order reaction.

Simple Hindi explanation

एक reactant बहुत अधिक मात्रा में हो तो उसकी concentration practically constant मान सकते हैं। तब उसका concentration rate law के constant में शामिल हो जाता है।

---

32. Effect of Temperature on Reaction Rate

In general:

Increasing temperature increases the rate of a chemical reaction.

A common observation is that many reactions become significantly faster when temperature is increased.

Why?

At higher temperature:

• molecules possess greater kinetic energy,
• collisions become more energetic,
• a greater fraction of molecules can cross the activation-energy barrier.

Therefore, the number of effective collisions increases.

---

33. Activation Energy

Activation energy EaE_a is the minimum additional energy required by reactant molecules to reach the activated state and undergo reaction.

In simple words:

Reactant molecules need to cross an energy barrier before products can form.

Hindi explanation

Reactants को products में बदलने के लिए एक minimum energy barrier पार करना पड़ता है। इस energy barrier को activation energy से समझा जाता है।

---

34. Activated Complex

The unstable high-energy arrangement formed during a chemical reaction is called the activated complex or transition state.

It exists only for a very short time.

A simplified energy profile is:$$\text{Reactants} \rightarrow \text{Activated Complex} \rightarrow \text{Products}$$

The energy difference between reactants and the activated complex corresponds to activation energy.

---

35. Effect of Catalyst

A catalyst increases the rate of a reaction by providing an alternative reaction pathway with lower activation energy.

A catalyst:

• lowers activation energy,
• increases the fraction of molecules capable of reacting,
• changes the reaction pathway,
• is not consumed overall in the reaction.

Important

A catalyst does not change the overall thermodynamic equilibrium constant simply by being present.

For a reversible reaction, a catalyst speeds up both forward and reverse processes, helping equilibrium to be reached faster.

---

36. Arrhenius Equation

The temperature dependence of the rate constant is represented by the Arrhenius equation:$$\boxed{k=Ae^{-E_a/RT}}$$

where:

• $k$ = rate constant
• $A$ = Arrhenius factor or frequency factor
• $E_a$ = activation energy
• $R$ = gas constant
• $T$ = absolute temperature in Kelvin

$$R=8.314\,J\,mol^{-1}K^{-1}$$

Meaning of $A$

$A$ is related to the frequency of collisions and, in collision-theory interpretation, the appropriate orientation of reacting molecules.

---

37. Logarithmic Form of Arrhenius Equation

Starting with:$$k=Ae^{-E_a/RT}$$

Taking natural logarithm:$$\ln k=\ln A-\frac{E_a}{RT}$$

Using common logarithm:$$\boxed{\log k=\log A-\frac{E_a}{2.303RT}}$$

This form is very useful in numerical problems.

---

38. Two-Temperature Arrhenius Equation

For two temperatures $T_1$ and $T_2$:$$\boxed{ \ln\frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1}-\frac{1}{T_2} \right) }$$

Using common logarithm:$$\boxed{ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1}-\frac{1}{T_2} \right) }$$

This is one of the most important formulas for Class 12 numerical questions.

---

39. Important Sign Convention in Arrhenius Numericals

If:$$T_2>T_1$$

then:$$\frac{1}{T_1}>\frac{1}{T_2}$$

Therefore:$$\ln\frac{k_2}{k_1}>0$$

which means:$$k_2>k_1$$

So increasing temperature generally increases $k$.

Common mistake

Always use Kelvin, not Celsius, in Arrhenius equations.$$T(K)=T(^\circ C)+273.15$$

---

40. Solved Numerical – Arrhenius Equation

The rate constant of a reaction is $1.0\times10^{-3}\,s^{-1}$ at 300 K and $2.0\times10^{-3}\,s^{-1}$ at 310 K. Calculate the activation energy.

Using:$$\ln\frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{1}{T_1}-\frac{1}{T_2} \right)$$

Given:$$k_1=1.0\times10^{-3}$$$$k_2=2.0\times10^{-3}$$$$T_1=300K,\quad T_2=310K$$

Therefore:$$\ln2 = \frac{E_a}{8.314} \left( \frac{1}{300}-\frac{1}{310} \right)$$

On solving:$$E_a\approx53.6\,kJ\,mol^{-1}$$

Hence:$$\boxed{E_a\approx53.6\,kJ\,mol^{-1}}$$

Exam tip

Keep units consistent. If $R=8.314\,J\,mol^{-1}K^{-1}$, calculate $E_a$ in joules per mole first and then convert to kJ mol⁻¹.

---

41. Arrhenius Plot

From:$$\ln k=\ln A-\frac{E_a}{RT}$$

compare with:$$y=c+mx$$

For a plot of:$$\ln k \text{ vs } \frac{1}{T}$$

the slope is:$$\boxed{-\frac{E_a}{R}}$$

and intercept is:$$\boxed{\ln A}$$

Therefore:$$E_a=-\text{slope}\times R$$

---

42. Collision Theory

Collision theory explains reaction rates in terms of collisions between reacting particles.

According to collision theory:

Molecules must collide with sufficient energy and suitable orientation for an effective reaction to occur.

Not every collision produces products.

Conditions for an effective collision

An effective collision requires:

1. sufficient energy,
2. proper orientation of molecules.

Why do most collisions not produce products?

Because many collisions either:

• do not have enough energy to cross $E_a$, or
• occur with unsuitable orientation.

---

43. Energy Distribution and Temperature

At a higher temperature, the distribution of molecular energies changes such that a greater fraction of molecules has energy equal to or greater than the activation energy.

Therefore:$$T\uparrow \Rightarrow \text{fraction of effective collisions}\uparrow \Rightarrow k\uparrow \Rightarrow \text{rate}\uparrow$$

This explains why even a moderate increase in temperature can significantly increase reaction rate.

---

44. Why Does a Catalyst Increase the Rate?

Without catalyst:$$\text{Reactants} \xrightarrow{\text{high }E_a} \text{Products}$$

With catalyst:$$\text{Reactants} \xrightarrow{\text{lower }E_a} \text{Products}$$

Since the activation-energy barrier is lower, more molecules can successfully cross it.

Remember

A catalyst does not provide energy to the reactants. It provides a different pathway requiring less activation energy.

---

45. Rate Constant and Temperature

From:$$k=Ae^{-E_a/RT}$$

when $T$ increases:$$-\frac{E_a}{RT}$$

becomes less negative.

Therefore $k$ increases.

Since rate depends on $k$:$$\boxed{T\uparrow\Rightarrow k\uparrow\Rightarrow Rate\uparrow}$$

---

46. Graph-Based Questions You Should Know

Zero-order

Plot:$$[A]\text{ vs }t$$

Straight line:

• slope = $-k$
• intercept = $[A]_0$

First-order

Plot:$$\ln[A]\text{ vs }t$$

Straight line:

• slope = $-k$
• intercept = $\ln[A]_0$

Arrhenius plot

Plot:$$\ln k\text{ vs }\frac{1}{T}$$

Straight line:

• slope = $-E_a/R$
• intercept = $\ln A$

---

47. Important Formula Sheet

Average rate

$$\boxed{\text{Rate}=-\frac{\Delta[R]}{\Delta t}}$$

or$$\boxed{\text{Rate}=\frac{\Delta[P]}{\Delta t}}$$

General rate expression

For:$$aA+bB\rightarrow cC+dD$$$$\boxed{ Rate= -\frac1a\frac{d[A]}{dt} = -\frac1b\frac{d[B]}{dt} = \frac1c\frac{d[C]}{dt} = \frac1d\frac{d[D]}{dt} }$$

Rate law

$$\boxed{Rate=k[A]^m[B]^n}$$

Overall order

$$\boxed{n=m+n}$$

Zero-order integrated equation

$$\boxed{[A]_t=[A]_0-kt}$$

Zero-order half-life

$$\boxed{t_{1/2}=\frac{[A]_0}{2k}}$$

First-order integrated equation

$$\boxed{\ln\frac{[A]_0}{[A]_t}=kt}$$

or$$\boxed{k=\frac{2.303}{t}\log\frac{[A]_0}{[A]_t}}$$

First-order half-life

$$\boxed{t_{1/2}=\frac{0.693}{k}}$$

Arrhenius equation

$$\boxed{k=Ae^{-E_a/RT}}$$

Logarithmic Arrhenius equation

$$\boxed{\log k=\log A-\frac{E_a}{2.303RT}}$$

Two-temperature form

$$\boxed{ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac1{T_1}-\frac1{T_2} \right) }$$

---

48. Important Units

Quantity	Common unit
Rate	$mol\,L^{-1}s^{-1}$
Concentration	$mol\,L^{-1}$ or M
Time	s, min, h
First-order $k$	$s^{-1}$
Zero-order $k$	$mol\,L^{-1}s^{-1}$
Activation energy	$J\,mol^{-1}$ or $kJ\,mol^{-1}$
Temperature in Arrhenius equation	K
Gas constant $R$	$8.314\,J\,mol^{-1}K^{-1}$

---

49. Common Student Mistakes

Mistake 1: Using the balanced equation to determine order

The balanced equation does not generally give the rate law.

Correct: Order is determined experimentally unless the reaction is specifically an elementary reaction.

Mistake 2: Forgetting the negative sign for reactants

Since reactant concentration decreases:$$Rate=-\frac{\Delta[R]}{\Delta t}$$

Mistake 3: Confusing order with molecularity

Order can be zero or fractional, while molecularity of an elementary step is a positive whole number.

Mistake 4: Using Celsius in Arrhenius equation

Always use:$$T(K)=T(^\circ C)+273.15$$

Mistake 5: Using the wrong half-life formula

Zero order:$$t_{1/2}=\frac{[A]_0}{2k}$$

First order:$$t_{1/2}=\frac{0.693}{k}$$

Mistake 6: Forgetting stoichiometric coefficients

For:$$2A\rightarrow B$$

the reaction rate is:$$Rate=-\frac12\frac{d[A]}{dt}$$

not simply $-d[A]/dt$.

Mistake 7: Mixing natural and common logarithms

Remember:$$\ln x=2.303\log x$$

Mistake 8: Thinking a catalyst increases equilibrium yield

A catalyst helps equilibrium to be reached faster. It does not change the equilibrium position merely by accelerating the forward and reverse reactions.

---

50. Important Conceptual Differences

Rate vs Rate Constant

Rate changes with concentration.

Rate constant kk is characteristic of a particular reaction at a given temperature and depends on the reaction conditions.

---

Order vs Molecularity

Order comes from the experimentally determined rate law.

Molecularity refers to the number of species involved in an elementary reaction step.

---

Average Rate vs Instantaneous Rate

Average rate: rate over a time interval.

Instantaneous rate: rate at a particular instant.

---

Activation Energy vs Energy of Reaction

Activation energy is the energy barrier required to reach the transition state.

It is not the same as the overall enthalpy change of the reaction.

---

51. Board Exam Important Points

For board examination preparation, make sure you can write and explain:

• definition of Chemical Kinetics,
• average and instantaneous rate,
• rate expression for a general reaction,
• rate law,
• rate constant,
• order of reaction,
• molecularity,
• difference between order and molecularity,
• units of rate constant,
• zero-order integrated rate equation,
• zero-order half-life,
• first-order integrated rate equation,
• first-order half-life,
• graphs for zero- and first-order reactions,
• pseudo first-order reaction,
• Arrhenius equation,
• activation energy,
• effect of temperature,
• effect of catalyst,
• collision theory,
• two-temperature Arrhenius equation,
• numerical problems based on $k$, concentration, half-life and $E_a$.

Derivations worth practising

1. Zero-order integrated rate equation
2. Zero-order half-life
3. First-order integrated rate equation
4. First-order half-life
5. Arrhenius two-temperature equation

---

52. NEET/JEE Important Points

For competitive examinations, pay particular attention to:

• identifying order from experimental data,
• units of rate constant,
• half-life relationships,
• concentration remaining after several half-lives,
• graph-based questions,
• Arrhenius calculations,
• activation-energy calculations,
• temperature dependence of $k$,
• pseudo first-order reactions,
• distinguishing order from molecularity,
• stoichiometric coefficients in rate expressions,
• logarithm-based numerical calculations.

Quick competitive-exam shortcut

For a first-order reaction:$$t_{1/2}=\frac{0.693}{k}$$

Therefore:$$k=\frac{0.693}{t_{1/2}}$$

This relationship can often solve a question within a few seconds.

---

53. Practice Questions

Try to solve these yourself before checking your notes.

Conceptual Questions

1. What is Chemical Kinetics?
2. Define average rate and instantaneous rate.
3. Why is a negative sign used while expressing the rate of disappearance of a reactant?
4. What is rate law?
5. Why cannot the order of a complex reaction generally be determined from its balanced equation?
6. What is molecularity?
7. Why can molecularity not be zero?
8. What is a pseudo first-order reaction?
9. Why does a catalyst increase reaction rate?
10. What is activation energy?

Short-Answer Questions

11. Differentiate between order and molecularity.
12. Write the integrated rate equation for a zero-order reaction.
13. Write the half-life equation for a first-order reaction.
14. What is the unit of rate constant for a first-order reaction?
15. What does the slope of a plot of $\ln[A]$ versus time represent?
16. What is the significance of the Arrhenius factor $A$?
17. Why is temperature expressed in Kelvin in the Arrhenius equation?

Numerical Practice

18. A zero-order reaction has $k=0.02\,M\,s^{-1}$ and initial concentration 0.50 M. Calculate its half-life.
19. A first-order reaction has $k=0.0693\,min^{-1}$. Calculate its half-life.
20. For a first-order reaction, concentration decreases from 0.80 M to 0.20 M. If $k=0.10\,min^{-1}$, calculate the time required.

Reaction-Based/Conceptual

21. Explain why the rate of a reaction generally increases when concentration increases.
22. Explain why all molecular collisions do not result in a chemical reaction.
23. Explain the effect of a catalyst on activation energy.
24. For:

$$2A+B\rightarrow C$$

write the correct relationship between the rates of disappearance of A and B and formation of C.

25. Explain why the half-life of a first-order reaction does not depend on initial concentration.

---

54. 20 Important MCQs on Chemical Kinetics

1. The branch of Chemistry that studies the rate of chemical reactions is:

(A) Thermodynamics
(B) Chemical Kinetics
(C) Electrochemistry
(D) Surface Chemistry

Answer: (B) Chemical Kinetics

Chemical Kinetics deals with reaction rates and factors affecting them.

---

2. The rate of disappearance of a reactant is represented by:

(A) $\frac{d[R]}{dt}$
(B) $-\frac{d[R]}{dt}$
(C) $\frac{d[P]}{dt}$
(D) $-\frac{d[P]}{dt}$

Answer: (B) −d[R]dt-\frac{d[R]}{dt}

Reactant concentration decreases with time, so the negative sign makes the rate positive.

---

3. For the reaction

$$2A\rightarrow B$$

the rate of reaction is:

(A) $-d[A]/dt$
(B) $d[B]/dt$
(C) $-\frac12d[A]/dt$
(D) $2d[B]/dt$

Answer: (C) −12d[A]/dt-\frac12d[A]/dt

The stoichiometric coefficient of A is 2.

---

4. If the rate law is

$$Rate=k[A]^2[B]$$

the overall order is:

(A) 1
(B) 2
(C) 3
(D) 4

Answer: (C) 3$$2+1=3$$

---

5. Which statement about order is correct?

(A) It is always equal to molecularity
(B) It is always a positive integer
(C) It is determined experimentally
(D) It can never be zero

Answer: (C) It is determined experimentally

Order is obtained from the experimentally determined rate law.

---

6. The unit of the rate constant for a first-order reaction is:

(A) $mol\,L^{-1}s^{-1}$
(B) $L\,mol^{-1}s^{-1}$
(C) $s^{-1}$
(D) $mol^{-1}L\,s^{-1}$

Answer: (C) s−1s^{-1}

---

7. The integrated rate equation for a zero-order reaction is:

(A) $[A]_t=[A]_0-kt$
(B) $\ln[A]_t=\ln[A]_0-kt$
(C) $1/[A]_t=1/[A]_0+kt$
(D) $k=0.693/t$

Answer: (A) [A]t=[A]0−kt[A]_t=[A]_0-kt

---

8. The half-life of a first-order reaction is:

(A) $[A]_0/2k$
(B) $0.693/k$
(C) $2k/[A]_0$
(D) $k/0.693$

Answer: (B) 0.693/k0.693/k

---

9. The half-life of a first-order reaction is:

(A) dependent on initial concentration
(B) inversely proportional to initial concentration
(C) independent of initial concentration
(D) directly proportional to initial concentration

Answer: (C) independent of initial concentration$$t_{1/2}=\frac{0.693}{k}$$

contains no $[A]_0$.

---

10. For a zero-order reaction, a plot of $[A]$ versus time is:

(A) a straight line
(B) a parabola
(C) a circle
(D) always horizontal

Answer: (A) a straight line

The slope is $-k$.

---

11. For a first-order reaction, the slope of the plot of $\ln[A]$ versus time is:

(A) $k$
(B) $1/k$
(C) $-k$
(D) $0.693/k$

Answer: (C) −k-k

---

12. According to Arrhenius equation:

$$k=Ae^{-E_a/RT}$$

an increase in temperature generally:

(A) decreases $k$
(B) increases $k$
(C) makes $k$ zero
(D) has no effect on $k$

Answer: (B) increases kk

At higher temperature, a larger fraction of molecules can overcome the activation-energy barrier.

---

13. A catalyst increases the rate of a reaction mainly by:

(A) increasing the enthalpy of reactants
(B) increasing equilibrium constant
(C) lowering activation energy through an alternative pathway
(D) increasing the concentration of products

Answer: (C) lowering activation energy through an alternative pathway

---

14. In the Arrhenius equation, temperature should be expressed in:

(A) Celsius
(B) Fahrenheit
(C) Kelvin
(D) any unit

Answer: (C) Kelvin

Absolute temperature is required.

---

15. The slope of a plot of $\ln k$ versus $1/T$ is:

(A) $E_a/R$
(B) $-E_a/R$
(C) $R/E_a$
(D) $-R/E_a$

Answer: (B) −Ea/R-E_a/R

From:$$\ln k=\ln A-\frac{E_a}{R}\frac1T$$

---

16. A reaction has a half-life of 10 minutes and follows first-order kinetics. Its rate constant is approximately:

(A) $0.0693\,min^{-1}$
(B) $0.693\,min^{-1}$
(C) $6.93\,min^{-1}$
(D) $10\,min^{-1}$

Answer: (A) 0.0693 min−10.0693\,min^{-1}$$k=\frac{0.693}{10} =0.0693\,min^{-1}$$

---

17. Which of the following can be zero?

(A) Molecularity
(B) Order of reaction
(C) Number of reacting species in an elementary step
(D) Both molecularity and reacting species

Answer: (B) Order of reaction

A zero-order reaction is possible, but molecularity cannot be zero.

---

18. A pseudo first-order reaction is one that:

(A) actually has only one reactant
(B) has zero activation energy
(C) behaves as first order because one reactant is present in large excess
(D) has molecularity equal to zero

Answer: (C)

The concentration of the excess reactant remains nearly constant.

---

19. Which condition is essential for an effective collision?

(A) Very low energy only
(B) Suitable orientation and sufficient energy
(C) High concentration only
(D) High pressure only

Answer: (B) Suitable orientation and sufficient energy

Both factors are important according to collision theory.

---

20. For a zero-order reaction, the half-life is:

(A) independent of initial concentration
(B) proportional to initial concentration
(C) inversely proportional to initial concentration only
(D) always equal to $0.693/k$

Answer: (B) proportional to initial concentration$$t_{1/2}=\frac{[A]_0}{2k}$$

---

55. 10 Frequently Asked Questions

1. What is the difference between rate and rate constant?

Rate depends on the concentration of reactants and changes during the reaction. Rate constant $k$ is the proportionality constant in the rate law and has a fixed value for a particular reaction at a specified temperature.

2. Can the order of a reaction be fractional?

Yes. Order is determined experimentally and can be zero, an integer or a fractional value.

3. Can molecularity be fractional?

No. Molecularity represents the number of reacting species involved in an elementary step, so it is a positive whole number.

4. Why is the order of a reaction not obtained from its balanced equation?

For a complex reaction, the overall equation may represent several elementary steps. The experimentally observed rate depends on the reaction mechanism, so the order must generally be determined experimentally.

5. Why is the half-life of a first-order reaction constant?

Because:$$t_{1/2}=\frac{0.693}{k}$$

and the equation contains no initial concentration term.

6. Why does a catalyst increase reaction rate?

A catalyst provides an alternative pathway with lower activation energy, allowing more molecules to undergo effective reactions.

7. Why does increasing temperature increase reaction rate?

Higher temperature increases molecular kinetic energy and increases the fraction of molecules having energy equal to or greater than activation energy.

8. What is a pseudo first-order reaction?

It is a reaction whose actual rate law involves more than one reactant but behaves as first order because one reactant is present in large excess and its concentration remains practically constant.

9. What is activation energy?

Activation energy is the energy barrier that reactant molecules must overcome to reach the activated state and proceed toward products.

10. Which Chemical Kinetics formulas should be memorised for Class 12?

The most important formulas are:$$[A]_t=[A]_0-kt$$$$t_{1/2}=\frac{[A]_0}{2k}$$$$\ln\frac{[A]_0}{[A]_t}=kt$$$$t_{1/2}=\frac{0.693}{k}$$$$k=Ae^{-E_a/RT}$$

and$$\ln\frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac1{T_1}-\frac1{T_2} \right)$$

---

56. Quick Revision – Chemical Kinetics

Rate

$$Rate=-\frac{\Delta[R]}{\Delta t}$$

or$$Rate=\frac{\Delta[P]}{\Delta t}$$

Rate Law

$$Rate=k[A]^m[B]^n$$

Order

$$\text{Overall order}=m+n$$

Zero Order

$$Rate=k$$$$[A]_t=[A]_0-kt$$$$t_{1/2}=\frac{[A]_0}{2k}$$

Half-life depends on initial concentration.

First Order

$$Rate=k[A]$$$$\ln\frac{[A]_0}{[A]_t}=kt$$$$t_{1/2}=\frac{0.693}{k}$$

Half-life is independent of initial concentration.

Molecularity

• Defined for elementary reactions.
• Positive whole number.
• Usually 1, 2 or 3.

Arrhenius Equation

$$k=Ae^{-E_a/RT}$$

Activation Energy

Higher $E_a$ generally means a smaller rate constant at the same temperature, all else being comparable.

Catalyst

Provides an alternative pathway with lower activation energy.

Collision Theory

Effective collision requires:

• sufficient energy,
• proper orientation.

---

57. Last-Minute Exam Checklist

Before the examination, make sure you can solve or explain all of these without looking at your notes:

• Average rate
• Instantaneous rate
• Rate expression using stoichiometric coefficients
• Rate law
• Order of reaction
• Molecularity
• Order vs molecularity
• Units of rate constant
• Zero-order integrated equation
• Zero-order half-life
• First-order integrated equation
• First-order half-life
• Graphs for zero and first order
• Experimental determination of order
• Pseudo first-order reaction
• Activation energy
• Arrhenius equation
• Two-temperature Arrhenius equation
• Catalyst and activation energy
• Collision theory
• Numerical problems involving $k$, $t$, concentration and $E_a$

One-line memory aid

Rate tells “how fast”, order tells “how concentration affects rate”, kk tells the proportionality, EaE_a tells the energy barrier, and Arrhenius explains how temperature changes kk.

---

You May Also Like :

1. Chapter 1 – Solutions
2. Chapter 2 – Electrochemistry
3. Chapter 3 – Chemical Kinetics