Chapter 5 · Thermodynamics

5.2 Enthalpy, Heat Capacity and Calorimetry

Most chemistry happens in open flasks at constant atmospheric pressure, where a reacting system may push back the air as well as give out heat. Enthalpy, $H=U+pV$, is the quantity whose change is exactly the heat exchanged at constant pressure. This section relates $\Delta H$ to $\Delta U$, introduces heat capacities and shows how calorimeters measure both.

Burn a spoonful of fuel in a sealed steel container and the heat you collect is the change in internal energy, $\Delta U$ — the container cannot expand, so no work is done. Burn the same fuel in an open dish and the products are free to push back the atmosphere. Now part of the energy may leave as work and only the rest as heat. Since almost every reaction in a laboratory, a kitchen or a living cell happens at constant pressure rather than constant volume, chemists use a quantity designed for those conditions: the enthalpy.

In this section you will meet enthalpy and its relationship with internal energy, learn how the heat needed to warm a substance is described by its heat capacity, and see how two kinds of calorimeter — a sealed "bomb" and a simple insulated cup — measure $\Delta U$ and $\Delta H$. These measurements are the source of all the thermochemical data used in Section 5.3.

Enthalpy: Heat at Constant Pressure

At constant volume no pressure–volume work is done, so the first law gives $\Delta U=q_{V}$: the heat absorbed at constant volume equals the change in internal energy. At constant pressure $p$, a system whose volume changes from $V_{i}$ to $V_{f}$ has $w=-p\Delta V$, so $\Delta U=q_{p}-p\Delta V$, or $$q_{p}=\Delta U+p\Delta V=(U_{f}+pV_{f})-(U_{i}+pV_{i}).$$ The heat at constant pressure is the change in the combination $U+pV$. Since $U$, $p$ and $V$ are all state functions, so is this combination.

📘 Definition 5.8 — Enthalpy
The enthalpy of a system is $$H=U+pV.$$ It is a state function and an extensive property, measured in joules. Its change is the heat absorbed by the system at constant pressure (when only pressure–volume work is possible): $\Delta H=q_{p}$. A process with $\Delta H\lt0$ gives out heat at constant pressure and is exothermic; a process with $\Delta H\gt0$ absorbs heat and is endothermic.

When a reaction involves only solids and liquids, the volume hardly changes, $p\Delta V$ is tiny, and $\Delta H\approx\Delta U$. When gases are produced or consumed, the volume change can be large. If $n_{A}$ mol of gaseous reactants become $n_{B}$ mol of gaseous products at constant $T$ and $p$, the ideal gas equation gives $p\Delta V=(n_{B}-n_{A})RT=\Delta n_{g}RT$.

📐 Rule 5.3 — ΔH and ΔU
  1. At constant volume $q_{V}=\Delta U$; at constant pressure $q_{p}=\Delta H$, and $\Delta H=\Delta U+p\Delta V$.
  2. For a reaction at constant temperature $T$ in which the gases behave ideally, $$\Delta H=\Delta U+\Delta n_{g}RT$$ where $\Delta n_{g}$ = (moles of gaseous products) − (moles of gaseous reactants) in the equation as written. Solids and liquids are not counted.
  3. If $\Delta n_{g}=0$, then $\Delta H=\Delta U$; if $\Delta n_{g}\gt0$, $\Delta H\gt\Delta U$; if $\Delta n_{g}\lt0$, $\Delta H\lt\Delta U$.
⚠️ Common slips with ΔngRT
(1) Count only the gases — $\mathrm{H_{2}O(l)}$ and $\mathrm{CaCO_{3}(s)}$ contribute nothing. Read the state symbols. (2) Use $R$ in the same energy unit as $\Delta H$: $R=8.314\times10^{-3}\ \mathrm{kJ\,mol^{-1}\,K^{-1}}$ for kJ; at $298\ \mathrm{K}$, $RT=2.478\ \mathrm{kJ\,mol^{-1}}$. Mixing J and kJ gives answers that are wrong by a factor of 1000. (3) $\Delta n_{g}$ belongs to the equation as written: doubling the equation doubles $\Delta H$, $\Delta U$ and $\Delta n_{g}$.
✏️ Example 1 — Is ΔH larger or smaller than ΔU?
Find $\Delta n_{g}$ and decide whether $\Delta H$ is greater than, less than or equal to $\Delta U$ at constant temperature for: (a) $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)}$; (b) $\mathrm{C(s)+O_{2}(g)\rightarrow CO_{2}(g)}$; (c) $\mathrm{CH_{4}(g)+2O_{2}(g)\rightarrow CO_{2}(g)+2H_{2}O(l)}$; (d) $\mathrm{PCl_{5}(g)\rightarrow PCl_{3}(g)+Cl_{2}(g)}$. (e) For (a), find $\Delta H-\Delta U$ at $298\ \mathrm{K}$.
(a), (b) (a) $\Delta n_{g}=2-4=-2$: $\Delta H\lt\Delta U$. (b) $\Delta n_{g}=1-1=0$ (carbon is a solid): $\Delta H=\Delta U$.
(c) Gaseous products: $1$ ($\mathrm{CO_{2}}$; the water is liquid). Gaseous reactants: $3$. $\Delta n_{g}=-2$: $\Delta H\lt\Delta U$.
(d) $\Delta n_{g}=2-1=+1$: $\Delta H\gt\Delta U$ — the system must also push back the atmosphere to make room for the extra gas.
(e) $\Delta H-\Delta U=\Delta n_{g}RT=(-2)(8.314\times10^{-3})(298)=-4.96\ \mathrm{kJ}$ per mole of reaction as written.
✏️ Example 2 — Converting between ΔH and ΔU
(a) The enthalpy of vaporisation of water at $373\ \mathrm{K}$ and $1\ \mathrm{atm}$ is $40.66\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta U$ for $\mathrm{H_{2}O(l)\rightarrow H_{2}O(g)}$ at $373\ \mathrm{K}$, treating the vapour as ideal. (b) For the combustion of liquid ethanol at $298\ \mathrm{K}$, $\mathrm{C_{2}H_{5}OH(l)+3O_{2}(g)\rightarrow 2CO_{2}(g)+3H_{2}O(l)}$, a bomb calorimeter gives $\Delta U=-1364.2\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta H$.
(a) $\Delta n_{g}=1-0=+1$. $\Delta U=\Delta H-\Delta n_{g}RT=40.66-(1)(8.314\times10^{-3})(373)=40.66-3.10=37.56\ \mathrm{kJ\,mol^{-1}}$.
Meaning Of the $40.66\ \mathrm{kJ}$ absorbed, $37.56\ \mathrm{kJ}$ separates the molecules (raises $U$) and $3.10\ \mathrm{kJ}$ is spent pushing back the atmosphere as the vapour forms.
(b) $\Delta n_{g}=2-3=-1$. $\Delta H=\Delta U+\Delta n_{g}RT=-1364.2+(-1)(2.478)=-1366.7\ \mathrm{kJ\,mol^{-1}}$.
Check The gas volume shrinks, the atmosphere does work on the system, and so slightly more heat is given out at constant pressure than at constant volume: $|\Delta H|\gt|\Delta U|$ ✓.

Heat Capacity

Supply heat to a substance and its temperature rises. For a small temperature change the rise is proportional to the heat: $q=C\,\Delta T$. The constant $C$ depends on the substance, on its amount and on the conditions — whether it is heated at constant volume or at constant pressure.

📘 Definition 5.9 — Heat capacity, specific and molar heat capacity
  • The heat capacity $C$ of a body is the heat needed to raise its temperature by $1\ \mathrm{K}$: $C=q/\Delta T$, unit $\mathrm{J\,K^{-1}}$ (extensive).
  • The specific heat capacity $c$ is the heat capacity per unit mass, unit $\mathrm{J\,g^{-1}\,K^{-1}}$ (intensive); $q=mc\,\Delta T$. For liquid water $c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$.
  • The molar heat capacity $C_{m}=C/n$ is the heat capacity per mole, unit $\mathrm{J\,mol^{-1}\,K^{-1}}$ (intensive); $q=nC_{m}\,\Delta T$.
  • Measured at constant volume they are written $C_{V}$ (or $C_{v}$), and at constant pressure $C_{p}$: $\;q_{V}=C_{V}\Delta T=\Delta U$, $\;q_{p}=C_{p}\Delta T=\Delta H$.

Heated at constant pressure, a gas expands and does work on its surroundings, so part of the heat supplied leaves again as work and the temperature rises less. More heat is needed for the same $\Delta T$: $C_{p}\gt C_{V}$. For an ideal gas the difference is exactly calculable.

📐 Rule 5.4 — Cp − Cv = R for an ideal gas
For one mole of an ideal gas, $H=U+pV=U+RT$, so for a temperature change $\Delta T$ $$\Delta H=\Delta U+R\,\Delta T.$$ Substituting $\Delta H=C_{p}\Delta T$ and $\Delta U=C_{v}\Delta T$ (molar values) and dividing by $\Delta T$: $$C_{p}-C_{v}=R$$ For a monatomic ideal gas such as helium or argon, $C_{v}=\tfrac32R=12.47\ \mathrm{J\,mol^{-1}\,K^{-1}}$ and $C_{p}=\tfrac52R=20.79\ \mathrm{J\,mol^{-1}\,K^{-1}}$ (measured values agree closely). For solids and liquids $C_{p}\approx C_{v}$.
✏️ Example 3 — Heating water
(a) How much heat is needed to warm $250\ \mathrm{g}$ of water from $25.0\ ^{\circ}\mathrm{C}$ to $80.0\ ^{\circ}\mathrm{C}$? ($c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$) (b) Find the molar heat capacity of water ($M=18.02\ \mathrm{g\,mol^{-1}}$) and repeat (a) with it.
(a) $\Delta T=80.0-25.0=55.0\ \mathrm{K}$ (a temperature difference is the same in K and °C). $q=mc\Delta T=250\times4.18\times55.0=5.75\times10^{4}\ \mathrm{J}=57.5\ \mathrm{kJ}$.
(b) $C_{m}=cM=4.18\times18.02=75.3\ \mathrm{J\,mol^{-1}\,K^{-1}}$. $n=250/18.02=13.87\ \mathrm{mol}$; $q=13.87\times75.3\times55.0=5.75\times10^{4}\ \mathrm{J}$ ✓.
✏️ Example 4 — Heating a gas at constant volume and at constant pressure
$2.00\ \mathrm{mol}$ of argon (an ideal monatomic gas) is warmed from $300\ \mathrm{K}$ to $350\ \mathrm{K}$, (a) in a rigid container; (b) at constant pressure. Find $q$, $w$, $\Delta U$ and $\Delta H$ in each case.
State functions $\Delta U$ and $\Delta H$ depend only on the temperature change for an ideal gas, so they are the same in (a) and (b): $\Delta U=nC_{v}\Delta T=2.00\times12.47\times50=1247\ \mathrm{J}$; $\Delta H=nC_{p}\Delta T=2.00\times20.79\times50=2079\ \mathrm{J}$.
(a) Constant volume: $w=0$, $q_{V}=\Delta U=+1247\ \mathrm{J}$ ($1.25\ \mathrm{kJ}$).
(b) Constant pressure: $q_{p}=\Delta H=+2079\ \mathrm{J}$ ($2.08\ \mathrm{kJ}$); $w=\Delta U-q=1247-2079=-832\ \mathrm{J}$ (the gas does work as it expands).
Check $w=-p\Delta V=-nR\Delta T=-2.00\times8.314\times50=-831\ \mathrm{J}$ ✓ (the 1 J difference is rounding).
Interactive 1 — Same Heat, Two Containers: Why Cₚ > Cᵥ

Two identical samples of $1.00\ \mathrm{mol}$ of an ideal gas start at $300\ \mathrm{K}$. The left one is in a rigid box (constant volume); the right one is under a free piston (constant pressure). Both receive the same heat $q$. Press ▶ Play to feed the heat in gradually. On the right some of the heat leaves again as work lifting the piston, so the temperature rises less; the extra heat per kelvin is exactly $R$.

1500 J
ΔT, constant V—
ΔT, constant p—
work by gas (const p)—
q/ΔT_p − q/ΔT_V—

Calorimetry: Measuring ΔU and ΔH

A calorimeter measures the heat of a process from the temperature change it causes in a surrounding body of known heat capacity — usually water and the vessel itself. The heat given out by the reaction is taken up by the calorimeter: $q_{\text{reaction}}=-q_{\text{calorimeter}}=-C_{cal}\Delta T$.

📘 Definition 5.10 — Bomb and coffee-cup calorimeters
  • A bomb calorimeter is a sealed steel vessel (the "bomb") in which a weighed sample is burned in excess oxygen, immersed in a known mass of stirred water inside an insulated jacket. The volume is constant, so the heat measured is $q_{V}=\Delta U$. Its heat capacity $C_{cal}$ (bomb + water) is found by calibration, usually by burning a known mass of benzoic acid.
  • A constant-pressure (coffee-cup) calorimeter is an insulated vessel open to the atmosphere, often two nested polystyrene cups, used for reactions in solution (neutralisation, dissolution). The pressure is constant, so the heat measured is $q_{p}=\Delta H$. Usually the solution is treated as water: $q=mc\Delta T$ with $c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$.
🧭 Problem-solving strategy — Calorimetry calculations
  1. Heat absorbed by the calorimeter: $q_{cal}=C_{cal}\Delta T$ (bomb) or $q_{cal}=m_{\text{soln}}c\,\Delta T$ (cup). A temperature rise means the reaction released heat.
  2. Heat of the reaction: $q_{r}=-q_{cal}$ (negative for an exothermic reaction).
  3. Divide by the amount that reacted to get the molar value: $\Delta U=q_{r}/n$ (bomb) or $\Delta H=q_{r}/n$ (cup), in $\mathrm{kJ\,mol^{-1}}$.
  4. If you need the other quantity, use $\Delta H=\Delta U+\Delta n_{g}RT$ with the balanced equation.
✏️ Example 5 — Calibrating and using a bomb calorimeter
(a) Burning $1.000\ \mathrm{g}$ of benzoic acid ($\Delta_{c}U=-26.43\ \mathrm{kJ\,g^{-1}}$) in a bomb calorimeter raises its temperature by $2.62\ \mathrm{K}$. Find the heat capacity of the calorimeter. (b) In the same calorimeter $0.600\ \mathrm{g}$ of graphite is burned completely: $\mathrm{C(s)+O_{2}(g)\rightarrow CO_{2}(g)}$. The temperature rises by $1.95\ \mathrm{K}$. Find $\Delta U$ and $\Delta H$ for the reaction per mole of carbon ($M=12.01\ \mathrm{g\,mol^{-1}}$).
(a) Heat released $=26.43\ \mathrm{kJ}$, all absorbed by the calorimeter: $C_{cal}=\dfrac{26.43\ \mathrm{kJ}}{2.62\ \mathrm{K}}=10.09\ \mathrm{kJ\,K^{-1}}$.
(b) heat $q_{cal}=10.09\times1.95=19.67\ \mathrm{kJ}$, so $q_{r}=-19.67\ \mathrm{kJ}$ for $0.600\ \mathrm{g}$.
(b) per mole $n=0.600/12.01=0.04996\ \mathrm{mol}$; $\Delta U=\dfrac{-19.67}{0.04996}=-394\ \mathrm{kJ\,mol^{-1}}$ (3 s.f.).
ΔH $\Delta n_{g}=1-1=0$, so $\Delta H=\Delta U=-394\ \mathrm{kJ\,mol^{-1}}$ — close to the accepted $-393.5\ \mathrm{kJ\,mol^{-1}}$.
✏️ Example 6 — A coffee-cup calorimeter: enthalpy of neutralisation
$50.0\ \mathrm{mL}$ of $1.00\ \mathrm{M}$ hydrochloric acid and $50.0\ \mathrm{mL}$ of $1.00\ \mathrm{M}$ sodium hydroxide, both at $25.0\ ^{\circ}\mathrm{C}$, are mixed in a polystyrene cup. The temperature rises to $31.7\ ^{\circ}\mathrm{C}$. Taking the density of the solution as $1.00\ \mathrm{g\,mL^{-1}}$, its specific heat capacity as $4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$ and ignoring the heat absorbed by the cup, find $\Delta H$ for $\mathrm{H^{+}(aq)+OH^{-}(aq)\rightarrow H_{2}O(l)}$ per mole of water formed.
Heat Mass of solution $=100.0\ \mathrm{g}$; $\Delta T=6.7\ \mathrm{K}$; $q_{cal}=100.0\times4.18\times6.7=2.80\times10^{3}\ \mathrm{J}$.
Amount $n(\mathrm{H^{+}})=n(\mathrm{OH^{-}})=0.0500\ \mathrm{L}\times1.00\ \mathrm{mol\,L^{-1}}=0.0500\ \mathrm{mol}$ → $0.0500\ \mathrm{mol}$ of water.
ΔH $\Delta H=\dfrac{-2.80\ \mathrm{kJ}}{0.0500\ \mathrm{mol}}=-56\ \mathrm{kJ\,mol^{-1}}$ (2 s.f., limited by $\Delta T=6.7\ \mathrm{K}$).
Check The accepted value for a strong acid with a strong base is about $-57\ \mathrm{kJ\,mol^{-1}}$; our value is slightly smaller in magnitude because the cup and thermometer also absorb a little heat.
Interactive 2 — Calorimeter Lab: Bomb (ΔU) vs Coffee-Cup (ΔH)

Choose an experiment and the amount of sample, then press ▶ Play to run it: the thermometer settles at the final temperature. The lab turns $\Delta T$ into the heat absorbed by the calorimeter, then into the molar quantity it measures — $\Delta U$ for the sealed bomb ($C_{cal}=10.0\ \mathrm{kJ\,K^{-1}}$), $\Delta H$ for the open cup ($100\ \mathrm{g}$ of solution, $c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$) — and converts it with $\Delta H=\Delta U+\Delta n_{g}RT$ at $298\ \mathrm{K}$.

—
ΔT—
heat absorbed by calorimeter—
measured—
converted—

Exercises

I
Enthalpy, ΔH and ΔU. Use $R=8.314\times10^{-3}\ \mathrm{kJ\,mol^{-1}\,K^{-1}}$ and treat gases as ideal.
  • 1 Define enthalpy. Explain why it is a state function and why $\Delta H=q_{p}$.
  • 2 Find $\Delta n_{g}$ for: (a) $\mathrm{2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)}$; (b) $\mathrm{CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)}$; (c) $\mathrm{2SO_{2}(g)+O_{2}(g)\rightarrow 2SO_{3}(g)}$; (d) $\mathrm{H_{2}(g)+Cl_{2}(g)\rightarrow 2HCl(g)}$.
  • 3 For which reactions in part 2 is $\Delta H\gt\Delta U$, and for which is $\Delta H=\Delta U$?
  • 4 For the decomposition of calcium carbonate (part 2b), $\Delta H=+178.3\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$. Find $\Delta U$.
  • 5 For reaction 2a, $\Delta H=-571.6\ \mathrm{kJ}$ at $298\ \mathrm{K}$. Find $\Delta U$.
  • 6 For $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)}$, $\Delta U=-87.2\ \mathrm{kJ}$ at $298\ \mathrm{K}$. Find $\Delta H$.
  • 7 Explain why $\Delta H\approx\Delta U$ for reactions that involve only solids and liquids.
  • 8 For $\mathrm{I_{2}(s)\rightarrow I_{2}(g)}$ at $298\ \mathrm{K}$, $\Delta H=62.4\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta U$.
1 $H=U+pV$. $U$, $p$ and $V$ are state functions, so $H$ is too. At constant $p$ with only pV work: $\Delta U=q_{p}-p\Delta V$, so $q_{p}=\Delta U+p\Delta V=\Delta(U+pV)=\Delta H$.
2 (a) $0-3=-3$; (b) $1-0=+1$; (c) $2-3=-1$; (d) $2-2=0$.
3 $\Delta H\gt\Delta U$: (b) only. $\Delta H=\Delta U$: (d). (a) and (c) have $\Delta H\lt\Delta U$.
4 $\Delta U=178.3-(1)(2.478)=+175.8\ \mathrm{kJ\,mol^{-1}}$.
5 $\Delta U=\Delta H-\Delta n_{g}RT=-571.6-(-3)(2.478)=-571.6+7.43=-564.2\ \mathrm{kJ}$.
6 $\Delta H=-87.2+(-2)(2.478)=-92.2\ \mathrm{kJ}$.
7 Their molar volumes are small and change very little, so $p\Delta V$ is negligible compared with $\Delta U$ (typically a few joules against kilojoules).
8 $\Delta n_{g}=+1$: $\Delta U=62.4-2.48=59.9\ \mathrm{kJ\,mol^{-1}}$.
II
Heat capacity. Use $c(\text{water})=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$, $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$.
  • 1 Define heat capacity, specific heat capacity and molar heat capacity, with units. Which are intensive?
  • 2 How much heat is needed to warm $500\ \mathrm{g}$ of water from $20.0\ ^{\circ}\mathrm{C}$ to $100.0\ ^{\circ}\mathrm{C}$?
  • 3 A $200\ \mathrm{g}$ copper block absorbs $3.08\ \mathrm{kJ}$ and warms by $40.0\ \mathrm{K}$. Find the specific heat capacity of copper.
  • 4 Find the molar heat capacity of liquid water ($M=18.02\ \mathrm{g\,mol^{-1}}$).
  • 5 A $150\ \mathrm{g}$ iron bolt at $95.0\ ^{\circ}\mathrm{C}$ ($c=0.449\ \mathrm{J\,g^{-1}\,K^{-1}}$) is dropped into $200\ \mathrm{g}$ of water at $25.0\ ^{\circ}\mathrm{C}$ in an insulated cup. Find the final temperature.
  • 6 The molar heat capacity of nitrogen at constant pressure is $29.1\ \mathrm{J\,mol^{-1}\,K^{-1}}$. Estimate $C_{v}$.
  • 7 $2.00\ \mathrm{mol}$ of helium ($C_{v}=12.47$, $C_{p}=20.79\ \mathrm{J\,mol^{-1}\,K^{-1}}$) is heated from $300\ \mathrm{K}$ to $400\ \mathrm{K}$ at constant pressure. Find $q$, $\Delta U$, $\Delta H$ and $w$.
  • 8 Explain, without equations, why $C_{p}\gt C_{v}$ for a gas.
1 $C=q/\Delta T$ ($\mathrm{J\,K^{-1}}$); $c=C/m$ ($\mathrm{J\,g^{-1}\,K^{-1}}$); $C_{m}=C/n$ ($\mathrm{J\,mol^{-1}\,K^{-1}}$). The specific and molar heat capacities are intensive.
2 $q=500\times4.18\times80.0=1.67\times10^{5}\ \mathrm{J}=167\ \mathrm{kJ}$.
3 $c=\dfrac{3080}{200\times40.0}=0.385\ \mathrm{J\,g^{-1}\,K^{-1}}$.
4 $C_{m}=4.18\times18.02=75.3\ \mathrm{J\,mol^{-1}\,K^{-1}}$.
5 Heat lost by iron = heat gained by water: $150(0.449)(95.0-T)=200(4.18)(T-25.0)$ → $67.35(95.0-T)=836(T-25.0)$ → $T=\dfrac{6398+20\,900}{903.4}=30.2\ ^{\circ}\mathrm{C}$.
6 $C_{v}=C_{p}-R=29.1-8.3=20.8\ \mathrm{J\,mol^{-1}\,K^{-1}}$.
7 $q=\Delta H=2.00\times20.79\times100=4158\ \mathrm{J}$ ($4.16\ \mathrm{kJ}$); $\Delta U=2.00\times12.47\times100=2494\ \mathrm{J}$; $w=\Delta U-q=-1664\ \mathrm{J}$ ($=-nR\Delta T=-1663\ \mathrm{J}$ to rounding).
8 At constant pressure the gas expands as it warms and does work on the surroundings; that energy is not available to raise the temperature, so more heat is needed per kelvin.
III
Calorimetry. Take the density of dilute solutions as $1.00\ \mathrm{g\,mL^{-1}}$, $c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$, and ignore the heat capacity of the cup unless told otherwise.
  • 1 Which quantity, $\Delta U$ or $\Delta H$, does each type of calorimeter measure? Why?
  • 2 $1.000\ \mathrm{g}$ of benzoic acid ($\Delta_{c}U=-26.4\ \mathrm{kJ\,g^{-1}}$) raises the temperature of a bomb calorimeter by $2.64\ \mathrm{K}$. Find $C_{cal}$.
  • 3 In the calorimeter of part 2, $0.500\ \mathrm{g}$ of sucrose ($\mathrm{C_{12}H_{22}O_{11}}$, $M=342.3\ \mathrm{g\,mol^{-1}}$) raises the temperature by $0.825\ \mathrm{K}$. Find $\Delta_{c}U$ per mole. Is $\Delta_{c}H$ different? ($\mathrm{C_{12}H_{22}O_{11}(s)+12O_{2}(g)\rightarrow 12CO_{2}(g)+11H_{2}O(l)}$)
  • 4 $2.00\ \mathrm{g}$ of solid NaOH ($M=40.00\ \mathrm{g\,mol^{-1}}$) dissolves in $100.0\ \mathrm{g}$ of water and the temperature rises by $5.22\ \mathrm{K}$. Find the molar enthalpy of solution (mass of solution $102\ \mathrm{g}$).
  • 5 $100\ \mathrm{mL}$ of $0.500\ \mathrm{M}$ HCl and $100\ \mathrm{mL}$ of $0.500\ \mathrm{M}$ NaOH, both at $24.0\ ^{\circ}\mathrm{C}$, are mixed; the temperature reaches $27.4\ ^{\circ}\mathrm{C}$. Find the enthalpy of neutralisation per mole of water.
  • 6 When ethanoic acid (a weak acid) is neutralised by NaOH, the enthalpy of neutralisation is less negative than $-57\ \mathrm{kJ\,mol^{-1}}$. Suggest why.
  • 7 $0.0400\ \mathrm{mol}$ of CO is burned in a bomb calorimeter ($C_{cal}=12.0\ \mathrm{kJ\,K^{-1}}$) and the temperature rises by $0.939\ \mathrm{K}$. For $\mathrm{2CO(g)+O_{2}(g)\rightarrow 2CO_{2}(g)}$ find $\Delta U$ and $\Delta H$ at $298\ \mathrm{K}$.
  • 8 Applied. $1.00\ \mathrm{g}$ of a dried cereal bar burns in a bomb calorimeter ($C_{cal}=10.0\ \mathrm{kJ\,K^{-1}}$) and the temperature rises by $2.05\ \mathrm{K}$. Find the energy released per gram, in kJ and in kcal ($1\ \mathrm{kcal}=4.184\ \mathrm{kJ}$), and per $40\ \mathrm{g}$ bar.
1 Bomb: sealed, constant volume, no pV work, so $q_{V}=\Delta U$. Coffee cup: open to the atmosphere, constant pressure, so $q_{p}=\Delta H$.
2 $C_{cal}=26.4/2.64=10.0\ \mathrm{kJ\,K^{-1}}$.
3 $q_{cal}=10.0\times0.825=8.25\ \mathrm{kJ}$; $n=0.500/342.3=1.461\times10^{-3}\ \mathrm{mol}$; $\Delta_{c}U=-8.25/1.461\times10^{-3}=-5.65\times10^{3}\ \mathrm{kJ\,mol^{-1}}$. $\Delta n_{g}=12-12=0$, so $\Delta_{c}H=\Delta_{c}U$.
4 $q=102\times4.18\times5.22=2226\ \mathrm{J}$; $n=0.0500\ \mathrm{mol}$; $\Delta_{sol}H=-2.226/0.0500=-44.5\ \mathrm{kJ\,mol^{-1}}$.
5 $q=200\times4.18\times3.4=2842\ \mathrm{J}$; $n(\mathrm{H_{2}O})=0.100\times0.500=0.0500\ \mathrm{mol}$; $\Delta H=-2.842/0.0500=-57\ \mathrm{kJ\,mol^{-1}}$ (2 s.f.).
6 The weak acid is only slightly ionised; some of the heat released by $\mathrm{H^{+}+OH^{-}}$ is used to complete the ionisation of the acid (an endothermic step), so less heat is given out overall.
7 $q_{cal}=12.0\times0.939=11.27\ \mathrm{kJ}$ for $0.0400\ \mathrm{mol}$ CO $=0.0200\ \mathrm{mol}$ of reaction as written. $\Delta U=-11.27/0.0200=-563\ \mathrm{kJ}$. $\Delta n_{g}=2-3=-1$: $\Delta H=-563.4-2.48=-566\ \mathrm{kJ}$.
8 $q=10.0\times2.05=20.5\ \mathrm{kJ}$ per gram $=20.5/4.184=4.90\ \mathrm{kcal\,g^{-1}}$. A $40\ \mathrm{g}$ bar: $820\ \mathrm{kJ}\approx196\ \mathrm{kcal}$ (nutrition labels write this as 196 "Calories").
IV
Putting it together (harder).
  • 1 Starting from the first law, show that the heat absorbed at constant pressure equals $\Delta H$. State the assumption about the kinds of work.
  • 2 Derive $\Delta H=\Delta U+\Delta n_{g}RT$ and list its assumptions.
  • 3 $1.00\ \mathrm{mol}$ of an ideal monatomic gas is heated at constant pressure from $300\ \mathrm{K}$ to $500\ \mathrm{K}$. Find $\Delta H$, $\Delta U$, $q$ and $w$, and check that $\Delta U=q+w$.
  • 4 $1.00\ \mathrm{kJ}$ of heat is given to $1.00\ \mathrm{mol}$ of nitrogen ($C_{v}=20.8$, $C_{p}=29.1\ \mathrm{J\,mol^{-1}\,K^{-1}}$), once at constant volume and once at constant pressure. Find the temperature rise in each case.
  • 5 At what temperature is $\Delta H-\Delta U=10.0\ \mathrm{kJ\,mol^{-1}}$ for a reaction with $\Delta n_{g}=+1$?
  • 6 Why must the heat capacity of a bomb calorimeter be found by calibration rather than calculated from the mass of water in it?
1 Only pV work: $w=-p\Delta V$, so $\Delta U=q_{p}-p\Delta V$ and $q_{p}=(U_{f}+pV_{f})-(U_{i}+pV_{i})=H_{f}-H_{i}=\Delta H$. (If other work, e.g. electrical, is done, $q_{p}\ne\Delta H$.)
2 $\Delta H=\Delta U+p\Delta V$ at constant $p$. With ideal gases at constant $T$: $pV_{A}=n_{A}RT$, $pV_{B}=n_{B}RT$, so $p\Delta V=(n_{B}-n_{A})RT=\Delta n_{g}RT$. Assumptions: constant $T$ and $p$; ideal gases; volumes of solids and liquids neglected.
3 $\Delta H=20.79\times200=4158\ \mathrm{J}=q$; $\Delta U=12.47\times200=2494\ \mathrm{J}$; $w=-R\Delta T=-8.314\times200=-1663\ \mathrm{J}$. Check: $4158-1663=2495\approx2494\ \mathrm{J}$ ✓ (rounding).
4 Constant V: $\Delta T=1000/20.8=48.1\ \mathrm{K}$. Constant p: $\Delta T=1000/29.1=34.4\ \mathrm{K}$.
5 $\Delta n_{g}RT=10.0\ \mathrm{kJ\,mol^{-1}}$ → $T=\dfrac{10\,000}{8.314}=1.20\times10^{3}\ \mathrm{K}$.
6 The steel bomb, the stirrer, the thermometer and the container also absorb heat, and some heat leaks; burning a standard of known $\Delta_{c}U$ measures the effective heat capacity of the whole assembly at once.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. NEET For which reaction is $\Delta H=\Delta U$ at constant temperature? (A) $\mathrm{H_{2}(g)+I_{2}(g)\rightarrow 2HI(g)}$ (B) $\mathrm{PCl_{5}(g)\rightarrow PCl_{3}(g)+Cl_{2}(g)}$ (C) $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)}$ (D) $\mathrm{2SO_{2}(g)+O_{2}(g)\rightarrow 2SO_{3}(g)}$

Idea $\Delta H=\Delta U$ when $\Delta n_{g}=0$. (A) $2-2=0$; (B) $+1$; (C) $-2$; (D) $-1$.
Answer (A).

E2. JEE Main Numerical answer. For the combustion of liquid benzene at $298\ \mathrm{K}$, $\mathrm{C_{6}H_{6}(l)+\tfrac{15}{2}O_{2}(g)\rightarrow 6CO_{2}(g)+3H_{2}O(l)}$, $\Delta U=-3263.9\ \mathrm{kJ\,mol^{-1}}$. The magnitude of $\Delta H$, in $\mathrm{kJ\,mol^{-1}}$, to the nearest integer is ____ ($R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$).

Δng $6-7.5=-1.5$ (benzene and water are liquids).
Compute $\Delta H=-3263.9+(-1.5)(8.314\times10^{-3})(298)=-3263.9-3.72=-3267.6\ \mathrm{kJ\,mol^{-1}}$.
Answer 3268.

E3. Board · 3 marks Show that for one mole of an ideal gas $C_{p}-C_{v}=R$.

1 mark $q_{V}=C_{v}\Delta T=\Delta U$ and $q_{p}=C_{p}\Delta T=\Delta H$.
1 mark For one mole of an ideal gas $H=U+pV=U+RT$, so $\Delta H=\Delta U+R\Delta T$.
1 mark Substitute: $C_{p}\Delta T=C_{v}\Delta T+R\Delta T$; divide by $\Delta T$: $C_{p}-C_{v}=R$.

E4. JEE Main $8.00\ \mathrm{g}$ of ammonium nitrate ($M=80.04\ \mathrm{g\,mol^{-1}}$, $\Delta_{sol}H=+25.7\ \mathrm{kJ\,mol^{-1}}$) dissolves in water in a coffee-cup calorimeter whose total heat capacity, with the solution, is $450\ \mathrm{J\,K^{-1}}$. The temperature (A) falls by $5.7\ \mathrm{K}$ (B) rises by $5.7\ \mathrm{K}$ (C) falls by $0.057\ \mathrm{K}$ (D) falls by $57\ \mathrm{K}$

Heat $n=8.00/80.04=0.09995\ \mathrm{mol}$; heat absorbed by the dissolving salt $=0.09995\times25.7=2.57\ \mathrm{kJ}$, taken from the solution.
ΔT $\Delta T=-\dfrac{2570\ \mathrm{J}}{450\ \mathrm{J\,K^{-1}}}=-5.71\ \mathrm{K}$: endothermic, so the solution cools.
Answer (A). (This is how an instant cold pack works.)

E5. JEE Advanced One or more options correct. (A) A bomb calorimeter measures $\Delta U$ of a combustion. (B) For a gas-phase reaction with $\Delta n_{g}\lt0$, $\Delta H\lt\Delta U$ at constant temperature. (C) For every ideal gas the molar $C_{p}$ exceeds the molar $C_{v}$ by $R$. (D) Enthalpy is an intensive property.

(A) True: constant volume, so $q_{V}=\Delta U$.
(B) True: $\Delta H-\Delta U=\Delta n_{g}RT\lt0$.
(C) True: Rule 5.4 does not depend on the kind of molecule, only on ideal behaviour.
(D) False: $H=U+pV$ doubles when the system is doubled — extensive (molar enthalpy is intensive).
Answer (A), (B), (C).

E6. Board · Case study Naphthalene, $\mathrm{C_{10}H_{8}}$ ($M=128.17\ \mathrm{g\,mol^{-1}}$), burns as $\mathrm{C_{10}H_{8}(s)+12O_{2}(g)\rightarrow 10CO_{2}(g)+4H_{2}O(l)}$. A $0.640\ \mathrm{g}$ sample is burned in a bomb calorimeter of heat capacity $10.0\ \mathrm{kJ\,K^{-1}}$ at about $298\ \mathrm{K}$, and the temperature rises by $2.57\ \mathrm{K}$. (i) Calculate $\Delta_{c}U$ per mole. (2 marks) (ii) Find $\Delta n_{g}$ for the reaction. (1 mark) (iii) Calculate $\Delta_{c}H$. (1 mark)

(i) $q_{cal}=10.0\times2.57=25.7\ \mathrm{kJ}$; $n=0.640/128.17=4.993\times10^{-3}\ \mathrm{mol}$; $\Delta_{c}U=-25.7/4.993\times10^{-3}=-5.15\times10^{3}\ \mathrm{kJ\,mol^{-1}}$ ($-5147$ before rounding).
(ii) $\Delta n_{g}=10-12=-2$.
(iii) $\Delta_{c}H=\Delta_{c}U+\Delta n_{g}RT=-5147+(-2)(2.478)=-5152\ \mathrm{kJ\,mol^{-1}}\approx-5.15\times10^{3}\ \mathrm{kJ\,mol^{-1}}$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. The energy printed on a food label is not measured by burning that food in a bomb calorimeter. How is it found, and how different would the bomb value be?

Why it matters. Food-energy values underpin dietary advice from the World Health Organization, national food regulators (India's FSSAI, the European Food Safety Authority, the US FDA) and the planning of meals for athletes, patients and astronauts. A bomb calorimeter burns food completely to $\mathrm{CO_{2}}$, water and $\mathrm{N_{2}}$; the body cannot digest fibre fully and excretes nitrogen as urea, so it gets less.

What to do.

  1. Typical bomb-calorimeter values are about $17.3\ \mathrm{kJ\,g^{-1}}$ for carbohydrate, $39.5\ \mathrm{kJ\,g^{-1}}$ for fat and $23.6\ \mathrm{kJ\,g^{-1}}$ for protein. The "Atwater factors" used on labels are $17$, $37$ and $17\ \mathrm{kJ\,g^{-1}}$. Work out the percentage of the bomb value that the label assumes the body obtains for each.
  2. Choose a packaged food with a nutrition table. Use the Atwater factors to calculate its energy per $100\ \mathrm{g}$ from the grams of carbohydrate, fat and protein, and compare with the printed value.
  3. Use the bomb values to estimate the temperature rise in a calorimeter with $C_{cal}=10.0\ \mathrm{kJ\,K^{-1}}$ when $1.00\ \mathrm{g}$ of your food is burned (Interactive 2 shows the idea).
  4. Explain, in terms of $\Delta U$ and $\Delta H$, whether it matters that the body works at constant pressure while the bomb works at constant volume (estimate $\Delta n_{g}$ for burning glucose).

Tools: Definition 5.10, Rule 5.3, a nutrition label, a calculator.

Hand in: Your three percentages, the label check, the predicted $\Delta T$ and a short paragraph on why label values are lower than bomb values.

Going further. Protein loses the most. Write a balanced equation for burning urea, $\mathrm{CO(NH_{2})_{2}}$, and explain why the energy still locked in urea is "lost" to the body.

Instructor note: carbohydrate $17/17.3\approx98\%$, fat $37/39.5\approx94\%$, protein $17/23.6\approx72\%$. For glucose, $\mathrm{C_{6}H_{12}O_{6}(s)+6O_{2}(g)\rightarrow 6CO_{2}(g)+6H_{2}O(l)}$, $\Delta n_{g}=0$, so $\Delta H=\Delta U$; for fats $\Delta n_{g}$ is slightly negative and the correction is under 0.1%. Label energies are rounded and the factors are averages, so a 5–10% difference from the student's estimate is normal.

🧪 Self-check quiz

0/5 correct

Q1. The heat absorbed by a system at constant pressure (only pV work possible) equals

Q2. For $\mathrm{C(s)+O_{2}(g)\rightarrow CO_{2}(g)}$ at constant temperature

Q3. A bomb calorimeter measures

Q4. For one mole of an ideal gas, $C_{p}-C_{v}$ equals

Q5. The heat needed to warm $100\ \mathrm{g}$ of water by $10\ \mathrm{K}$ ($c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$) is

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