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.

All Chemical Thermodynamics and Energetics notes MHT-CET Chemistry strategy

Why this matters

23 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

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

$H = U + PV$ ; at constant P, $q_P = \Delta H = \Delta U + P\Delta V$ . A gas doing 200 J of expansion work while U rises 432 J has $\Delta H = 632$ J.
$\Delta H_{\text{freezing}} = -\Delta H_{\text{fusion}}$ ; $\Delta H_{\text{condensation}} = -\Delta H_{\text{vaporisation}}$ . Melting ice is ENDOTHERMIC; freezing, condensation and deposition are exothermic.
$\Delta H_{\text{vap}}$ 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: $\Delta H_{\text{vap}}(\text{H}_2\text{O}) = \Delta H_f(g) - \Delta H_f(l) = -57 - (-68.3) = 11.3$ kcal mol⁻¹; for 9 g, 5.65 kcal.
Isothermal process of an ideal gas: $\Delta H = nC_P\Delta T = 0$ , whatever the volume change.

Enthalpy

H = 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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q88Easy

Example 1 · Chemical Thermodynamics and Energetics · Enthalpy and Relation Between ΔH and ΔU

Under similar conditions enthalpy of freezing is exactly opposite to

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

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

$\Delta H = \Delta U + \Delta n_g RT$ ; $\Delta H - \Delta U = \Delta n_g RT$ . Significant only for systems with GASES.
$\text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O}(l)$ : $\Delta n_g = 3 - 6 = -3$ , $\Delta H - \Delta U = -3RT$ . $2\text{C}(s) + 3\text{H}_2 \to \text{C}_2\text{H}_6$ : −2RT. $\text{CO} + \tfrac{1}{2}\text{O}_2 \to \text{CO}_2$ : $\Delta n_g = -\tfrac{1}{2}$ , so $\Delta H < \Delta U$ .
$\Delta H = \Delta U$ : constant volume, or $\Delta n_g = 0$ — $\text{H}_2 + \text{Br}_2 \to 2\text{HBr}$ .
Numbers: $\text{NH}_2\text{CN} + \tfrac{3}{2}\text{O}_2 \to \text{N}_2 + \text{CO}_2 + \text{H}_2\text{O}(l)$ , $\Delta U = -740.5$ kJ, $\Delta n_g = -\tfrac{1}{2}$ : $\Delta H = -740.5 - 0.5 \times 8.314 \times 10^{-3} \times 298 = -741.7$ kJ.
Reverse: $\Delta U = \Delta H - \Delta n_g RT$ . $\text{OF}_2 + \text{H}_2\text{O}(g) \to 2\text{HF} + \text{O}_2$ : $\Delta H = -310$ kJ, $\Delta n_g = +1$ , $\Delta U = -310 - 2.5 = -312.5$ kJ.

ΔH and ΔU

\Delta H = \Delta U + \Delta n_g\,RT

Worked example

For

\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)

at 300 K, ΔH = −92.0 kJ. Find ΔU.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q52Easy

Example 2 · Chemical Thermodynamics and Energetics · Enthalpy and Relation Between ΔH and ΔU

For the reaction,

\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,

\Delta H - \Delta U

Counting liquid water as a gas

\text{H}_2\text{O}(l)

contributes nothing to

\Delta n_g

. Propane combustion is −3RT because the four waters are liquid; counting them gives +1RT, an offered option.

Drill 9 more on δh = δu + δn_g rt

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Enthalpy, Heat at Constant Pressure and Phase Changes
Enthalpy
$H = U + PV,\qquad \Delta H = q_P,\qquad \Delta H_{\text{vap}} = \frac{q}{n}$
ΔH = ΔU + Δn_g RT
ΔH and ΔU
$\Delta H = \Delta U + \Delta n_g\,RT$

Watch out for (2)

Dividing the heat by the grams
→ Enthalpy, Heat at Constant Pressure and Phase Changes
Counting liquid water as a gas
→ ΔH = ΔU + Δn_g RT

Test yourself on Chemical Thermodynamics and Energetics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Chemical Thermodynamics and Energetics chapter test Drill all 23 questions