Thermodynamics: Enthalpy Change & Hess's Law | Class 11 Chemistry + JEE Main
Enthalpy Change aur Hess's Law, Class 11 Chemistry ke Thermodynamics chapter ke important concepts hain. JEE Main preparation mein inka use reaction ki heat calculate karne, standard enthalpy of formation se reaction enthalpy find karne aur multiple chemical equations ko combine karke unknown enthalpy change nikalne mein hota hai.
Is article mein hum enthalpy ka meaning, important formulas, exothermic aur endothermic reactions, standard enthalpy of formation, Hess's Law aur different types of enthalpy changes ko simple Hinglish mein samjhenge. Saath hi, solved examples aur practice questions se concepts revise karenge.
1. Enthalpy Kya Hoti Hai?
Enthalpy ek thermodynamic state function hai jo system ki internal energy aur pressure-volume contribution ko combine karti hai. Isse symbol H se represent kiya jata hai.
Enthalpy Formula
H = U + PV
Where:
- H: Enthalpy.
- U: Internal energy.
- P: Pressure.
- V: Volume.
Enthalpy ek state function hai. Iska matlab hai ki enthalpy change sirf initial aur final states par depend karta hai, system un states ke beech kis path se gaya, is par nahi.
Constant Pressure Par Enthalpy Change
Jab pressure constant ho aur sirf pressure-volume work ho, toh system ko di gayi heat constant-pressure enthalpy change ke equal hoti hai.
ΔH = qp
Iska matlab hai ki constant pressure par reaction mein absorb ya release hone wali heat enthalpy change ko represent karti hai, provided stated conditions apply.
JEE Trick: Constant pressure par heat change se related questions mein ΔH = qp yaad rakho. Constant volume par heat change generally ΔU se related hota hai.
2. Enthalpy Change Ka Formula
Chemical reaction ke during products aur reactants ki enthalpy different ho sakti hai. Reaction ka enthalpy change final state ki enthalpy aur initial state ki enthalpy ka difference hota hai.
ΔH = H(products) − H(reactants)
Reaction enthalpy calculate karte waqt products ki total enthalpy se reactants ki total enthalpy subtract ki jati hai.
Exothermic Reaction: ΔH Negative
Jab reaction surroundings mein heat release karti hai, toh use exothermic reaction kaha jata hai.
ΔH < 0
Example:
A → B, ΔH = −40 kJ
Iska matlab hai ki reaction as written complete hone par 40 kJ heat release hoti hai. Agar reaction equation mein 1 mole A react karta hai, toh yeh heat 1 mole A ke reaction ke corresponding hogi.
Endothermic Reaction: ΔH Positive
Jab reaction surroundings se heat absorb karti hai, toh use endothermic reaction kaha jata hai.
ΔH > 0
Is case mein products ki enthalpy reactants ki enthalpy se greater hoti hai.
Zero Enthalpy Change
ΔH = 0
Iska matlab hai ki specified initial aur final states ki enthalpy equal hai.
| Enthalpy Change | Reaction Type | Heat Transfer |
|---|---|---|
| ΔH < 0 | Exothermic | Heat released |
| ΔH > 0 | Endothermic | Heat absorbed |
| ΔH = 0 | Zero enthalpy change | Initial and final enthalpy equal |
Important: Enthalpy change reaction equation ke according hota hai. Agar reaction ko reverse karte ho, toh ΔH ka sign bhi reverse ho jata hai.
3. Standard Enthalpy of Formation
Standard enthalpy of formation, jise ΔH°f se represent karte hain, woh enthalpy change hai jab kisi compound ka 1 mole uske constituent elements ke standard states se form hota hai, specified standard conditions ke under.
Is quantity ki help se hum chemical reaction ka enthalpy change calculate kar sakte hain.
Reaction Enthalpy Formula
ΔH°reaction = ÎŁ ΔH°f(products) − ÎŁ ΔH°f(reactants)
Is formula mein har substance ki standard enthalpy of formation ko uske balanced chemical equation ke stoichiometric coefficient se multiply karna zaroori hai.
Important Rule: Elements in Their Standard States
Kisi element ki standard enthalpy of formation uski standard state mein zero hoti hai.
- H₂(g): ΔH°f = 0
- O₂(g): ΔH°f = 0
- C(graphite): ΔH°f = 0
Yaad rakho, yeh rule element ki standard state ke liye hai. Element ki kisi doosri physical form ya allotrope ke liye enthalpy of formation necessarily zero nahi hoti.
Solved Example: Standard Enthalpy of Formation
Consider the reaction:
C(graphite) + O₂(g) → CO₂(g)
Given:
- ΔH°f[C(graphite)] = 0
- ΔH°f[O₂(g)] = 0
- ΔH°f[CO₂(g)] = −394 kJ/mol
Formula apply karo:
ΔH°reaction = [−394] − [0 + 0]
ΔH°reaction = −394 kJ/mol
Reaction exothermic hai, kyunki enthalpy change negative hai.
JEE Trick: Formula mein products minus reactants hota hai, reactants minus products nahi. Balanced equation ke coefficients ko include karna mat bhoolna.
4. Hess's Law Kya Hai?
Hess's Law ke according, kisi reaction ka total enthalpy change sirf initial aur final states par depend karta hai. Reaction ek step mein ho ya multiple steps mein, same initial aur final states ke liye total enthalpy change same rehta hai.
ΔH(total) = ΔH₁ + ΔH₂ + ΔH₃ + ...
Hess's Law ka use tab hota hai jab kisi reaction ka enthalpy change directly measure karna difficult ho, lekin related reactions ke enthalpy changes known hon.
Hess's Law Ke Important Rules
- Reaction reverse karo: Enthalpy change ka sign reverse ho jayega. Agar ΔH = −100 kJ hai, toh reverse reaction ke liye ΔH = +100 kJ hoga.
- Reaction ko n times multiply karo: Enthalpy change bhi n times ho jayega. Agar equation ko 2 se multiply kiya, toh ΔH ko bhi 2 se multiply karo.
- Reactions add karo: Individual enthalpy changes ko add karo.
- Common substances cancel karo: Dono sides par present intermediate substances cancel karne se required net reaction milti hai.
Exam Shortcut: Hess's Law questions mein sabse pehle target equation likho. Phir given equations ko reverse ya multiply karke target equation ke reactants aur products match karao.
5. Hess's Law – Solved Example
Given reaction 1:
C(graphite) + O₂(g) → CO₂(g)
ΔH₁ = −394 kJ/mol
Given reaction 2:
C(graphite) + ½O₂(g) → CO(g)
ΔH₂ = −111 kJ/mol
Find the enthalpy change for:
CO(g) + ½O₂(g) → CO₂(g)
Step 1: Second Equation Reverse Karo
Original equation:
C(graphite) + ½O₂(g) → CO(g)
Reverse karne par:
CO(g) → C(graphite) + ½O₂(g)
Reverse reaction ka enthalpy change:
ΔH = +111 kJ/mol
Step 2: Dono Equations Add Karo
Reaction 1:
C(graphite) + O₂(g) → CO₂(g), ΔH = −394 kJ/mol
Reversed reaction 2:
CO(g) → C(graphite) + ½O₂(g), ΔH = +111 kJ/mol
Step 3: Common Substances Cancel Karo
Dono equations add karne par graphite left aur right side par cancel ho jata hai. Oxygen ke ½ mole ko bhi cancel karne par required reaction milti hai:
CO(g) + ½O₂(g) → CO₂(g)
Step 4: Enthalpy Changes Add Karo
ΔH = −394 + 111
ΔH = −283 kJ/mol
Final Answer: −283 kJ/mol
Negative sign indicate karta hai ki reaction exothermic hai aur heat release karti hai.
6. Important Types of Enthalpy Changes
Chemical Thermodynamics mein different processes ke enthalpy changes ko specific names diye jate hain. JEE Main ke liye inki definitions aur sign conventions samajhna useful hai.
1. Enthalpy of Formation
Jab constituent elements ke standard states se 1 mole compound form hota hai, toh enthalpy change ko enthalpy of formation kehte hain.
2. Enthalpy of Combustion
Jab 1 mole substance specified conditions mein oxygen ke saath completely burn karta hai, toh enthalpy change ko enthalpy of combustion kehte hain.
Complete combustion generally exothermic hoti hai, isliye enthalpy of combustion commonly negative hoti hai.
3. Bond Dissociation Enthalpy
Gaseous molecules mein specified bond ko homolytically break karke gaseous fragments form karne ke liye required enthalpy change ko bond dissociation enthalpy kehte hain. Ek specific bond dissociation process ke liye yeh positive hoti hai.
4. Enthalpy of Atomization
Jab 1 mole gaseous atoms form hote hain, toh corresponding enthalpy change ko enthalpy of atomization kaha jata hai.
5. Enthalpy of Solution
Jab solute specified amount of solvent mein dissolve hota hai, toh process ka enthalpy change enthalpy of solution kehlata hai.
Yeh positive ya negative ho sakti hai, depending on the balance between energy required to separate particles and energy released during solvation.
6. Enthalpy of Neutralization
Aqueous acid-base neutralization mein 1 mole water form hone ke corresponding enthalpy change ko enthalpy of neutralization kaha jata hai.
Strong acid aur strong base ki dilute aqueous solutions ke reaction mein yeh generally negative hoti hai.
| Enthalpy Type | Meaning |
|---|---|
| Formation | 1 mole compound form hone par enthalpy change |
| Combustion | 1 mole substance ke complete combustion par enthalpy change |
| Bond dissociation | Specified gaseous bond break karne ke liye enthalpy change |
| Atomization | 1 mole gaseous atoms form hone par enthalpy change |
| Solution | Solute dissolve hone par enthalpy change |
| Neutralization | 1 mole water form hone par enthalpy change |
7. Solved JEE Main-Level Question
Question: Given the following reactions:
C(graphite) + O₂(g) → CO₂(g), ΔH = −394 kJ/mol
CO(g) + ½O₂(g) → CO₂(g), ΔH = −283 kJ/mol
Calculate ΔH for:
C(graphite) + ½O₂(g) → CO(g)
Options:
- A) −677 kJ/mol
- B) −111 kJ/mol
- C) +111 kJ/mol
- D) −394 kJ/mol
Step 1: Hess's Law Apply Karo
Target reaction mein CO product side par hai, jabki second given equation mein CO reactant side par hai. Isliye second equation ko reverse karne ki zaroorat nahi hai; instead, given equations ko algebraically compare karte hain.
Let target reaction ka enthalpy change ΔH₃ ho.
Given:
ΔH₁ = −394 kJ/mol
CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283 kJ/mol
Step 2: Enthalpy Equation Banao
Target reaction aur CO oxidation reaction ko add karne par overall reaction C(graphite) + O₂(g) → CO₂(g) milti hai.
ΔH₁ = ΔH₃ + ΔH₂
−394 = ΔH₃ + (−283)
Step 3: Unknown Value Calculate Karo
ΔH₃ = −394 + 283
ΔH₃ = −111 kJ/mol
Correct Answer
Option B) −111 kJ/mol
Quick Trick: Target equation ko given equations se match karo. Reaction reverse karne par sign change hota hai aur reactions add karne par enthalpy values add hoti hain.
8. Common Mistakes and JEE Main Exam Tricks
- Sign of ΔH: Negative enthalpy change exothermic reaction ko show karta hai; positive enthalpy change endothermic reaction ko.
- Products minus reactants: Reaction enthalpy calculate karte waqt products ki total enthalpy se reactants ki total enthalpy subtract karo.
- Formation enthalpy: Balanced equation ke coefficients ko include karna mat bhoolo.
- Standard state: Element ki standard state mein standard enthalpy of formation zero hoti hai.
- Reverse reaction: Reaction reverse karne par ΔH ka sign reverse ho jata hai.
- Multiplication: Reaction ko 3 times multiply karo, toh enthalpy change bhi 3 times hoga.
- Hess's Law: Intermediate species ko cancel karke target equation obtain karo.
- Units: Check karo ki answer kJ/mol mein hai ya reaction equation ke as-written extent ke liye kJ mein.
9. Enthalpy Change and Hess's Law Formula Sheet
| Concept | Formula or Rule |
|---|---|
| Enthalpy | H = U + PV |
| Constant-pressure heat | ΔH = qp, under the stated work conditions |
| Reaction enthalpy | ΔH = H(products) − H(reactants) |
| Exothermic reaction | ΔH < 0 |
| Endothermic reaction | ΔH > 0 |
| Standard formation enthalpy method | ΔH°reaction = ΣΔH°f(products) − ΣΔH°f(reactants) |
| Element in standard state | ΔH°f = 0 |
| Hess's Law | ΔH(total) = ΣΔH(step) |
| Reverse a reaction | ΔH changes sign |
| Multiply a reaction by n | Multiply ΔH by n |
10. One-Minute Revision
- Enthalpy formula: H = U + PV.
- Constant pressure par ΔH = qp, under the appropriate work conditions.
- Reaction enthalpy = products ki enthalpy minus reactants ki enthalpy.
- ΔH negative ho toh reaction exothermic hai.
- ΔH positive ho toh reaction endothermic hai.
- Formation enthalpy method: products minus reactants, with coefficients.
- Elements in their standard states have ΔH°f = 0.
- Hess's Law ke according total enthalpy change path-independent hota hai.
- Reaction reverse karo toh ΔH ka sign reverse hota hai.
- Reaction ko n times multiply karo toh ΔH bhi n times hota hai.
11. Practice Questions for JEE Main
Question 1
Ek reaction ka enthalpy change −75 kJ hai. Reaction exothermic hai ya endothermic?
Answer: Exothermic, because ΔH is negative.
Question 2
Agar forward reaction ka ΔH = +120 kJ hai, toh reverse reaction ka ΔH kya hoga?
Answer: −120 kJ.
Question 3
Agar kisi reaction ko 3 times multiply kiya jaye aur original ΔH = −40 kJ ho, toh new ΔH kya hoga?
Solution: New ΔH = 3 × (−40).
Answer: −120 kJ.
Question 4
Standard enthalpy of formation of O₂(g) kitni hoti hai?
Answer: Zero, because O₂(g) oxygen ki standard state hai.
Question 5
Given: A → B, ΔH = −100 kJ and B → C, ΔH = +30 kJ. Find ΔH for A → C.
Solution: ΔH(total) = −100 + 30.
Answer: −70 kJ.
Question 6
Hess's Law ke according reaction ka total enthalpy change kis par depend karta hai?
Answer: Initial aur final states par, reaction ke path par nahi.
12. Download Thermodynamics Notes PDF
Agar aap Enthalpy Change aur Hess's Law ke notes ko PDF format mein blog par show karna chahte ho, toh neeche diye gaye iframe mein apne hosted PDF ka direct viewing URL add kar do.
13. Frequently Asked Questions (FAQs)
Q1. What is enthalpy?
Enthalpy ek thermodynamic state function hai jise H = U + PV se define kiya jata hai.
Q2. What is the formula for enthalpy change?
Reaction ke liye ΔH = H(products) − H(reactants) hota hai.
Q3. What is Hess's Law?
Hess's Law ke according, reaction ka total enthalpy change initial aur final states par depend karta hai, reaction ke path par nahi.
Q4. Why does the sign of enthalpy change reverse when a reaction is reversed?
Reverse reaction mein initial aur final states interchange ho jati hain. Isliye enthalpy difference ka sign reverse ho jata hai.
Q5. Why is the standard enthalpy of formation of an element in its standard state zero?
Standard enthalpy of formation ko reference convention ke according define kiya jata hai. Element ki standard state mein uski formation enthalpy zero assign hoti hai.
Q6. Is Hess's Law important for JEE Main?
Hess's Law reaction enthalpy calculate karne ke liye useful concept hai. Iski practice karne se students reverse reactions, enthalpy addition aur standard formation enthalpy based numerical questions solve kar sakte hain.
Final Thoughts
Enthalpy Change aur Hess's Law ko master karne ke liye sabse pehle H = U + PV aur ΔH = H(products) − H(reactants) ko understand karo. Uske baad exothermic aur endothermic reactions ke signs, standard enthalpy of formation aur Hess's Law ke rules revise karo.
JEE Main numerical questions mein target reaction ko clearly likhna, equations ko correctly reverse ya multiply karna aur intermediate substances cancel karna sabse important steps hain. In concepts ki regular practice se calculation errors kam hongi aur Thermodynamics chapter ki understanding strong hogi.