Easy Practice — Set 1

Quick Recap — System, First Law & Enthalpy

• System types: open (matter + energy), closed (energy only), isolated (neither).
• State functions depend only on state ($U$, $H$, $S$, $G$, $T$, $P$, $V$); path functions are $q$ and $w$.
• First law: $\Delta U=q+w$; work done on the system is positive ($w=-P_{ext}\Delta V$ for expansion).
• $H=U+PV$; at constant $P$, $q_p=\Delta H$; at constant $V$, $q_v=\Delta U$. Intensive: $T$, $P$, density; extensive: $V$, $U$, $H$, mass.

Beyond-NCERT JEE Formulae

Use this as a JEE-oriented formula sheet: each entry says WHEN to reach for the relation, and a [JEE Tip] flags the classic exam trap.

1. Enthalpy vs Internal Energy

• Master link: $\Delta H=\Delta U+\Delta n_g RT$, where $\Delta n_g=$ (moles of gaseous products) $-$ (moles of gaseous reactants). Reach for it when a bomb calorimeter gives $\Delta U$ (constant volume) and the question wants $\Delta H$ (constant pressure), or the reverse.
• Only GASES count in $\Delta n_g$; solids and liquids are dropped. If $\Delta n_g=0$ then $\Delta H=\Delta U$.
• [JEE Tip] With $R=8.314$ J/mol/K the term $\Delta n_g RT$ comes out in joules, so convert it to kJ before adding to a kJ enthalpy.

2. Work, Heat and the Sign Convention

• First law $\Delta U=q+w$ (IUPAC): $q>0$ for heat absorbed BY the system and $w>0$ for work done ON the system.
• Reversible isothermal (ideal gas): $w=-nRT\ln\dfrac{V_2}{V_1}=-2.303\,nRT\log\dfrac{V_2}{V_1}$, and since $\Delta U=0$ the heat is $q=-w$.
• Irreversible against constant pressure: $w=-P_{ext}\Delta V$. Free expansion into vacuum has $P_{ext}=0$, so $w=0$ however large $\Delta V$ becomes.
• [JEE Tip] Reversible isothermal expansion delivers the MAXIMUM work; the irreversible value is smaller in magnitude. For an isothermal ideal gas swap ratios using $\dfrac{V_2}{V_1}=\dfrac{P_1}{P_2}$.

3. Hess's Law, Formation and Bond Enthalpies

• Formation route: $\Delta H_{rxn}=\sum\Delta H_f(\text{products})-\sum\Delta H_f(\text{reactants})$, each weighted by its coefficient; $\Delta H_f$ of an element in its standard state is zero.
• Bond-enthalpy route: $\Delta H_{rxn}=\sum(\text{bonds broken})-\sum(\text{bonds formed})$, that is reactant-bond energies minus product-bond energies.
• [JEE Tip] The bond-enthalpy method holds ONLY when every species is gaseous; unchanged bonds cancel, so you may count just the bonds that break and form.

4. Gibbs Energy, Spontaneity and K

• $\Delta G=\Delta H-T\Delta S$: spontaneous when $\Delta G<0$, at equilibrium when $\Delta G=0$, non-spontaneous when $\Delta G>0$.
• Mixed-sign cases flip at the crossover temperature $T=\dfrac{\Delta H}{\Delta S}$: for $\Delta H>0,\ \Delta S>0$ spontaneity begins ABOVE it, while for $\Delta H<0,\ \Delta S<0$ it holds only BELOW it.
• Bridges: $\Delta G^\circ=-2.303\,RT\log K$ links to equilibrium, and $\Delta G^\circ=-nFE^\circ$ (with $F=96500$ C/mol) links to electrochemistry.
• [JEE Tip] At equilibrium it is $\Delta G$, not $\Delta G^\circ$, that vanishes; a negative $\Delta G^\circ$ merely forces $K>1$. Always put $\Delta S$ in kJ/K before subtracting from a kJ $\Delta H$.

5. Entropy and Heat Capacities

• Entropy toolbox: $\Delta S=\dfrac{q_{rev}}{T}$; isothermal ideal gas $\Delta S=nR\ln\dfrac{V_2}{V_1}$; phase change $\Delta S=\dfrac{\Delta H_{trans}}{T_{trans}}$; surroundings $\Delta S_{surr}=-\dfrac{\Delta H_{sys}}{T}$; and $\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}\ge 0$.
• Heat capacities: $C_p-C_v=R$ per mole, with $\gamma=\dfrac{C_p}{C_v}$ equal to $\tfrac{5}{3}$ (monatomic) or $\tfrac{7}{5}$ (diatomic).
• Adiabatic ($q=0$): $\Delta U=w=nC_{v,m}\Delta T$ with $PV^{\gamma}=$ constant and $TV^{\gamma-1}=$ constant. Isothermal ideal gas: $\Delta U=0$ and $\Delta H=0$, so $q=-w$.
• [JEE Tip] For an ideal gas $\Delta U=nC_{v,m}\Delta T$ and $\Delta H=nC_{p,m}\Delta T$ on ANY path, because $U$ and $H$ depend on temperature alone.

Solved Examples — Beyond-NCERT Formulae

Example 1 - $\Delta U$ from $\Delta H$ using $\Delta n_g$. For ammonia synthesis, N2(g) + 3H2(g) -> 2NH3(g), the enthalpy change at 298 K is $\Delta H=-92.2$ kJ. Find $\Delta U$. (R = 8.314 J/mol/K)

• Gaseous moles: $\Delta n_g=2-(1+3)=-2$.
• Correction term: $\Delta n_g RT=(-2)(8.314)(298)=-4955$ J $=-4.955$ kJ.
• Rearrange the master link: $\Delta U=\Delta H-\Delta n_g RT=-92.2-(-4.955)=-87.2$ kJ. Because the gas moles fall, the surroundings do work on the gas, so $\Delta U$ is slightly less negative than $\Delta H$.

Example 2 - Reversible isothermal work and heat. Five moles of an ideal gas expand isothermally and reversibly at 300 K from 10 L to 100 L. Find the work done on the gas and the heat absorbed. (ln 10 = 2.303)

• Volume ratio: $\dfrac{V_2}{V_1}=\dfrac{100}{10}=10$.
• Work: $w=-nRT\ln\dfrac{V_2}{V_1}=-(5)(8.314)(300)(2.303)=-2.87\times10^{4}$ J $=-28.7$ kJ.
• Isothermal ideal gas has $\Delta U=0$, so $q=-w=+28.7$ kJ; heat is absorbed as the gas expands.

Example 3 - Hess's law. Given C(gr) + O2(g) -> CO2(g), $\Delta H_1=-393.5$ kJ; H2(g) + 1/2 O2(g) -> H2O(l), $\Delta H_2=-285.8$ kJ; and CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l), $\Delta H_3=-890.3$ kJ. Find $\Delta H_f$ of CH4(g).

• Target equation: C(gr) + 2H2(g) -> CH4(g).
• Assemble it as (i) + 2(ii) - (iii): $\Delta H_f=\Delta H_1+2\Delta H_2-\Delta H_3$.
• $\Delta H_f=-393.5+2(-285.8)-(-890.3)=-393.5-571.6+890.3=-74.8$ kJ/mol.

Example 4 - $\Delta H$ from bond enthalpies. For the hydrogenation C2H4(g) + H2(g) -> C2H6(g), take E(C=C) = 615, E(C-H) = 414, E(H-H) = 436 and E(C-C) = 347 kJ/mol.

• Bonds broken (reactants): one C=C, four C-H and one H-H, giving $615+4(414)+436=2707$ kJ.
• Bonds formed (products): one C-C and six C-H, giving $347+6(414)=2831$ kJ.
• $\Delta H=\sum(\text{broken})-\sum(\text{formed})=2707-2831=-124$ kJ. Only the C=C and H-H bonds truly break while a C-C and two C-H form, so the untouched C-H bonds cancel and give the same $-124$ kJ.

Example 5 - $\Delta G$ and spontaneity. At 298 K a reaction has $\Delta H=-57.2$ kJ/mol and $\Delta S=-175.8$ J/K/mol. Find $\Delta G$ and judge spontaneity.

• Entropy term: $T\Delta S=(298)(-175.8)=-52388$ J/mol $=-52.4$ kJ/mol.
• Gibbs relation: $\Delta G=\Delta H-T\Delta S=-57.2-(-52.4)=-4.8$ kJ/mol.
• $\Delta G<0$, so the reaction is (just) spontaneous at 298 K; the unfavourable entropy nearly cancels the exothermic drive.

Example 6 - Temperature of equilibrium. For CaCO3(s) -> CaO(s) + CO2(g), $\Delta H=+178.3$ kJ/mol and $\Delta S=+160.5$ J/K/mol. Find the temperature above which the decomposition becomes spontaneous.

• At the crossover $\Delta G=0$, so $T=\dfrac{\Delta H}{\Delta S}$.
• Keep both in joules: $T=\dfrac{178300}{160.5}=1111$ K.
• With $\Delta H>0$ and $\Delta S>0$, the reaction is spontaneous ABOVE 1111 K.

Example 7 - $K$ from $\Delta G^\circ$. A reaction at 298 K has $\Delta G^\circ=-22.8$ kJ/mol. Find its equilibrium constant. (Take $2.303\,RT=5.706$ kJ/mol.)

• Logarithm: $\log K=\dfrac{-\Delta G^\circ}{2.303\,RT}=\dfrac{22.8}{5.706}=4.00$.
• Antilog: $K=10^{4.00}\approx1.0\times10^{4}$.
• A negative $\Delta G^\circ$ gives $K\gg1$, so products are strongly favoured.

Example 8 - $\Delta G^\circ$ from cell EMF. The Daniell cell Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s) has $E^\circ=+1.10$ V with $n=2$. Find $\Delta G^\circ$. (F = 96500 C/mol)

• Apply the electrochemical bridge: $\Delta G^\circ=-nFE^\circ=-(2)(96500)(1.10)$.
• $\Delta G^\circ=-212300$ J/mol $=-212.3$ kJ/mol.
• $E^\circ>0$ forces $\Delta G^\circ<0$, confirming the cell reaction is spontaneous.