Enthalpy — Definition and the Relation Between Delta H and Delta U

Why Internal Energy Alone Is Not Enough

The first law gives a clean result at constant volume. With $\Delta V = 0$ there is no expansion work, so $w = 0$ and

$$\Delta U = q_V$$

The heat measured in a rigid sealed vessel is exactly the change in internal energy. That is a genuinely useful result, and it is how a bomb calorimeter works.

The trouble is that almost no chemistry is done that way. A reaction in a test tube, a beaker, a round-bottom flask or a conical flask is open to the room. Gas evolved during the reaction does not stay put — it pushes the atmosphere back and escapes. The volume of the system changes freely while the pressure stays pinned at whatever the atmosphere happens to be, roughly $1\,\mathrm{bar}$.

So a laboratory reaction runs at constant pressure, not constant volume. Under that condition part of the energy released by the reaction is spent pushing the atmosphere out of the way, and only the rest shows up as heat in the thermometer. The heat measured is no longer $\Delta U$.

Start from the first law at constant pressure. The only work is expansion work, $w = -p\,\Delta V$, so

$$\Delta U = q_p - p\,\Delta V$$

The measured heat $q_p$ is therefore $\Delta U + p\Delta V$, and that quantity — not $\Delta U$ by itself — is what a chemist actually measures in a flask. Chemistry needs a state function whose change equals $q_p$, in the same way that $\Delta U$ equals $q_V$. That function is enthalpy.

How much energy is lost to the atmosphere

The size of the effect is easy to estimate. Suppose a reaction releases one mole of gas at $298\,\mathrm{K}$ against an atmospheric pressure of $1\,\mathrm{bar}$. The gas produced occupies about $24.8\,\mathrm{L}$, so the system pushes the atmosphere back through that volume:

$$p\,\Delta V = nRT = 1 \times 8.314 \times 298 = 2478\,\mathrm{J} \approx 2.5\,\mathrm{kJ}$$

Two and a half kilojoules per mole of gas released never reaches the thermometer. Against a combustion enthalpy of several hundred kilojoules that is a small slice, but it is not zero, and it is exactly the gap between the two quantities this section is about.

Key Point: At constant volume the heat exchanged is $\Delta U$. At constant pressure it is not, because some energy goes into pushing the surroundings. A second state function is defined to handle the constant-pressure case.

Enthalpy, H = U + pV

Rearrange the constant-pressure first law with initial state 1 and final state 2:

$$U_2 - U_1 = q_p - p(V_2 - V_1)$$

Collecting the state-2 terms on one side and the state-1 terms on the other,

$$q_p = (U_2 + pV_2) - (U_1 + pV_1)$$

Both brackets have the same form, $U + pV$. That combination is given a name.

Key Point (Definition): Enthalpy is defined as $H = U + pV$. Its SI unit is the joule; chemists usually quote it in $\mathrm{kJ\,mol^{-1}}$.

With that definition the two brackets are $H_2$ and $H_1$, so

$$q_p = H_2 - H_1 = \Delta H$$

Key Point: At constant pressure, $\Delta H = q_p$. The heat absorbed by a system at constant pressure is its enthalpy change.

Why H is a state function

$U$ is a state function, $p$ is a state variable and $V$ is a state variable. $H$ is built only from these three, so its value depends only on the present state of the system and not on how the system got there. $\Delta H$ between two given states is fixed, whatever the route.

This has a consequence worth pausing on. Heat is a path function — the heat exchanged between two states depends on the route taken. But once the pressure is held constant, $q_p$ is forced to equal $\Delta H$, and $\Delta H$ is route-independent. Pinning the pressure removes the freedom that made $q$ path-dependent. The same thing happened at constant volume, where $q_V$ was forced to equal $\Delta U$.

The finite-change form

Written for a finite change,

$$\Delta H = \Delta U + \Delta(pV)$$

When the pressure is constant this collapses to

$$\Delta H = \Delta U + p\,\Delta V$$

Read the three terms physically. $\Delta U$ is the energy actually stored inside the system. $p\Delta V$ is the energy the system spent shoving the atmosphere aside to make room for itself. $\Delta H$ is the sum, and it is the quantity a thermometer in an open flask reports.

Only changes in enthalpy can be measured

The absolute enthalpy of a substance is unknown, because the absolute internal energy $U$ is unknown — nobody can count the total kinetic and potential energy of every particle in a mole of a substance. What can be measured is the difference between two states, and that is all chemistry ever needs:

$$\Delta H = H_{final} - H_{initial}$$

For a reaction, "final" means the products and "initial" means the reactants, so

$$\Delta_r H = \sum H_{products} - \sum H_{reactants}$$

The old name "heat content" for enthalpy survives in some books and is misleading. A system does not contain a stock of heat that it hands out; heat is energy in transit across the boundary. Enthalpy is a property of the state, and it happens to change by exactly the amount of heat transferred when, and only when, the pressure is held constant.

[Board] State the condition with the result. $\Delta H = q_p$ is true at constant pressure only, and $\Delta H = \Delta U + p\Delta V$ needs constant pressure as well. A bare "$\Delta H = q$" loses the mark.

The Sign of Delta H

An exothermic change gives out heat. Heat leaves the system, so $q_p$ is negative, and

$$\Delta H < 0 \quad \text{for an exothermic change}$$

An endothermic change takes in heat. Heat enters the system, $q_p$ is positive, and

$$\Delta H > 0 \quad \text{for an endothermic change}$$

An enthalpy level diagram makes this visual. Enthalpy is plotted upward. For an exothermic reaction the products sit lower than the reactants: the system has shed enthalpy into the surroundings and $\Delta H = H_{products} - H_{reactants}$ comes out negative. Combustion of methane, neutralisation of an acid by an alkali and the setting of cement are all of this kind, and all warm their surroundings.

For an endothermic reaction the products sit higher. The system has drawn enthalpy in from the surroundings, so $\Delta H$ is positive and the flask feels cold. Dissolving ammonium chloride in water and decomposing limestone both behave this way.

One caution about the word "system". The flask feels warm in an exothermic reaction because the heat has gone into the surroundings — your hand is part of the surroundings, not part of the system. The system itself has lost enthalpy. Students who reason from the temperature they feel often get the sign backwards.

Key Point: $\Delta H$ negative means heat released, products lower in enthalpy. $\Delta H$ positive means heat absorbed, products higher in enthalpy.

Writing the value with the equation

A balanced equation carrying its enthalpy change is a thermochemical equation:

$$\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)} \qquad \Delta H = -890.3\,\mathrm{kJ\,mol^{-1}}$$

Two conventions travel with it. Every species carries its physical state, because the value depends on it. And the value belongs to the equation exactly as written — for the amounts shown by the coefficients, not per mole of any one species picked at random.

Reversing the equation reverses the sign. Condensation of water gives out the same $40.79\,\mathrm{kJ\,mol^{-1}}$ that vaporisation at $373\,\mathrm{K}$ absorbed:

$$\mathrm{H_2O(l) \rightarrow H_2O(g)} \qquad \Delta H = +40.79\,\mathrm{kJ\,mol^{-1}}$$

$$\mathrm{H_2O(g) \rightarrow H_2O(l)} \qquad \Delta H = -40.79\,\mathrm{kJ\,mol^{-1}}$$

This follows from $H$ being a state function: swapping the initial and final states can only flip the sign of the difference.

Relating Delta H to Delta U for Reactions Involving Gases

$\Delta H = \Delta U + p\Delta V$ is correct but awkward, because $\Delta V$ for a reaction is not something anyone measures. For gases the ideal gas equation converts it into something countable.

Take a reaction run at constant temperature $T$ and constant pressure $p$. Let $n_A$ be the total moles of gaseous reactants occupying total volume $V_A$, and $n_B$ the total moles of gaseous products occupying $V_B$. Applying $pV = nRT$ to each side,

$$pV_A = n_A RT \qquad pV_B = n_B RT$$

Subtracting,

$$pV_B - pV_A = (n_B - n_A)RT$$

$$p(V_B - V_A) = (n_B - n_A)RT$$

The volume difference $V_B - V_A$ is the volume change of the reaction, and $n_B - n_A$ is given the symbol $\Delta n_g$:

$$p\,\Delta V = \Delta n_g RT$$

Substituting into $\Delta H = \Delta U + p\Delta V$ gives the working relation of this section.

Key Point: $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the number of moles of gaseous products minus the number of moles of gaseous reactants.

Rearranged the other way, $\Delta U = \Delta H - \Delta n_g RT$. Bomb calorimetry delivers $\Delta U$ and the relation converts it to $\Delta H$; an open-flask measurement delivers $\Delta H$ and the relation converts it back. Both directions appear in exams.

Counting Delta n_g

Only species labelled $(g)$ are counted. Solids, liquids and aqueous species contribute nothing, because their volumes are thousands of times smaller than the volume of a gas and are treated as negligible.

$$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$$

The moles are the stoichiometric coefficients of the balanced equation as it is written, so $\Delta n_g$ belongs to that equation, not to the reaction in the abstract. Doubling every coefficient doubles $\Delta n_g$, and it doubles $\Delta H$ and $\Delta U$ with it.

Fractional coefficients give fractional values, and that is fine. $\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g)}$ has $\Delta n_g = 1 - 1.5 = -0.5$.

What the derivation assumed

Three assumptions went into $\Delta H = \Delta U + \Delta n_g RT$, and each one is worth knowing.

The gases were treated as ideal, so that $pV = nRT$ could be applied to both sides. Real gases at ordinary pressures are close enough for the correction term, which is small anyway.

The temperature was the same before and after. If a reaction heats itself up, the relation applies to the reactants and products compared at the same temperature.

The volumes of solids, liquids and solutions were taken as negligible against the volumes of gases. At $298\,\mathrm{K}$ a mole of gas occupies about $24.8\,\mathrm{L}$ and a mole of water about $18\,\mathrm{mL}$, a ratio of roughly $1400$ to $1$, so the approximation is a safe one.

Working Through Delta n_g, and When the Two Are Equal

Three reactions, one of each kind.

Positive $\Delta n_g$. Limestone decomposing:

$$\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}$$

Gaseous products: $1$. Gaseous reactants: $0$ — the solids are not counted. So $\Delta n_g = 1 - 0 = +1$. Gas is created, the system expands, it spends energy pushing the atmosphere back, and $\Delta H$ is larger than $\Delta U$.

Negative $\Delta n_g$. Ammonia synthesis:

$$\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$$

$\Delta n_g = 2 - 4 = -2$. Gas is consumed, the system shrinks, the atmosphere does work on it, and $\Delta H$ is smaller than $\Delta U$ — more negative, in this exothermic case.

Zero $\Delta n_g$. Burning graphite:

$$\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$$

One mole of gas in, one mole of gas out. The solid carbon is not counted at all, so $\Delta n_g = 1 - 1 = 0$, and $\Delta H = \Delta U$ exactly.

The two cases where Delta H equals Delta U

The first is $\Delta n_g = 0$, as above. The gas volume created equals the gas volume destroyed, no net expansion work is done, and the two quantities coincide.

The second is a reaction involving only solids and liquids. Condensed phases barely change volume on reacting or on heating, so $p\Delta V$ is tiny compared with the energies of chemical change, and

$$\Delta H \approx \Delta U \quad \text{for solids and liquids only}$$

A number makes the size of the neglect clear. One mole of a gas at $298\,\mathrm{K}$ and $1\,\mathrm{bar}$ occupies about $25\,\mathrm{L}$; one mole of a liquid or solid occupies tens of millilitres. Expanding by one mole of gas costs $RT = 2.48\,\mathrm{kJ}$, while the volume changes of condensed phases cost a few joules at most.

Key Point: $\Delta H = \Delta U$ when $\Delta n_g = 0$, and $\Delta H \approx \Delta U$ when no gases take part at all.

The mistakes that cost marks

• Counting a solid or a liquid in $\Delta n_g$. $\mathrm{H_2(g) + I_2(s) \rightarrow 2HI(g)}$ has $\Delta n_g = 2 - 1 = +1$, not $2 - 2 = 0$.
• Counting an aqueous species. $\mathrm{Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)}$ has $\Delta n_g = 1 - 0 = +1$.
• Missing the phase label on water. $\mathrm{2H_2(g)+O_2(g) \rightarrow 2H_2O(l)}$ has $\Delta n_g = -3$, but writing $\mathrm{2H_2O(g)}$ instead makes it $-1$. The phase in the equation decides it.
• Reversing the subtraction. It is products minus reactants, in that order.

[JEE/NEET] When a question offers "$\Delta H = \Delta U$" as an option, count $\Delta n_g$ before dismissing it. Combustion of graphite, $\mathrm{H_2(g)+Cl_2(g) \rightarrow 2HCl(g)}$ and the esterification of ethanoic acid by ethanol all qualify.

Using the Relation Without Slipping on Units

Three habits keep $\Delta H = \Delta U + \Delta n_g RT$ from going wrong.

Match the energy units. $R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$ produces $\Delta n_g RT$ in joules, while $\Delta H$ and $\Delta U$ are almost always quoted in $\mathrm{kJ\,mol^{-1}}$. Divide the correction by $1000$ before adding it. At $298\,\mathrm{K}$,

$$RT = 8.314 \times 298 = 2478\,\mathrm{J\,mol^{-1}} = 2.48\,\mathrm{kJ\,mol^{-1}}$$

That single number handles most room-temperature problems: the correction is $2.48\,\Delta n_g$ kilojoules.

Use kelvin. $T$ in the relation is absolute temperature. A question quoting $25\,{}^\circ\mathrm{C}$ means $298\,\mathrm{K}$, and $100\,{}^\circ\mathrm{C}$ means $373\,\mathrm{K}$.

Fix the direction of the conversion. Going from $\Delta U$ to $\Delta H$, add $\Delta n_g RT$. Going from $\Delta H$ to $\Delta U$, subtract it. Writing the equation down before substituting prevents the sign flip.

A physical check catches most sign errors without any arithmetic. If gas is created, $\Delta n_g$ is positive, the system had to spend energy pushing the atmosphere back, and $\Delta H$ comes out greater than $\Delta U$. If gas is consumed, the atmosphere pushed inward and did work on the system, so $\Delta H$ is less than $\Delta U$. An answer that violates that pattern has a sign error somewhere.

The other $R$ value, $0.08206\,\mathrm{L\,atm\,K^{-1}\,mol^{-1}}$, is for work in litre-atmospheres and has no business in this relation unless the enthalpies themselves are quoted in litre-atmospheres.

A size check

For most reactions the correction is small. Combustion of methane has $\Delta H = -890.3\,\mathrm{kJ\,mol^{-1}}$ and a correction of about $5\,\mathrm{kJ\,mol^{-1}}$ — under one per cent. This is why the difference between $\Delta H$ and $\Delta U$ can be ignored in rough work but never in an exam answer that asks for both.

The correction matters most for phase changes and for reactions with a large $\Delta n_g$. Vaporising water at $373\,\mathrm{K}$ has $\Delta H = 40.79\,\mathrm{kJ\,mol^{-1}}$ and a correction of $3.10\,\mathrm{kJ\,mol^{-1}}$, which is nearly eight per cent of the answer.

Key Point: The correction term $\Delta n_g RT$ is $2.48\,\Delta n_g$ kilojoules at $298\,\mathrm{K}$. Its sign follows the sign of $\Delta n_g$.

Solved Examples

Question 1: Vaporising one mole of water

Treating water vapour as an ideal gas, the molar enthalpy change for vaporisation of $1\,\mathrm{mol}$ of water at $1\,\mathrm{bar}$ and $100\,{}^\circ\mathrm{C}$ ($373\,\mathrm{K}$) is $40.79\,\mathrm{kJ\,mol^{-1}}$. Calculate the internal energy change for the same process.

Answer:

The change is $\mathrm{H_2O(l) \rightarrow H_2O(g)}$. Liquid water is not counted, so the gaseous moles go from $0$ to $1$:

$$\Delta n_g = 1 - 0 = +1$$

The temperature is $100\,{}^\circ\mathrm{C} = 373\,\mathrm{K}$. I want $\Delta U$, so I subtract the correction:

$$\Delta U = \Delta H - \Delta n_g RT$$

$$\Delta n_g RT = 1 \times 8.314 \times 373 = 3101\,\mathrm{J\,mol^{-1}} = 3.10\,\mathrm{kJ\,mol^{-1}}$$

$$\Delta U = 40.79 - 3.10 = 37.69\,\mathrm{kJ\,mol^{-1}}$$

$\Delta U$ is smaller because part of the $40.79\,\mathrm{kJ}$ supplied was spent making room for the vapour, not on separating the molecules.

Ans: $\Delta U = 37.69\,\mathrm{kJ\,mol^{-1}}$

Watch out: $373\,\mathrm{K}$, not $373\,{}^\circ\mathrm{C}$ and not $100$. Using $100$ in place of $373$ gives a correction of $0.83\,\mathrm{kJ}$ and a wrong answer.

Question 2: Counting the gaseous moles

Find $\Delta n_g$ for each reaction.

(a) $\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)}$ (b) $\mathrm{NH_4NO_3(s) \rightarrow N_2O(g) + 2H_2O(g)}$ (c) $\mathrm{H_2(g) + I_2(s) \rightarrow 2HI(g)}$ (d) $\mathrm{Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)}$

Answer:

Each time I list only the species carrying a $(g)$ label and subtract reactants from products.

(a) Products: liquid water, so $0$ moles of gas. Reactants: $2 + 1 = 3$. $\Delta n_g = 0 - 3 = -3$.

(b) Products: $1 + 2 = 3$. Reactants: the solid contributes nothing, so $0$. $\Delta n_g = +3$.

(c) Products: $2$. Reactants: hydrogen only, since iodine is a solid here, so $1$. $\Delta n_g = 2 - 1 = +1$.

(d) Products: $3$. Reactants: $3$. The two solids drop out. $\Delta n_g = 0$.

Ans: (a) $-3$ (b) $+3$ (c) $+1$ (d) $0$

Watch out: In (c) the temptation is to write $\Delta n_g = 0$ by counting $\mathrm{I_2}$. It is a solid, and solids never enter $\Delta n_g$.

Question 3: Combustion of graphite

For the combustion of $1\,\mathrm{mol}$ of graphite at $298\,\mathrm{K}$, $\Delta U = -393.5\,\mathrm{kJ\,mol^{-1}}$. Find $\Delta H$.

Answer:

The equation is $\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}$.

Gaseous products $= 1$, gaseous reactants $= 1$, and graphite is not counted.

$$\Delta n_g = 1 - 1 = 0$$

$$\Delta H = \Delta U + (0)RT = \Delta U = -393.5\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta H = -393.5\,\mathrm{kJ\,mol^{-1}}$

Question 4: Ammonia synthesis

For $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$, $\Delta H = -92.4\,\mathrm{kJ}$ at $298\,\mathrm{K}$. Calculate $\Delta U$.

Answer:

Gaseous products $= 2$; gaseous reactants $= 1 + 3 = 4$.

$$\Delta n_g = 2 - 4 = -2$$

$$\Delta U = \Delta H - \Delta n_g RT = -92.4 - (-2)(8.314)(298)/1000$$

$$\Delta n_g RT = -2 \times 2.478 = -4.96\,\mathrm{kJ}$$

$$\Delta U = -92.4 + 4.96 = -87.4\,\mathrm{kJ}$$

The system contracted, so the surroundings did $4.96\,\mathrm{kJ}$ of work on it. That energy stayed inside, which is why $\Delta U$ is less negative than $\Delta H$.

Ans: $\Delta U = -87.4\,\mathrm{kJ}$

Watch out: Subtracting a negative correction adds. Writing $-92.4 - 4.96 = -97.4\,\mathrm{kJ}$ is the commonest slip in this question.

Question 5: Cyanamide burnt in a bomb

The reaction

$$\mathrm{NH_2CN(s) + \tfrac{3}{2}O_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(l)}$$

has $\Delta U = -742.7\,\mathrm{kJ\,mol^{-1}}$ at $298\,\mathrm{K}$. Calculate $\Delta H$.

Answer:

Gaseous products: nitrogen and carbon dioxide, $1 + 1 = 2$. Liquid water is not counted. Gaseous reactants: oxygen only, $1.5$; the solid cyanamide is not counted.

$$\Delta n_g = 2 - 1.5 = +0.5$$

$$\Delta n_g RT = 0.5 \times 8.314 \times 298 = 1239\,\mathrm{J} = 1.24\,\mathrm{kJ}$$

$$\Delta H = \Delta U + \Delta n_g RT = -742.7 + 1.24 = -741.5\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta H = -741.5\,\mathrm{kJ\,mol^{-1}}$

Watch out: A fractional $\Delta n_g$ is perfectly legal. The half comes from the $\tfrac{3}{2}$ in front of oxygen and must not be rounded to $1$ or $2$.

Question 6: Combustion of benzene

Liquid benzene burns according to

$$\mathrm{C_6H_6(l) + \tfrac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)}$$

with $\Delta U = -3263.9\,\mathrm{kJ\,mol^{-1}}$ at $298\,\mathrm{K}$. Find $\Delta H$.

Answer:

Gaseous products $= 6$. Gaseous reactants $= 7.5$. Both benzene and water are liquids and are ignored.

$$\Delta n_g = 6 - 7.5 = -1.5$$

$$\Delta n_g RT = -1.5 \times 2.478 = -3.72\,\mathrm{kJ}$$

$$\Delta H = -3263.9 + (-3.72) = -3267.6\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta H = -3267.6\,\mathrm{kJ\,mol^{-1}}$

Question 7: Decomposing limestone

$\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}$ has $\Delta H = +178.3\,\mathrm{kJ\,mol^{-1}}$ at $298\,\mathrm{K}$. Calculate $\Delta U$ and say whether the reaction is exothermic or endothermic.

Answer:

Only carbon dioxide is a gas, and there are no gaseous reactants.

$$\Delta n_g = 1 - 0 = +1$$

$$\Delta U = \Delta H - \Delta n_g RT = 178.3 - 2.48 = 175.8\,\mathrm{kJ\,mol^{-1}}$$

$\Delta H$ is positive, so the reaction absorbs heat and is endothermic.

Ans: $\Delta U = 175.8\,\mathrm{kJ\,mol^{-1}}$; the reaction is endothermic.

Question 8: The phase of the water changes the answer

Methane burns as

$$\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$$

with $\Delta H = -890.3\,\mathrm{kJ\,mol^{-1}}$ at $298\,\mathrm{K}$.

(a) Find $\Delta U$. (b) What would $\Delta n_g$ have been if the water had been produced as vapour?

Answer:

(a) Gaseous products: carbon dioxide only, $1$, because the water is liquid. Gaseous reactants: $1 + 2 = 3$.

$$\Delta n_g = 1 - 3 = -2$$

$$\Delta U = \Delta H - \Delta n_g RT = -890.3 - (-2)(2.478) = -890.3 + 4.96 = -885.3\,\mathrm{kJ\,mol^{-1}}$$

(b) With $\mathrm{2H_2O(g)}$ the gaseous products become $1 + 2 = 3$, and

$$\Delta n_g = 3 - 3 = 0$$

so $\Delta H$ and $\Delta U$ would have been equal.

Ans: (a) $\Delta U = -885.3\,\mathrm{kJ\,mol^{-1}}$ (b) $\Delta n_g = 0$

Watch out: The same reaction gives two different $\Delta n_g$ values depending on the state of the water. Read the phase labels before counting anything.

Question 9: Heat measured in an open beaker

Burning $0.500\,\mathrm{mol}$ of a compound in an open vessel at $1\,\mathrm{bar}$ releases $24.6\,\mathrm{kJ}$ of heat. What is $\Delta H$ per mole of the compound?

Answer:

The vessel is open, so the pressure is constant and the heat measured is $q_p = \Delta H$.

Heat is released, so $q_p$ is negative: $q_p = -24.6\,\mathrm{kJ}$ for $0.500\,\mathrm{mol}$.

$$\Delta H = \frac{-24.6}{0.500} = -49.2\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta H = -49.2\,\mathrm{kJ\,mol^{-1}}$

Watch out: Had the same reaction been run in a sealed rigid bomb, the measured heat would have been $q_V = \Delta U$ instead, and the two answers would differ by $\Delta n_g RT$.

Question 10: Working backwards to Delta n_g

For a certain reaction at $298\,\mathrm{K}$, $\Delta H - \Delta U = -4.95\,\mathrm{kJ}$. Find $\Delta n_g$ and suggest a reaction that fits.

Answer:

From the working relation, $\Delta H - \Delta U = \Delta n_g RT$.

$$\Delta n_g = \frac{\Delta H - \Delta U}{RT} = \frac{-4950}{8.314 \times 298} = \frac{-4950}{2478} = -2.0$$

Two moles of gas disappear. Ammonia synthesis, $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$, has exactly this $\Delta n_g$.

Ans: $\Delta n_g = -2$

Watch out: Convert $-4.95\,\mathrm{kJ}$ to $-4950\,\mathrm{J}$ before dividing, because $R$ carries joules.

Question 11: Zinc dissolving in acid

$\mathrm{Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)}$ releases $154.0\,\mathrm{kJ}$ per mole of zinc when carried out in an open flask at $298\,\mathrm{K}$. Find $\Delta H$ and $\Delta U$.

Answer:

Open flask means constant pressure, so the measured heat is $\Delta H$, and it is released:

$$\Delta H = -154.0\,\mathrm{kJ\,mol^{-1}}$$

For $\Delta n_g$, the only gas anywhere in the equation is hydrogen. Zinc is a solid; the acid and the zinc chloride are aqueous.

$$\Delta n_g = 1 - 0 = +1$$

$$\Delta U = \Delta H - \Delta n_g RT = -154.0 - 2.48 = -156.5\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta H = -154.0\,\mathrm{kJ\,mol^{-1}}$, $\Delta U = -156.5\,\mathrm{kJ\,mol^{-1}}$

Watch out: Aqueous species carry no entry in $\Delta n_g$. Counting $\mathrm{2HCl(aq)}$ as gas would give $\Delta n_g = -1$ and reverse the sign of the correction.

Question 12: A reaction with no gases at all

The neutralisation $\mathrm{HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)}$ has $\Delta H = -57.1\,\mathrm{kJ\,mol^{-1}}$. What is $\Delta U$, and why?

Answer:

No species in the equation carries a $(g)$ label, so $\Delta n_g = 0$ and the correction term vanishes:

$$\Delta H = \Delta U + (0)RT$$

Beyond that, everything present is a liquid or a solution, and condensed phases change volume by only a few millilitres during a reaction. The $p\Delta V$ term is a few joules against an enthalpy change of $57100\,\mathrm{J}$, which is negligible.

$$\Delta U = -57.1\,\mathrm{kJ\,mol^{-1}}$$

Ans: $\Delta U = -57.1\,\mathrm{kJ\,mol^{-1}}$; with no gases involved, $\Delta H$ and $\Delta U$ are equal to the accuracy quoted.