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2.5.8Thermodynamics (Chemical)

Hess's law — enthalpy is a state function; enthalpy cycles

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WHY does Hess's law exist?

WHY is HH a state function? Because H=U+PVH = U + PV, and UU (internal energy), PP, and VV are all state functions. A sum of state functions is a state function.

WHAT this implies: if you go from reactants to products, the change ΔH=Hfinal−Hinitial\Delta H = H_{\text{final}} - H_{\text{initial}} is fixed. Whether you go there in one violent step or ten gentle steps, the endpoints are the same, so the difference is the same.

HOW to state Hess's law

Two operational rules (derived from the same fact)

  1. Reverse a reaction ⟹ flip the sign of ΔH\Delta H. Why? ΔHZ→A=HA−HZ=−(HZ−HA)=−ΔHA→Z\Delta H_{Z\to A} = H_A - H_Z = -(H_Z - H_A) = -\Delta H_{A\to Z}.
  2. Multiply a reaction by nn ⟹ multiply ΔH\Delta H by nn. Why? Enthalpy is extensive — double the amount of stuff, double the enthalpy content, double the change.
Figure — Hess's law — enthalpy is a state function; enthalpy cycles

WORKED EXAMPLE 1 — Classic algebra of reactions

Goal: find ΔH\Delta H for   C(s)+12O2(g)→CO(g)\;C(s) + \tfrac12 O_2(g) \to CO(g) (hard to measure directly — CO always over-oxidizes).

Given:

  • (i) C(s)+O2(g)→CO2(g)C(s) + O_2(g) \to CO_2(g), ΔH1=−393.5 kJ\Delta H_1 = -393.5\ \text{kJ}
  • (ii) CO(g)+12O2(g)→CO2(g)CO(g) + \tfrac12 O_2(g) \to CO_2(g), ΔH2=−283.0 kJ\Delta H_2 = -283.0\ \text{kJ}

Target = (i) −- (ii): reverse (ii) and add to (i).

Step Reaction ΔH\Delta H
(i) C+O2→CO2C + O_2 \to CO_2 −393.5-393.5
(ii) reversed CO2→CO+12O2CO_2 \to CO + \tfrac12 O_2 +283.0+283.0
Sum C+12O2→COC + \tfrac12 O_2 \to CO −110.5\mathbf{-110.5}

Why reverse (ii)? We need COCO as a product, but in (ii) it is a reactant. Reversing puts it on the product side and flips the sign. Why does CO2CO_2 vanish? It appears as a product in (i) and a reactant in the reversed (ii) — same species, same state, so it cancels (telescoping).

ΔH=−393.5+283.0=−110.5 kJ\boxed{\Delta H = -393.5 + 283.0 = -110.5\ \text{kJ}}

WORKED EXAMPLE 2 — Formation enthalpies

Combust methane: CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH_4(g) + 2O_2(g)\to CO_2(g) + 2H_2O(l) Data (ΔfH∘\Delta_f H^\circ, kJ/mol): CH4=−74.8, CO2=−393.5, H2O(l)=−285.8, O2=0CH_4=-74.8,\ CO_2=-393.5,\ H_2O(l)=-285.8,\ O_2=0.

ΔrH∘=[(−393.5)+2(−285.8)]−[(−74.8)+2(0)]\Delta_r H^\circ = [(-393.5) + 2(-285.8)] - [(-74.8) + 2(0)] Why the coefficient 2 on water? Extensive property × stoichiometric coefficient. =(−965.1)−(−74.8)=−890.3 kJ/mol= (-965.1) - (-74.8) = \mathbf{-890.3\ \text{kJ/mol}}

WORKED EXAMPLE 3 — A Born–Haber style cycle (mini)

Find ΔH\Delta H for A→DA \to D if: A→B (+50)A\to B\ (+50), B→C (−120)B\to C\ (-120), D→C (+30)D\to C\ (+30).

Path: A→B→C→DA\to B\to C\to D. Note C→D=−(D→C)=−30C\to D = -(D\to C) = -30. ΔH=(+50)+(−120)+(−30)=−100 kJ\Delta H = (+50) + (-120) + (-30) = \mathbf{-100\ \text{kJ}} Why flip the last one? Given data goes D→CD\to C; our path needs C→DC\to D, so reverse the sign.

Common mistakes

Feynman check

Recall Explain to a 12-year-old (click to reveal)

Imagine you have ₹100. Whether you spend it as one ₹100 note or as coins little by little, you still end up ₹100 poorer. The amount you lose only depends on start and end money, not on how you spent. Chemistry's "money" is enthalpy (heat content). To find how much heat a hard reaction gives out, we take easier reactions and add/subtract them like adding and subtracting money receipts — and we land on the exact same total.

Active recall

Why is enthalpy a state function?
Because H=U+PVH=U+PV and U,P,VU,P,V are all state functions; a combination of state functions is one, so ΔH\Delta H depends only on initial/final states.
State Hess's law.
The total ΔH\Delta H of a reaction equals the sum of ΔH\Delta H of any set of steps that add up to it: ΔHrxn=∑ΔHi\Delta H_{rxn}=\sum \Delta H_i.
What happens to ΔH\Delta H when you reverse a reaction?
Its sign flips: ΔHrev=−ΔHfwd\Delta H_{rev}=-\Delta H_{fwd} (because ΔH=Hfinal−Hinitial\Delta H = H_{final}-H_{initial}).
What happens to ΔH\Delta H when you multiply a reaction by nn?
It multiplies by nn (enthalpy is extensive).
Formula for ΔrH∘\Delta_r H^\circ from formation enthalpies?
ΔrH∘=∑νpΔfHprod∘−∑νrΔfHreact∘\Delta_r H^\circ=\sum\nu_p\Delta_f H^\circ_{prod}-\sum\nu_r\Delta_f H^\circ_{react}.
Why is ΔfH∘\Delta_f H^\circ of an element in its standard state zero?
Forming an element from itself is no change; endpoints identical ⟹ ΔH=0\Delta H=0.
Compute ΔH\Delta H for C+12O2→COC+\tfrac12O_2\to CO given C+O2→CO2 (−393.5)C+O_2\to CO_2\,(-393.5) and CO+12O2→CO2 (−283.0)CO+\tfrac12O_2\to CO_2\,(-283.0).
−393.5−(−283.0)=−110.5-393.5-(-283.0)=-110.5 kJ.
Why can't we measure ΔH\Delta H of forming CO directly?
CO readily oxidizes further to CO2CO_2, so pure single-step data is hard to obtain — Hess's law lets us get it indirectly.

Connections

Concept Map

sum gives

is a

depends only on

path independent

telescoping cancellation

sum of steps

rule 1

rule 2, extensive

applied to

used in

analogy

U, P, V state functions

Enthalpy H = U + PV

State function

Initial and final states

Delta H fixed

Hess's Law

Delta H reaction = sum Delta Hi

Reverse reaction flips sign

Multiply by n scales Delta H

CO formation example

Altitude on mountain

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Hess's law ka core idea bahut simple hai: enthalpy ek state function hai. Matlab kisi reaction mein jitni heat nikalti ya lagti hai, wo sirf starting materials aur final products par depend karti hai — beech mein kaunsa raasta liya, isse koi farak nahi padta. Jaise pahaad chadhte waqt aap seedha jao ya ghum-ghum ke, altitude ka change same rahega. Bas wahi funda enthalpy pe lagta hai.

Isse bada fayda ye hai ki jo reactions directly measure karna mushkil hai (jaise C+12O2→COC + \tfrac12 O_2 \to CO, kyunki CO aage jaake CO2CO_2 ban jaati hai), unko hum aasan reactions ko jodh-ghata ke nikaal lete hain. Rule yaad rakho: reaction ko ulta karo to sign ulta, reaction ko nn se multiply karo to ΔH\Delta H bhi nn se multiply, aur phir align karke add kar do — beech ke common species khud cancel ho jaate hain.

Ek chhoti si galti bacho: given ΔH\Delta H ko seedhe add mat kar dena. Pehle har equation ko target ke hisaab se orient karo (product side pe sahi species aaye), coefficient match karo, tabhi add karo. Aur states ka dhyaan — H2O(l)H_2O(l) aur H2O(g)H_2O(g) alag hote hain, mix mat karna.

Exam mein isse formation enthalpy wale problems, Born–Haber cycle, aur combustion enthalpy sab nikal jaate hain. Ek baar telescoping wali derivation samajh li (HZ−HAH_Z-H_A = sum of small steps), to poora chapter easy lagega.

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Connections

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