Follow Us
माध्यम चुनें / Select Medium:
Eng (English) Hindi (हिन्दी)
ICSE • कक्षा XII • Chemistry • अध्याय 3
अनुमानित समय: 90 Mins
प्रगति: अध्ययनरत

रासायनिक बलगतिकी (Chemical Kinetics)

In Class 12 Chemistry, "Chemical Kinetics" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

⏱️ Have You Ever Wondered?

Why do diamond gemstones take billions of years to turn into pencil graphite even though thermodynamics says they want to, while fireworks explode in a microsecond? Reaction rate, activation energy, and catalysts govern chemical speed.

यह अध्याय क्यों महत्वपूर्ण है

In Class 12 Chemistry, "Chemical Kinetics" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

  • Rates of change from Mathematics.
  • Thermodynamic equilibrium from Class 11.
  • Concentration units.

इस अध्याय के लक्ष्य

  • Define Average Rate and Instantaneous Rate of a chemical reaction: $r = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt}$.
  • Distinguish between Order of a Reaction (experimentally determined) and Molecularity (theoretical collision count).
  • Derive integrated rate equations and half-life expressions for Zero Order ($k = \frac{[A]_0 - [A]}{t}, t_{1/2} = \frac{[A]_0}{2k}$) and First Order ($k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}, t_{1/2} = \frac{0.693}{k}$).
  • State and apply the Arrhenius Equation: $k = A e^{-E_a/RT}$ and $\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)$.
  • Explain the role of a Catalyst in lowering Activation Energy ($E_a$) without altering $\Delta H$.

अध्याय रूपरेखा एवं प्रगति

1 1. Order vs. Molecularity & Rate La...
2 2. Zero & First Order Integrated Ra...
3 3. Arrhenius Equation & Activation...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Order vs. Molecularity & Rate Law

The Rate Law expresses rate in terms of molar concentrations: $\text{Rate} = k [A]^x [B]^y$ where Overall Order $= x + y$.
• Order: Can be zero, fractional, or negative; strictly experimental.
• Molecularity: Number of reacting species colliding simultaneously in an elementary step; must be an integer ($1, 2, 3$), never zero or fractional!

2. Zero & First Order Integrated Rates

  • Zero Order: Rate independent of concentration ($[A] = [A]_0 - kt$). Half-life: $\mathbf{t_{1/2} = \frac{[A]_0}{2k}}$ (proportional to initial concentration).
  • First Order: $$\mathbf{k = \frac{2.303}{t}\log_{10}\left(\frac{[A]_0}{[A]}\right)} \quad \text{and} \quad \mathbf{t_{1/2} = \frac{0.693}{k}}$$ Key Property: Half-life of a first-order reaction is completely independent of initial concentration!

3. Arrhenius Equation & Activation Energy

Molecules must collide with minimum energy (Threshold Energy) to react. The energy barrier is the Activation Energy ($E_a$): $$\mathbf{k = A e^{-E_a/RT}} \implies \mathbf{\log_{10}\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)}$$ A catalyst provides an alternative pathway with a lower $E_a$, accelerating forward and backward rates equally!

महत्वपूर्ण रासायनिक सूत्र, अभिक्रियाएँ एवं समीकरण

Answer architecture
$$Concept \to Evidence \to Application \to Evaluation$$
Use the chapter principle, show the working or evidence, and state the conclusion.
Revision loop
$$Learn \to Practise \to Check \to Correct \to Reattempt$$
Keep an error log and revisit questions that exposed a misconception.

अवधारणात्मक हल उदाहरण एवं अनुप्रयोग (Solved Examples)

उदाहरण 1
A first-order reaction has a rate constant of $1.15 \times 10^{-3}\text{ s}^{-1}$. How long will $5\text{ g}$ of this reactant take to reduce to $3\text{ g}$?
विस्तृत समाधान / उत्तर:
$t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{2.303}{1.15 \times 10^{-3}}\log\left(\frac{5}{3}\right) = 2002.6 \times \log(1.667) = 2002.6 \times 0.2219 \approx 444.4\text{ seconds}$.
उदाहरण 2
Show that for a first-order reaction, the time required for 99% completion is twice the time required for 90% completion.
विस्तृत समाधान / उत्तर:
$t_{90\%} = \frac{2.303}{k}\log\frac{100}{10} = \frac{2.303}{k}(1)$. $t_{99\%} = \frac{2.303}{k}\log\frac{100}{1} = \frac{2.303}{k}(2)$. Therefore, $t_{99\%} = 2 \times t_{90\%}$.
उदाहरण 3
Differentiate between Order of a Reaction and Molecularity.
विस्तृत समाधान / उत्तर:
Order is the sum of powers of concentrations in the experimental rate law (can be zero, fractional, or negative); Molecularity is the theoretical number of reacting species in an elementary step (always a non-zero positive integer, max 3).

सामान्य गलतियाँ एवं परीक्षक के जाल (Examiner Traps)

सामान्य भ्रम / गलत उत्तर

Reciting a definition without applying it to the question or data.

सही वैज्ञानिक तथ्य

Identify the concept, show the relevant evidence or calculation, and explain the final implication.

सामान्य भ्रम / गलत उत्तर

Skipping conditions, units, domain restrictions, or adjustment effects.

सही वैज्ञानिक तथ्य

State assumptions, preserve units, check boundary cases, and verify the answer against the original problem.

सामान्य भ्रम / गलत उत्तर

Treating a correct intermediate result as proof that the whole solution is correct.

सही वैज्ञानिक तथ्य

Perform an independent reasonableness check and connect the result back to the chapter principle.

Chemical Kinetics - Key Molecular Architecture & Reaction Mechanism Model

Chemical Kinetics - Molecular Architecture Electronic & Orbital Mechanisms Stereochemistry, reaction kinetics & pathways Thermodynamic & Coordination Frameworks Crystal field splitting, cell potentials & free energy High-Stakes Examination & Industrial Synthesis CISCE Class 12 Board criteria, JEE/NEET diagnostic applications & conversions

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1
Rate Constant ($k$): Characteristic reaction velocity invariant with concentration.
मुख्य बिंदु 2
First-Order Invariance: Half-life $t_{1/2} = 0.693/k$ remaining constant regardless of starting concentration.
मुख्य बिंदु 3
Activation Barrier ($E_a$): Energetic threshold required to form the transition state activated complex.
मुख्य बिंदु 4
Arrhenius Temperature Sensitivity: Exponential rate acceleration with increasing absolute temperature.
मुख्य बिंदु 5
Catalytic Acceleration: Lowering $E_a$ without shifting equilibrium constant $K_{eq}$.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

1
A first-order reaction has a rate constant of $1.15 \times 10^{-3}\text{ s}^{-1}$. How long will $5\text{ g}$ of this reactant take to reduce to $3\text{ g}$?
उत्तर एवं व्याख्या देखें
उत्तर: $t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{2.303}{1.15 \times 10^{-3}}\log\left(\frac{5}{3}\right) = 2002.6 \times \log(1.667) = 2002.6 \times 0.2219 \approx 444.4\text{ seconds}$.
444.4 seconds.
2
Show that for a first-order reaction, the time required for 99% completion is twice the time required for 90% completion.
उत्तर एवं व्याख्या देखें
उत्तर: $t_{90\%} = \frac{2.303}{k}\log\frac{100}{10} = \frac{2.303}{k}(1)$. $t_{99\%} = \frac{2.303}{k}\log\frac{100}{1} = \frac{2.303}{k}(2)$. Therefore, $t_{99\%} = 2 \times t_{90\%}$.
Proved t_99% = 2 × t_90%.
3
Differentiate between Order of a Reaction and Molecularity.
उत्तर एवं व्याख्या देखें
उत्तर: Order is the sum of powers of concentrations in the experimental rate law (can be zero, fractional, or negative); Molecularity is the theoretical number of reacting species in an elementary step (always a non-zero positive integer, max 3).
Experimental sum of powers vs theoretical colliding species count.
4
The rate of a reaction quadruples when the temperature changes from $293\text{ K}$ to $313\text{ K}$. Calculate the activation energy of the reaction ($R = 8.314\text{ J/mol}\cdot\text{K}$).
उत्तर एवं व्याख्या देखें
उत्तर: $\log\frac{k_2}{k_1} = \log 4 = 0.6021$. $\frac{E_a}{2.303 \times 8.314}\left(\frac{313 - 293}{293 \times 313}\right) = 0.6021 \implies \frac{E_a}{19.147}\left(\frac{20}{91709}\right) = 0.6021 \implies E_a = \frac{0.6021 \times 19.147 \times 91709}{20} \approx 52,863\text{ J/mol} \approx 52.86\text{ kJ/mol}$.
E_a ≈ 52.86 kJ/mol.
5
What is a Pseudo First-Order Reaction? Give an example.
उत्तर एवं व्याख्या देखें
उत्तर: A reaction that is bimolecular in mechanism but follows first-order kinetics because one of the reactants is present in large excess such that its concentration remains practically constant. Example: Acid-catalyzed hydrolysis of ethyl acetate: $\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}(\text{excess}) \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$.
Bimolecular reaction behaving as first order due to excess reactant.
अध्याय का अध्ययन पूर्ण हुआ?
अभ्यास के लिए तैयार?

ऑनलाइन CBT टेस्ट देकर तैयारी का मूल्यांकन करें

ICSE परीक्षा पैटर्न पर आधारित बहुविकल्पीय प्रश्नों का ऑनलाइन टेस्ट दें। तुरंत परिणाम, समय विश्लेषण और प्रत्येक प्रश्न का विस्तृत हल प्राप्त करें।

AI अध्ययन मित्र

त्वरित शंका समाधान

रासायनिक बलगतिकी (Chemical Kinetics) में कोई संदेह या प्रश्न है? हमारे AI अध्ययन मित्र से तुरंत समझें।

STAY UPDATED

Get the Latest Updates

New test series, syllabus updates, exam notifications and more.

Syllabus
Updates
Exam
Alerts
Study
Resources