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JEE Mains Chemistry · Formula sheet

Chemical Thermodynamics formulas

18 formulas, 2 reference tables and 39 common traps for JEE Mains Chemistry Chemical Thermodynamics, grouped by subtopic.

Full notes with worked examples

Systems, State Functions and the First Law

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First law sign convention: ΔU = q + w

First law of thermodynamics

ΔU=q+w\Delta U = q + w

State functions, intensive properties and the standard relations

QuantityIntensive or extensiveState or path function
Temperature, pressure, densityIntensiveState function
Molarity, molar heat capacity, standard cell potentialIntensiveState function
A per-mole or per-litre quantity is intensive, even though it is a ratio of two extensive ones.
Volume, amount in moles, massExtensiveState function
Internal energy U, enthalpy H, entropy S, Gibbs energy GExtensiveState function
Take less of a solution and G falls, even though its concentration and density stay the same.
Heat capacity of a whole sampleExtensiveState function
Heat q, work wExtensive (they scale with the amount)Path function
Among U, V, q and H, only q is not a state variable.
Halve the sample and ask what changes.

Common traps

Sign-reversed textbook relations

Distractors write ΔU=q+pΔV\Delta U = q + p\Delta V, ΔH=ΔU−ΔngRT\Delta H = \Delta U - \Delta n_g RT or ΔH=ΔU−pΔV\Delta H = \Delta U - p\Delta V. With work on the system positive, expansion work is −pΔV-p\Delta V, so ΔU=q−pΔV\Delta U = q - p\Delta V and ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT.

Same concentration, different Gibbs energy

Two solutions with the same concentration have the same density, molar heat capacity and concentration, because those are intensive. Their Gibbs energies differ if they hold different amounts, because GG is extensive.

Adding the magnitudes

If a system does 200 J of work and absorbs 150 J of heat, ΔU=150−200=−50\Delta U = 150 - 200 = -50 J. Adding them (350 J) or flipping the sign (+50 J) are the planted options.

Boiling water does work

Water heated to boiling takes in heat (q>0q > 0) and stores more energy (ΔU>0\Delta U > 0). The steam it makes pushes back the atmosphere, so the system does work: w<0w < 0, not zero.

Work of Expansion

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Reversible isothermal and adiabatic work

Reversible isothermal work

wrev=−2.303 nRTlog⁡V2V1=−2.303 nRTlog⁡p1p2w_{\mathrm{rev}} = -2.303\,nRT\log\frac{V_2}{V_1} = -2.303\,nRT\log\frac{p_1}{p_2}

Work against a constant external pressure and free expansion

Work against a constant external pressure

w=−pext (V2−V1)w = -p_{\mathrm{ext}}\,(V_2 - V_1)

Work as an area on a p–V graph

Work around a cycle

wnet=−∮p dV,∣wnet∣=area enclosedw_{\mathrm{net}} = -\oint p\,dV,\qquad |w_{\mathrm{net}}| = \text{area enclosed}

Common traps

log for ln

nRTlog⁡V2V1nRT\log\frac{V_2}{V_1} without the 2.303 is 2.303 times too small. For nRT=5nRT = 5 kJ and a volume ratio of 4 it gives 5×0.602=3.05 \times 0.602 = 3.0 kJ against the correct 5×1.386=6.95 \times 1.386 = 6.9 kJ, and the small value is always an option.

q and w with the same sign

In an isothermal change of an ideal gas, q=−wq = -w. An option where qq and ww have the same sign, or where ΔU\Delta U equals the work, is wrong.

A change in volume does not mean work

In a free expansion the volume does change. The work is still zero because the external pressure is zero, and ww is −pextΔV-p_{\mathrm{ext}}\Delta V, not −pgasΔV-p_{\mathrm{gas}}\Delta V.

Reversible is the extreme in two different senses

A reversible path gives the MOST work out of an expansion but needs the LEAST work for a compression. So single-stage compression needs the most work of all, and single-stage expansion gives the least.

V on the vertical axis

When a graph plots V upwards, read each point as (p, V) and work the steps out one by one. The rule 'clockwise means work done by the gas' reverses when the axes are swapped.

A cycle is not always the biggest area

A cycle's net work is only the area between its two paths. One long expansion sweeps the whole area under its curve down to the V axis, which is often bigger than any enclosed loop.

Heat Capacity, Calorimetry and Enthalpy vs Internal Energy

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Heat capacity: Cp, Cv and q = nCΔT

Heat and heat capacity

q=nCmΔT,Cp−Cv=Rq = nC_m\Delta T,\qquad C_p - C_v = R

Bomb calorimeter: heat at constant volume

Bomb calorimeter

ΔcU=−Ccal ΔTn\Delta_c U = -\frac{C_{\mathrm{cal}}\,\Delta T}{n}

ΔH and ΔU through the change in gas moles

Enthalpy and internal energy

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT

Common traps

Cv for heat added at constant pressure

Heat supplied at constant pressure is nCpΔTnC_p\Delta T. Dividing it by nCvnC_v overstates the temperature rise. And ΔU\Delta U is less than the heat supplied, because part of the heat did expansion work.

Bomb heat is ΔU, not ΔH

When a question quotes heat measured in a bomb calorimeter and asks for an enthalpy, it needs the ΔngRT\Delta n_g RT step. Skipping it is a planted option, a few kJ away from the answer.

Positive heat, negative ΔU

The calorimeter warms because the reaction gives out heat. CΔTC\Delta T is positive, but ΔcU\Delta_c U of the reaction is negative.

Counting liquid water as a gas

In a combustion that makes H2O(l)\mathrm{H_2O(l)}, water is not a gas. For C2H6(g)+72O2(g)→2CO2(g)+3H2O(l)\mathrm{C_2H_6(g) + \tfrac{7}{2}O_2(g) \to 2CO_2(g) + 3H_2O(l)}, Δng=2−4.5=−2.5\Delta n_g = 2 - 4.5 = -2.5, not +0.5+0.5.

R in joules beside ΔH in kilojoules

8.314×300=24948.314 \times 300 = 2494 J must be written as 2.494 kJ before it is added to a ΔH\Delta H in kJ. Otherwise the correction is a thousand times too big.

Hess's Law: Formation and Combustion Enthalpies

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Reaction enthalpy from formation enthalpies

ΔrH∘=∑ν ΔfHproducts∘−∑ν ΔfHreactants∘\Delta_r H^\circ = \sum \nu\,\Delta_f H^\circ_{\mathrm{products}} - \sum \nu\,\Delta_f H^\circ_{\mathrm{reactants}}

Formation enthalpy from combustion enthalpies

ΔfH=∑ν ΔcHelements−ΔcHcompound\Delta_f H = \sum \nu\,\Delta_c H_{\mathrm{elements}} - \Delta_c H_{\mathrm{compound}}

Hess's law: adding, reversing and scaling equations

Hess's law

ΔHtarget=∑iki ΔHi\Delta H_{\mathrm{target}} = \sum_i k_i\,\Delta H_i

Common traps

The standard state is not 0 °C

'Standard' fixes the pressure at 1 bar and says nothing about temperature. So ΔfH500∘\Delta_f H^\circ_{500} of O2(g)\mathrm{O_2(g)} is zero too, and options that say 273 K are wrong.

One mole of everything

Multiply each ΔfH∘\Delta_f H^\circ by its coefficient in the balanced equation before adding. Using one mole of each species is the most common slip on this page.

Flipping the combustion rule

With combustion enthalpies it is reactants minus products. Using products minus reactants gives the right size with the wrong sign, and that sign-flipped value is always an option.

Which water?

ΔcH(H2)\Delta_c H(\mathrm{H_2}) is about −286 kJ mol⁻¹ when the water is liquid and about −242 kJ mol⁻¹ when it is vapour. Use the value the question gives and match the water's state in the target equation.

Heat on the product side

'C(s)+O2(g)→CO2(g)+400\mathrm{C(s) + O_2(g) \to CO_2(g)} + 400 kJ' means ΔH=−400\Delta H = -400 kJ. Taking it as +400 flips every sign that follows.

22.4 L at room temperature

At 25 °C and 1 atm one mole of gas fills 24.47 L, not 22.4 L. With 22.4 the moles, and so the heat, come out about 9% too large.

Enthalpies of Phase Change, Solution and Neutralisation

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Enthalpies of phase change, solution, dilution and hydration

Hess cycles for phase changes and hydration

ΔsubH=ΔfusH+ΔvapH,ΔhydH=ΔsolHanhydrous−ΔsolHhydrate\Delta_{\mathrm{sub}}H = \Delta_{\mathrm{fus}}H + \Delta_{\mathrm{vap}}H,\qquad \Delta_{\mathrm{hyd}}H = \Delta_{\mathrm{sol}}H_{\mathrm{anhydrous}} - \Delta_{\mathrm{sol}}H_{\mathrm{hydrate}}

Enthalpy of neutralisation and the temperature rise

Temperature rise on neutralisation

ΔT=nH2O ∣ΔneutH∣m c\Delta T = \frac{n_{\mathrm{H_2O}}\,|\Delta_{\mathrm{neut}}H|}{m\,c}

Common traps

The sign of a heat of dilution

Heat of dilution is the more dilute value minus the more concentrated one. Going from −60 to −65 kJ mol⁻¹ gives −5 kJ mol⁻¹, not +5.

Kilojoules beside joules in a heating path

ΔfusH\Delta_{\mathrm{fus}}H is quoted in kJ mol⁻¹ but CpC_p in J K⁻¹ mol⁻¹. Convert one of them before adding, or the latent heat comes out a thousand times too small.

Only one solution's volume

The heat warms the whole mixture. 100 mL of acid plus 100 mL of base is 200 g of solution, not 100 g.

More acid is not more heat

Heat follows the limiting reagent. 50 mL of acid with 20 mL of base neutralises less than 30 mL with 30 mL, and it warms a larger volume, so its temperature rise is smaller.

Bond Enthalpy and Atomisation Cycles

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Reaction enthalpy from bond enthalpies

ΔrH=∑BEbroken−∑BEformed\Delta_r H = \sum BE_{\mathrm{broken}} - \sum BE_{\mathrm{formed}}

Average bond enthalpy and atomisation cycles

Average bond enthalpy from formation enthalpies

BE‾=ΔfH(A,g)+n ΔfH(B,g)−ΔfH(ABn,g)n\overline{BE} = \frac{\Delta_f H(\mathrm{A},g) + n\,\Delta_f H(\mathrm{B},g) - \Delta_f H(\mathrm{AB}_n,g)}{n}

Common traps

Formed minus broken

The order is broken minus formed. Reversing it gives the right size with the wrong sign, and that value is offered beside the right one.

Bond enthalpies are for gases

The broken-minus-formed rule applies only when every species is a gas. A liquid or solid needs its vaporisation or sublimation enthalpy added first.

The sign on a lattice enthalpy

If lattice enthalpy is quoted as forming the lattice (−z), breaking it costs +z. With hydration enthalpies −x and −y, the heat of solution is z−(x+y)z - (x + y).

Per mole of compound

ΔfH\Delta_f H is per mole of the compound. For A2+B2→2AB\mathrm{A_2 + B_2 \to 2AB}, the equation's ΔH\Delta H is twice ΔfH\Delta_f H(AB); set up the bond balance with that doubled value.

Entropy, Gibbs Energy and Spontaneity

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Entropy change and Gibbs energy of a reaction

Entropy and Gibbs energy of reaction

ΔrS∘=∑νSproducts∘−∑νSreactants∘,ΔrG∘=ΔrH∘−TΔrS∘\Delta_r S^\circ = \sum \nu S^\circ_{\mathrm{products}} - \sum \nu S^\circ_{\mathrm{reactants}},\qquad \Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ

Temperature at which ΔG changes sign

ΔG=0 ⇒ T=ΔHΔS\Delta G = 0\ \Rightarrow\ T = \frac{\Delta H}{\Delta S}

Spontaneity from the signs of ΔH and ΔS

ΔHΔSSign of ΔGSpontaneous
NegativePositiveNegative at every temperatureAt all temperatures
PositiveNegativePositive at every temperatureAt no temperature
PositivePositiveNegative above ΔH/ΔSAt high temperature
An endothermic change that goes at 373 K but not at 273 K belongs in this row.
NegativeNegativeNegative below ΔH/ΔSAt low temperature
The sign of ΔS decides which way a rise in temperature pushes ΔG.

Common traps

Exothermic is not enough

An exothermic change with ΔS<0\Delta S < 0 stops being spontaneous above ΔH/ΔS\Delta H/\Delta S. Only ΔH<0\Delta H < 0 with ΔS>0\Delta S > 0 is spontaneous at every temperature.

Swapping the two derivatives of G

G falls as temperature rises, at the rate S, and rises with pressure, at the rate V: (∂G/∂T)p=−S\left(\partial G/\partial T\right)_p = -S and (∂G/∂p)T=V\left(\partial G/\partial p\right)_T = V. Matching lists swap them.

ΔS in joules, ΔH in kilojoules

This is the most common slip in the chapter. With ΔH=50\Delta H = 50 kJ, ΔS=100\Delta S = 100 J K⁻¹ and T = 400 K, 50−400×10050 - 400 \times 100 is nonsense; 50−400×0.100=1050 - 400 \times 0.100 = 10 kJ is right.

Heating entropy needs the 1/T

The entropy of warming is ∫Cp dT/T\int C_p\,dT/T, not ∫Cp dT\int C_p\,dT (that is the enthalpy). Each phase change on the way adds its own ΔH/T\Delta H/T, at its own temperature.

A factor of a thousand

With ΔH=60\Delta H = 60 kJ and ΔS=150\Delta S = 150 J K⁻¹, T=60 000/150=400T = 60\,000/150 = 400 K, not 0.4 K. Convert before dividing.

Below, not above

For an exothermic change with falling entropy, ΔH/ΔS\Delta H/\Delta S is the temperature to stay BELOW. Quoting it as a minimum temperature reverses the answer.

Gibbs Energy and the Equilibrium Constant

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Standard Gibbs energy and the equilibrium constant

Gibbs energy and the equilibrium constant

ΔG∘=−RTln⁡K=−2.303 RTlog⁡K\Delta G^\circ = -RT\ln K = -2.303\,RT\log K

Graphs of log K against 1/T and of G against extent

log K against 1/T

log⁡K=−ΔH∘2.303R⋅1T+ΔS∘2.303R\log K = -\frac{\Delta H^\circ}{2.303R}\cdot\frac{1}{T} + \frac{\Delta S^\circ}{2.303R}

Common traps

A positive ΔG° is not 'no reaction'

Some product always forms. ΔG∘>0\Delta G^\circ > 0 only means the equilibrium lies on the reactant side, with K below 1. Statements that the reaction 'will not occur at all' are wrong.

Moles are not partial pressures

KpK_p uses partial pressures. Divide each equilibrium amount by the total moles to get a mole fraction, then multiply by the total pressure, before building KpK_p.

Dropping the minus sign

ln⁡K=ΔH∘−TΔS∘RT\ln K = \frac{\Delta H^\circ - T\Delta S^\circ}{RT} is the planted wrong form. Since ln⁡K=−ΔG∘/RT\ln K = -\Delta G^\circ/RT, the numerator is TΔS∘−ΔH∘T\Delta S^\circ - \Delta H^\circ.

Reading the slope

The slope is −ΔH∘/2.303R-\Delta H^\circ/2.303R. A line that falls from left to right means ΔH∘>0\Delta H^\circ > 0: K grows as the temperature rises.

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