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MHT-CET Chemistry · Chemical Thermodynamics and Energetics

Enthalpy and the Relation Between ΔH and ΔU

Enthalpy H = U + PV is the heat of a process at constant pressure; for a reaction with a change in gas moles, ΔH = ΔU + Δn_g RT, so ΔH and ΔU differ only when gases are made or consumed.

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Why this matters

18 PYQs, 2 HARD. Half are ΔH − ΔU = Δn_g RT — read the equation, count gas moles, multiply — including the reverse move from ΔU to ΔH with R = 8.314. The rest are an enthalpy of vaporisation from grams and kilojoules, and recall: freezing is minus fusion, melting is endothermic, ΔH of an isothermal ideal-gas step is zero.

Concept 1 of 2

Enthalpy, Heat at Constant Pressure and Phase Changes

Intuition

At constant pressure part of the heat supplied goes into pushing back the atmosphere, so the heat is ΔU plus PΔV — that sum is ΔH. A phase change at constant T and P is pure enthalpy: melting and boiling absorb heat (positive), freezing and condensing release the same amount (negative).

Definition

  • H=U+PVH = U + PV; at constant P, qP=ΔH=ΔU+PΔVq_P = \Delta H = \Delta U + P\Delta V. A gas doing 200 J of expansion work while U rises 432 J has ΔH=632\Delta H = 632 J.
  • ΔHfreezing=−ΔHfusion\Delta H_{\text{freezing}} = -\Delta H_{\text{fusion}}; ΔHcondensation=−ΔHvaporisation\Delta H_{\text{condensation}} = -\Delta H_{\text{vaporisation}}. Melting ice is ENDOTHERMIC; freezing, condensation and deposition are exothermic.
  • ΔHvap\Delta H_{\text{vap}} per mole = heat supplied ÷ moles vaporised: 13 g benzene (1/6 mol) by 5.1 kJ → 30.6 kJ mol⁻¹; 11.5 g ethanol (0.25 mol) by 11.8 kJ → 47.2; 1.8 g water (0.1 mol) by 4 kJ → 40.
  • Two formation equations give a phase change: ΔHvap(H2O)=ΔHf(g)−ΔHf(l)=−57−(−68.3)=11.3\Delta H_{\text{vap}}(\text{H}_2\text{O}) = \Delta H_f(g) - \Delta H_f(l) = -57 - (-68.3) = 11.3 kcal mol⁻¹; for 9 g, 5.65 kcal.
  • Isothermal process of an ideal gas: ΔH=nCPΔT=0\Delta H = nC_P\Delta T = 0, whatever the volume change.

Enthalpy

H=U+PV,ΔH=qP,ΔHvap=qnH = U + PV,\qquad \Delta H = q_P,\qquad \Delta H_{\text{vap}} = \frac{q}{n}

Worked example

4.6 g of ethanol (M = 46) absorbs 4.3 kJ to vaporise at its boiling point. Find ΔH_vap, and the heat released when 23 g of ethanol vapour condenses.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Chemical Thermodynamics and EnergeticsEASY
Under similar conditions enthalpy of freezing is exactly opposite to

[Q86 · 15th May Shift 1 · 2023]

Dividing the heat by the grams

5.1 kJ / 13 g is a per-gram figure and matches no option. Convert to moles first: ΔH_vap is per MOLE.
Drill 7 more on enthalpy, heat at constant pressure and phase changes

Concept 2 of 2

ΔH = ΔU + Δn_g RT

Intuition

The PΔV term of a reaction at constant T is the gas made or consumed: PΔV = Δn_g RT. So ΔH exceeds ΔU when gas is produced, falls short when gas is consumed, and equals it when Δn_g = 0 — or at constant volume, where there is no PΔV at all. Solids and liquids contribute nothing.

Definition

  • ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT; ΔH−ΔU=ΔngRT\Delta H - \Delta U = \Delta n_g RT. Significant only for systems with GASES.
  • C3H8+5O2→3CO2+4H2O(l)\text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O}(l): Δng=3−6=−3\Delta n_g = 3 - 6 = -3, ΔH−ΔU=−3RT\Delta H - \Delta U = -3RT. 2C(s)+3H2→C2H62\text{C}(s) + 3\text{H}_2 \to \text{C}_2\text{H}_6: −2RT. CO+12O2→CO2\text{CO} + \tfrac{1}{2}\text{O}_2 \to \text{CO}_2: Δng=−12\Delta n_g = -\tfrac{1}{2}, so ΔH<ΔU\Delta H < \Delta U.
  • ΔH=ΔU\Delta H = \Delta U: constant volume, or Δng=0\Delta n_g = 0 — H2+Br2→2HBr\text{H}_2 + \text{Br}_2 \to 2\text{HBr}.
  • Numbers: NH2CN+32O2→N2+CO2+H2O(l)\text{NH}_2\text{CN} + \tfrac{3}{2}\text{O}_2 \to \text{N}_2 + \text{CO}_2 + \text{H}_2\text{O}(l), ΔU=−740.5\Delta U = -740.5 kJ, Δng=−12\Delta n_g = -\tfrac{1}{2}: ΔH=−740.5−0.5×8.314×10−3×298=−741.7\Delta H = -740.5 - 0.5 \times 8.314 \times 10^{-3} \times 298 = -741.7 kJ.
  • Reverse: ΔU=ΔH−ΔngRT\Delta U = \Delta H - \Delta n_g RT. OF2+H2O(g)→2HF+O2\text{OF}_2 + \text{H}_2\text{O}(g) \to 2\text{HF} + \text{O}_2: ΔH=−310\Delta H = -310 kJ, Δng=+1\Delta n_g = +1, ΔU=−310−2.5=−312.5\Delta U = -310 - 2.5 = -312.5 kJ.

ΔH and ΔU

ΔH=ΔU+Δng RT\Delta H = \Delta U + \Delta n_g\,RT

Worked example

For N2(g)+3H2(g)→2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g) at 300 K, ΔH = −92.0 kJ. Find ΔU.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Chemical Thermodynamics and EnergeticsEASY
For the reaction, C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O(l)} at constant temperature, ΔH−ΔU\Delta H - \Delta U is

[Q52 · 9th May Shift 1 · 2023]

Counting liquid water as a gas

H2O(l)\text{H}_2\text{O}(l) contributes nothing to Δng\Delta n_g. Propane combustion is −3RT because the four waters are liquid; counting them gives +1RT, an offered option.
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