Solved Examples

How to Use This Section

The theory is finished. What follows is one long problem set: 35 questions worked out in full, arranged from the simplest to the hardest, covering the in-text problems and the end-of-chapter exercises that boards, JEE Main and NEET keep recycling.

Every solution is written the way an answer script should be written: the relation first, the substitution second, the number with its unit last. Where a question hides a trap, a Watch out line names it.

The working method

1. Cover the answer and attempt the question on paper. Reading a solution feels like learning and is not.
2. Compare your steps with mine, not only your final number. A right answer reached by a wrong route collapses the moment the data change.
3. Before pressing a calculator key, check that every energy term in the equation carries the same unit. Half the lost marks in this chapter are a stray factor of 1000.
4. Read the Watch out line even when your answer matched.
5. Two days later, redo from a blank page any question you could not finish.

The errors that cost the most marks in this chapter

Error	Where it bites	The fix
Writing $w = +p_{ex}\Delta V$	Every expansion question; the sign of $w$ and then of $\Delta U$ flips	The IUPAC form is $w = -p_{ex}\Delta V$. Expansion means $\Delta V > 0$, so $w$ comes out negative
Counting solids and liquids in $\Delta n_g$	Converting a bomb-calorimeter $\Delta U$ into $\Delta H$	$\Delta n_g$ counts gaseous moles only, products minus reactants. Water written as $\mathrm{H_2O(l)}$ contributes nothing
Leaving work in L atm when joules were asked	Expansion-work numericals	$1\ \mathrm{L\,atm} = 101.3\ \mathrm{J}$ and $1\ \mathrm{L\,bar} = 100\ \mathrm{J}$
Mixing J and kJ in $\Delta G = \Delta H - T\Delta S$	Crossover-temperature and spontaneity questions	$\Delta S$ almost always arrives in $\mathrm{J\,K^{-1}\,mol^{-1}}$ and $\Delta H$ in $\mathrm{kJ\,mol^{-1}}$. Convert one of them first
Flipping a thermochemical equation but not its sign	Hess cycles, Born-Haber, formation-from-combustion problems	Reverse the arrow, reverse the sign. Multiply the equation, multiply the value
Using the system's own sign for $\Delta S_{surr}$	Second-law questions	$\Delta S_{surr} = -\dfrac{\Delta H_{sys}}{T}$. An exothermic reaction raises the entropy of the surroundings

A last piece of advice about presentation. In a board answer script, a numerical earns marks in three places: the relation quoted, the substitution shown with units, and the final value with its sign and unit. An answer that jumps from the data straight to a boxed number forfeits two of those three even when the number is right. Every solution below is laid out in that order deliberately, so that copying the layout costs nothing extra.

[JEE/NEET] In a numerical, the marks sit on the sign and the unit far more often than on the arithmetic. Write $w$, $q$, $\Delta n_g$ and $\Delta S$ with their signs explicitly before substituting.

Question 1: Reading the first law off the wall

Express the change in internal energy of a system when

(i) no heat is absorbed by the system from the surroundings, but work $w$ is done on the system. What kind of wall does the system have? (ii) no work is done on the system, but $q$ amount of heat is taken out of the system and given to the surroundings. What kind of wall does the system have? (iii) $w$ amount of work is done by the system and $q$ amount of heat is supplied to the system. What kind of system is this?

Answer:

I start from the first law in the IUPAC convention, $\Delta U = q + w$, where both $q$ and $w$ are positive when energy enters the system.

In (i) no heat crosses the boundary, so $q = 0$ and

$$\Delta U = w$$

A boundary that blocks heat but still lets the piston move is an adiabatic wall. The change is an adiabatic compression, and every joule of work done on the gas stays in it as internal energy.

In (ii) no work is done, so $w = 0$. Heat leaves the system, so $q$ carries a negative sign: the heat entering it is $-q$, and

$$\Delta U = -q$$

Heat can cross the boundary, so the wall is thermally conducting, a diathermic wall.

In (iii) the symbol $w$ is the magnitude of the work done by the system, so the work done on it is $-w$, while heat enters, so $q$ is positive:

$$\Delta U = q - w$$

Both heat and work cross the boundary but matter does not, so this is a closed system with a diathermic wall.

A fourth case is worth naming even though it was not asked. If the walls block both heat and work, nothing at all crosses the boundary, $q = 0$ and $w = 0$, and $\Delta U = 0$. That is an isolated system. The first law then has nothing left to say about which way a change inside it will run, which is precisely the gap entropy is invented to fill later in the chapter.

The pattern to carry forward is that the first law never needs memorising in three versions. There is one equation, $\Delta U = q + w$, and the wall decides which term survives.

Ans: (i) $\Delta U = w$, adiabatic wall; (ii) $\Delta U = -q$, diathermic wall; (iii) $\Delta U = q - w$, closed system with a diathermic wall. Watch out: In part (iii) the symbol $w$ in the question already means the magnitude of work done by the system. Feeding $+w$ into $\Delta U = q + w$ would give the wrong sign. Decide the direction first, then attach the sign.

Question 2: Heat in, work out

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Answer:

Heat is absorbed by the system, so $q = +701\ \mathrm{J}$.

Work is done by the system on the surroundings, so energy leaves the system as work and $w = -394\ \mathrm{J}$.

$$\Delta U = q + w = (+701\ \mathrm{J}) + (-394\ \mathrm{J}) = +307\ \mathrm{J}$$

The internal energy rises by 307 J: the system took in more energy as heat than it gave out as work.

A physical check confirms the sign. The system collected 701 J and paid out 394 J, so it must end 307 J richer than it started. Whenever the heat absorbed exceeds the work done by the system, $\Delta U$ comes out positive.

One more remark, because it causes confusion between subjects. Many physics texts define $w$ as the work done by the system and write the first law as $\Delta U = q - w$. The physics is identical and the final number is identical; only the bookkeeping differs. Chemistry uses the IUPAC convention throughout, in which energy entering the system is positive, so keep $\Delta U = q + w$ and let the direction of flow supply the sign.

Ans: $\Delta U = +307\ \mathrm{J}$ Watch out: "394 J of work is done by the system" is not $w = +394\ \mathrm{J}$. Writing it as positive gives $+1095\ \mathrm{J}$, which is the commonest wrong answer to this question.

Question 3: Three one-line checks

(a) A thermodynamic state function is a quantity of what kind? (b) For a process to occur under adiabatic conditions, what must be true? (c) What are the standard enthalpies of all elements in their standard states?

Answer:

(a) A state function is fixed by the present state of the system alone. Its change between two states depends only on the initial and final states and not on the route taken. Internal energy, enthalpy, entropy, Gibbs energy, pressure, volume and temperature are state functions; heat and work are not, because they are quantities of energy in transit and their values depend on the path. Questions 4 to 6 make the distinction concrete: the same pair of initial and final states there gives three completely different values of $w$ and of $q$, and the same value of $\Delta U$ every time.

(b) Adiabatic means no heat is exchanged with the surroundings, so $q = 0$. Temperature is not fixed in an adiabatic change; in fact an adiabatic expansion cools the gas, so $\Delta T = 0$ is wrong. Nor is $w = 0$: an adiabatic process usually does work, and with $q = 0$ the first law reduces to $\Delta U = w_{ad}$.

(c) By convention the standard enthalpy of formation of an element in its reference state is taken as zero. That fixes the zero of the enthalpy scale, so $\Delta_f H^{\circ}$ of $\mathrm{O_2(g)}$, $\mathrm{C(graphite)}$, $\mathrm{H_2(g)}$ and $\mathrm{Br_2(l)}$ are all zero. It applies only to the reference form: $\mathrm{O_3(g)}$, diamond and $\mathrm{Br_2(g)}$ have non-zero values.

Ans: (a) a quantity whose value is independent of the path; (b) $q = 0$; (c) zero. Watch out: $\Delta T = 0$ is the isothermal condition, not the adiabatic one. A great many students tick it here.

Question 4: An expansion into a vacuum

Two litres of an ideal gas at a pressure of 12.2 atm expands isothermally at $25\ {}^\circ\mathrm{C}$ into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?

Answer:

Expansion into a vacuum means there is nothing outside pushing back, so the external pressure is zero:

$$p_{ex} = 0$$

Work of expansion is $w = -p_{ex}\Delta V$, and with $p_{ex} = 0$,

$$w = -0 \times (10 - 2)\ \mathrm{L} = 0$$

The gas is ideal and the temperature is constant, so the internal energy does not change, $\Delta U = 0$. The reason is worth stating once: the internal energy of an ideal gas depends on temperature alone, because its molecules exert no forces on one another, so moving them further apart costs no energy. From the first law,

$$q = \Delta U - w = 0 - 0 = 0$$

No work is done and no heat is absorbed. The gas simply spreads out, and because the molecules of an ideal gas do not attract one another, spreading out costs no energy.

Ans: $w = 0$ and $q = 0$. Watch out: The initial pressure of 12.2 atm plays no part in the calculation. Only $p_{ex}$ appears in $w = -p_{ex}\Delta V$, and here it is zero. Students who substitute 12.2 atm get $-97.6\ \mathrm{L\,atm}$, which is the work of a completely different process.

Question 5: The same expansion against 1 atm

Consider the same expansion, but this time against a constant external pressure of 1 atm.

Answer:

Now the gas has to push the surroundings back at a steady 1 atm.

$$\Delta V = 10\ \mathrm{L} - 2\ \mathrm{L} = 8\ \mathrm{L}$$

$$w = -p_{ex}\Delta V = -(1\ \mathrm{atm})(8\ \mathrm{L}) = -8\ \mathrm{L\,atm}$$

The negative sign says energy has left the system as work, which is right for an expansion.

The process is isothermal and the gas is ideal, so $\Delta U = 0$ and

$$q = -w = +8\ \mathrm{L\,atm}$$

Converting to joules with $1\ \mathrm{L\,atm} = 101.3\ \mathrm{J}$:

$$w = -8 \times 101.3\ \mathrm{J} = -810.4\ \mathrm{J}, \qquad q = +810.4\ \mathrm{J}$$

The gas absorbs 810 J of heat from the surroundings and hands out exactly the same energy as work, so its internal energy is unchanged.

Compare this with Question 4. The gas started at the same 2 L and finished at the same 10 L, at the same temperature, and yet this time it did 810 J of work and drew 810 J of heat. Nothing about the two end states changed. Only the route did.

Ans: $w = -8\ \mathrm{L\,atm} = -810.4\ \mathrm{J}$; $q = +8\ \mathrm{L\,atm} = +810.4\ \mathrm{J}$. Watch out: $p_{ex}$ is the pressure the gas pushes against, not the pressure of the gas. In an irreversible expansion the two are different at every instant, and only the external one belongs in $w = -p_{ex}\Delta V$.

Question 6: The same expansion carried out reversibly

Consider the expansion of Question 4 again, this time for 1 mol of an ideal gas conducted reversibly.

Answer:

For an isothermal reversible expansion of an ideal gas the external pressure is kept infinitesimally below the gas pressure at every instant, and the work is

$$w = -2.303\,nRT\log\frac{V_f}{V_i}$$

With $n = 1\ \mathrm{mol}$, $T = 298\ \mathrm{K}$, $V_i = 2\ \mathrm{L}$, $V_f = 10\ \mathrm{L}$, so $V_f/V_i = 5$ and $\log 5 = 0.6990$.

Taking $R = 0.08206\ \mathrm{L\,atm\,K^{-1}\,mol^{-1}}$:

$$w = -2.303 \times 1 \times 0.08206 \times 298 \times 0.6990 = -39.37\ \mathrm{L\,atm}$$

In joules, using $R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}$:

$$w = -2.303 \times 8.314 \times 298 \times 0.6990 = -3988\ \mathrm{J} \approx -3.99\ \mathrm{kJ}$$

Since $\Delta U = 0$ for the isothermal change,

$$q = -w = +3.99\ \mathrm{kJ}$$

Why the reversible number is the largest is worth seeing rather than memorising. In the irreversible run the gas pushed against a fixed 1 atm the whole way, even at the start when its own pressure was 12.2 atm and it could have pushed against nearly 12.2 atm. In the reversible run the external pressure is lowered in step with the gas pressure, so at every instant the gas is doing the most work the surroundings will let it do. Add up all those instants and the total is the maximum any path can deliver between these two states.

Ans: $w = -39.37\ \mathrm{L\,atm} = -3.99\ \mathrm{kJ}$; $q = +3.99\ \mathrm{kJ}$. Watch out: Line up the three answers. The same initial and final states give $w = 0$ into a vacuum, $-0.81\ \mathrm{kJ}$ against 1 atm and $-3.99\ \mathrm{kJ}$ reversibly. Work is a path function, and the reversible path extracts the maximum. $\Delta U$ is zero in all three, because $U$ is a state function.

Question 7: Compressing a gas in one step

One mole of an ideal gas held at 300 K is compressed from 20.0 L to 5.0 L in a single step against a constant external pressure of 5.0 bar. Calculate $w$, $\Delta U$ and $q$ in joules.

Answer:

$$\Delta V = 5.0\ \mathrm{L} - 20.0\ \mathrm{L} = -15.0\ \mathrm{L}$$

$$w = -p_{ex}\Delta V = -(5.0\ \mathrm{bar})(-15.0\ \mathrm{L}) = +75.0\ \mathrm{L\,bar}$$

Positive work means work has been done on the gas, which is what compression should give.

Converting with $1\ \mathrm{L\,bar} = 100\ \mathrm{J}$:

$$w = +75.0 \times 100\ \mathrm{J} = +7500\ \mathrm{J} = +7.5\ \mathrm{kJ}$$

The temperature is held at 300 K throughout and the gas is ideal, so $\Delta U = 0$. From the first law,

$$q = \Delta U - w = 0 - 7500\ \mathrm{J} = -7500\ \mathrm{J}$$

The 7.5 kJ pushed in as work leaves again as heat into the thermostat bath.

It is worth asking how this compares with a reversible compression between the same two volumes. That would need $w = -2.303\,nRT\log(V_f/V_i) = -2.303 \times 8.314 \times 300 \times \log(5/20) = +3458\ \mathrm{J}$, less than half the single-step figure. The rule for expansion inverts for compression: the reversible path takes the least work out of you, because the external pressure is never more than a whisker above the gas pressure. Slamming a piston down in one step against a fixed 5 bar wastes energy as heat.

One more check on the signs, because they are the whole difficulty in this question. Compression means $\Delta V$ is negative. A negative $\Delta V$ multiplied by the minus sign in $w = -p_{ex}\Delta V$ leaves $w$ positive, and positive $w$ means energy has gone into the system. Every one of those three statements agrees with the physical picture of someone pushing a piston in. If your answer for a compression comes out negative, the error is almost always that $\Delta V$ was written as $20 - 5$ instead of $5 - 20$.

Ans: $w = +7.5\ \mathrm{kJ}$, $\Delta U = 0$, $q = -7.5\ \mathrm{kJ}$. Watch out: $1\ \mathrm{L\,bar} = 100\ \mathrm{J}$ exactly, while $1\ \mathrm{L\,atm} = 101.3\ \mathrm{J}$. Read which unit the external pressure came in before choosing the factor.

Question 8: Two paths and a closed cycle

A system is taken from state A to state B along path I, absorbing 60 J of heat and doing 30 J of work on the surroundings. It is then returned from B to A along a different path II, along which 40 J of work is done on the system. Find $q$ for path II, and the totals of $q$ and $w$ for the complete cycle.

Answer:

For path I, $q = +60\ \mathrm{J}$ and $w = -30\ \mathrm{J}$ because the system does the work.

$$\Delta U_{A \rightarrow B} = 60 + (-30) = +30\ \mathrm{J}$$

Internal energy is a state function, so the return trip must undo exactly this change no matter which route it takes:

$$\Delta U_{B \rightarrow A} = -30\ \mathrm{J}$$

Along path II work is done on the system, so $w = +40\ \mathrm{J}$, and

$$q = \Delta U - w = -30 - 40 = -70\ \mathrm{J}$$

The system gives out 70 J of heat on the way back.

Over the whole cycle the system returns to its starting state, so

$$\Delta U_{cycle} = 0$$

$$q_{total} = 60 - 70 = -10\ \mathrm{J}, \qquad w_{total} = -30 + 40 = +10\ \mathrm{J}$$

and their sum is zero, as it must be.

Every step of this problem rests on one idea. I was never told what happens between A and B on either path, and I did not need to know. $\Delta U$ is fixed by the two end states, so once path I gave me $+30\ \mathrm{J}$, path II was forced to give $-30\ \mathrm{J}$ whatever it did in between. The individual $q$ and $w$ values, by contrast, are different on the two paths and could have taken any pair of values summing correctly.

The cycle result is also the reason a heat engine can exist at all. Over one complete cycle the working substance ends where it began, so its internal energy is unchanged, and every joule of net heat it takes in must leave as net work. Nothing accumulates inside the engine. What the second law adds later is that the heat cannot all be taken from one source; some must be dumped at a lower temperature.

Ans: $q_{\mathrm{II}} = -70\ \mathrm{J}$; over the cycle $\Delta U = 0$, $q_{total} = -10\ \mathrm{J}$, $w_{total} = +10\ \mathrm{J}$. Watch out: $\Delta U = 0$ round a cycle does not mean $q = 0$ and $w = 0$ round a cycle. It means they are equal and opposite. That single line is the whole idea behind every heat engine.

Question 9: Internal energy of vaporisation of water

Assuming water vapour to be a perfect gas, the molar enthalpy change for vaporisation of 1 mol of water at 1 bar and $100\ {}^\circ\mathrm{C}$ is $40.79\ \mathrm{kJ\,mol^{-1}}$. Calculate the internal energy change when 1 mol of water is vaporised at 1 bar and $100\ {}^\circ\mathrm{C}$.

Answer:

The change is

$$\mathrm{H_2O(l)} \rightarrow \mathrm{H_2O(g)}$$

Gaseous moles: one on the right, none on the left, so $\Delta n_g = 1 - 0 = +1$.

$$\Delta H = \Delta U + \Delta n_g RT \quad \Rightarrow \quad \Delta U = \Delta H - \Delta n_g RT$$

$$T = 100\ {}^\circ\mathrm{C} = 373\ \mathrm{K}$$

$$\Delta n_g RT = (1)(8.314 \times 10^{-3}\ \mathrm{kJ\,K^{-1}\,mol^{-1}})(373\ \mathrm{K}) = 3.10\ \mathrm{kJ\,mol^{-1}}$$

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

Of the 40.79 kJ supplied, about 3.1 kJ is spent pushing the atmosphere back to make room for the vapour, and only 37.7 kJ goes into the molecules themselves. One mole of liquid water occupies about 18 mL; as vapour at 373 K and 1 bar it occupies about 31 L. Something has to move the air out of the way, and that something is paid for out of the heat supplied.

For any process in which gas is created, $\Delta n_g$ is positive and $\Delta H$ is therefore larger than $\Delta U$. Vaporisation, sublimation and the decomposition of a carbonate all sit in that family.

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

Question 10: The bomb-calorimeter value for cyanamide

The reaction of cyanamide, $\mathrm{NH_2CN(s)}$, with dioxygen was carried out in a bomb calorimeter and $\Delta U$ was found to be $-742.7\ \mathrm{kJ\,mol^{-1}}$ at 298 K. Calculate the enthalpy change for the reaction at 298 K.

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

Answer:

A bomb calorimeter runs at constant volume, so what it measures is $q_V = \Delta U$. The vessel is sealed steel, $\Delta V = 0$, no expansion work is possible, and the first law collapses to $\Delta U = q_V$. Converting that measurement into the $\Delta H$ a chemist actually wants is the whole point of this question.

I count gaseous moles only. On the product side: $\mathrm{N_2}$ and $\mathrm{CO_2}$, that is 2 mol of gas. Water is liquid, so it contributes nothing. On the reactant side: $\tfrac{3}{2}$ mol of $\mathrm{O_2}$, since cyanamide is a solid.

$$\Delta n_g = 2 - \tfrac{3}{2} = +0.5$$

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

$$\Delta n_g RT = (0.5)(8.314 \times 10^{-3}\ \mathrm{kJ\,K^{-1}\,mol^{-1}})(298\ \mathrm{K}) = 1.24\ \mathrm{kJ\,mol^{-1}}$$

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

Ans: $\Delta H = -741.5\ \mathrm{kJ\,mol^{-1}}$ Watch out: $\Delta n_g$ here is a fraction, and fractions are allowed. Rounding $\tfrac{3}{2}$ to 2 makes $\Delta n_g = 0$ and hands back $\Delta H = \Delta U$, which is wrong.

Question 11: Counting gaseous moles, and the sign it forces

(a) Write $\Delta n_g$ for each reaction below and state which ones have $\Delta H = \Delta U$.

(i) $\mathrm{CH_4(g)} + 2\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\mathrm{H_2O(l)}$ (ii) $\mathrm{C(graphite)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}$ (iii) $\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$ (iv) $\mathrm{H_2(g)} + \mathrm{I_2(g)} \rightarrow 2\mathrm{HI(g)}$ (v) $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightarrow 2\mathrm{NH_3(g)}$

(b) $\Delta U^{\circ}$ of combustion of methane is $-X\ \mathrm{kJ\,mol^{-1}}$. What can be said about $\Delta H^{\circ}$?

Answer:

(a) Counting gases only, products minus reactants:

Reaction	Gaseous products	Gaseous reactants	$\Delta n_g$
(i)	1	3	$-2$
(ii)	1	1	$0$
(iii)	1	0	$+1$
(iv)	2	2	$0$
(v)	2	4	$-2$

$\Delta H = \Delta U$ whenever $\Delta n_g = 0$, so in (ii) and (iv).

(b) Methane combustion is reaction (i), with $\Delta n_g = -2$.

$$\Delta H^{\circ} = \Delta U^{\circ} + \Delta n_g RT = -X - 2RT$$

Subtracting a positive quantity $2RT$ from $-X$ makes the result more negative, so $\Delta H^{\circ} < \Delta U^{\circ}$.

Numerically at 298 K, $2RT = 2 \times 8.314 \times 10^{-3} \times 298 = 4.96\ \mathrm{kJ\,mol^{-1}}$, so $\Delta H^{\circ} = (-X - 4.96)\ \mathrm{kJ\,mol^{-1}}$.

Notice how small that correction is. At room temperature $RT$ is only about $2.5\ \mathrm{kJ\,mol^{-1}}$, so unless $\Delta n_g$ is large the gap between $\Delta H$ and $\Delta U$ is a few kilojoules on a reaction enthalpy of several hundred. That is why the two are often quoted as though they were interchangeable, and also why an exam question that asks for the difference is really asking whether you can count gaseous moles.

Ans: (a) $-2,\ 0,\ +1,\ 0,\ -2$; $\Delta H = \Delta U$ for (ii) and (iv). (b) $\Delta H^{\circ} < \Delta U^{\circ}$. Watch out: "More negative" means smaller, not larger. With $\Delta n_g$ negative the answer is always $\Delta H < \Delta U$, whichever way the arithmetic feels.

Question 12: Heating a block of aluminium

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from $35\ {}^\circ\mathrm{C}$ to $55\ {}^\circ\mathrm{C}$. Molar heat capacity of Al is $24\ \mathrm{J\,mol^{-1}\,K^{-1}}$.

Answer:

The capacity given is molar, so I need moles, not grams. The molar mass of aluminium is $27\ \mathrm{g\,mol^{-1}}$.

$$n = \frac{60.0\ \mathrm{g}}{27\ \mathrm{g\,mol^{-1}}} = 2.222\ \mathrm{mol}$$

A temperature difference is the same number in celsius and in kelvin:

$$\Delta T = 55 - 35 = 20\ {}^\circ\mathrm{C} = 20\ \mathrm{K}$$

$$q = n\,C_m\,\Delta T = (2.222\ \mathrm{mol})(24\ \mathrm{J\,mol^{-1}\,K^{-1}})(20\ \mathrm{K}) = 1066.7\ \mathrm{J}$$

$$q = 1.07\ \mathrm{kJ}$$

Three forms of the same relation are worth keeping straight, because the data decide which one to use. With a molar heat capacity, $q = n\,C_m\,\Delta T$. With a specific heat, $q = m\,c\,\Delta T$. With the heat capacity of a whole object such as a calorimeter, $q = C\,\Delta T$ with nothing else attached. Reading which of the three has been handed to you takes five seconds and saves the question.

It is instructive to convert the given value into a specific heat: $24 \div 27 = 0.89\ \mathrm{J\,g^{-1}\,K^{-1}}$ for aluminium, against $4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$ for water. Gram for gram, water needs almost five times as much heat for the same rise in temperature. That is why a metal spoon in a hot drink burns the fingers within seconds while the drink itself cools slowly, and why water is the standard coolant in radiators and calorimeters alike.

Ans: $q \approx 1.07\ \mathrm{kJ}$ Watch out: Do not convert $35\ {}^\circ\mathrm{C}$ and $55\ {}^\circ\mathrm{C}$ to kelvin separately and then subtract in a panic. The difference is unchanged; converting only wastes time.

Question 13: Heating the same gas two different ways

Two moles of an ideal gas whose molar heat capacity at constant pressure is $29.10\ \mathrm{J\,K^{-1}\,mol^{-1}}$ are warmed from 300 K to 350 K. Find the heat required (a) at constant pressure and (b) at constant volume, and account for the difference. Take $R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}$.

Answer:

For an ideal gas the two molar capacities differ by $R$:

$$C_{p,m} - C_{V,m} = R \quad \Rightarrow \quad C_{V,m} = 29.10 - 8.314 = 20.79\ \mathrm{J\,K^{-1}\,mol^{-1}}$$

(a) At constant pressure, $q_p = n\,C_{p,m}\Delta T$:

$$q_p = (2)(29.10)(50) = 2910\ \mathrm{J}$$

and since $q_p = \Delta H$, the enthalpy rises by 2910 J.

(b) At constant volume, $q_V = n\,C_{V,m}\Delta T$:

$$q_V = (2)(20.79)(50) = 2079\ \mathrm{J}$$

and since $q_V = \Delta U$, the internal energy rises by 2079 J.

The difference is

$$q_p - q_V = 2910 - 2079 = 831\ \mathrm{J}$$

which is $nR\Delta T = (2)(8.314)(50) = 831.4\ \mathrm{J}$, rounded to the $831\ \mathrm{J}$ above. That extra energy is the expansion work the gas does on the atmosphere while it warms at constant pressure. At constant volume it does no work, so all the heat stays inside as internal energy.

This is the whole content of $C_p - C_V = R$, and it is the same idea as $\Delta H = \Delta U + \Delta n_g RT$ seen from a different angle. Heat a gas in a sealed box and every joule raises the internal energy. Heat it under a free piston and part of every joule is spent shoving the atmosphere aside, so more heat is needed for the same rise in temperature. The extra, per mole per kelvin, is exactly $R$.

The value of $C_{p,m}$ given here, close to $29\ \mathrm{J\,K^{-1}\,mol^{-1}}$, is worth recognising. For a monatomic ideal gas $C_{V,m} = \tfrac{3}{2}R = 12.5$ and $C_{p,m} = \tfrac{5}{2}R = 20.8\ \mathrm{J\,K^{-1}\,mol^{-1}}$; for a diatomic gas at ordinary temperatures the figures are $\tfrac{5}{2}R = 20.8$ and $\tfrac{7}{2}R = 29.1$. The gas in this question is therefore behaving as a diatomic one. The difference $C_{p,m} - C_{V,m} = R$ holds for both, because the expansion work per mole per kelvin does not care how many atoms are in the molecule.

Ans: (a) $q_p = 2910\ \mathrm{J} = \Delta H$; (b) $q_V = 2079\ \mathrm{J} = \Delta U$; the difference $831\ \mathrm{J}$, which is $nR\Delta T = 831.4\ \mathrm{J}$ rounded, is expansion work. Watch out: $\Delta U$ is 2079 J in both cases, because for an ideal gas $\Delta U$ depends only on the temperature change. Only in the constant-volume run does it happen to equal the heat supplied.

Question 14: Burning graphite in a bomb calorimeter

1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation

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

During the reaction the temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is $32.8\ \mathrm{kJ\,K^{-1}}$, what is the enthalpy change for the reaction at 298 K and 1 atm?

Answer:

Heat gained by the calorimeter:

$$q_{cal} = C\,\Delta T = (32.8\ \mathrm{kJ\,K^{-1}})(299 - 298)\ \mathrm{K} = 32.8\ \mathrm{kJ}$$

Whatever the calorimeter gained, the reaction mixture lost. Same magnitude, opposite sign:

$$q = -32.8\ \mathrm{kJ}$$

The negative sign says the reaction is exothermic. The bomb is sealed, so the volume is fixed and no work is done; this heat is therefore $\Delta U$ for the burning of 1 g of graphite.

For one mole I scale by the molar mass, $12.0\ \mathrm{g\,mol^{-1}}$:

$$\Delta U = \frac{(12.0\ \mathrm{g\,mol^{-1}})(-32.8\ \mathrm{kJ})}{1\ \mathrm{g}} = -393.6\ \mathrm{kJ\,mol^{-1}}$$

One mole of gas is consumed and one mole is produced, so $\Delta n_g = 1 - 1 = 0$ and

$$\Delta H = \Delta U + \Delta n_g RT = \Delta U = -393.6\ \mathrm{kJ\,mol^{-1}}$$

Two sanity checks are worth running. First, the sign: the water bath got hotter, so the reaction gave out heat, so $\Delta U$ and $\Delta H$ must be negative. Second, the magnitude: the accepted enthalpy of combustion of graphite is $-393.5\ \mathrm{kJ\,mol^{-1}}$, and this calorimeter gives $-393.6\ \mathrm{kJ\,mol^{-1}}$, which is the tabulated value to the precision of the data.

Ans: $\Delta H = \Delta U = -393.6\ \mathrm{kJ\,mol^{-1}}$ Watch out: The heat capacity quoted is that of the whole calorimeter assembly, already in $\mathrm{kJ\,K^{-1}}$, so no mass and no specific heat is needed. Multiplying by 1 g or by 1000 is a common slip.

Question 15: A neutralisation in a coffee-cup calorimeter

50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH, both at $25.0\ {}^\circ\mathrm{C}$, are mixed in a coffee-cup calorimeter. The temperature rises to $31.7\ {}^\circ\mathrm{C}$. Take the density of the mixture as $1.00\ \mathrm{g\,mL^{-1}}$, its specific heat as $4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$, and the heat capacity of the cup as negligible. Find the enthalpy of neutralisation per mole of water formed.

Answer:

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

Mass of solution:

$$m = (50.0 + 50.0)\ \mathrm{mL} \times 1.00\ \mathrm{g\,mL^{-1}} = 100\ \mathrm{g}$$

$$\Delta T = 31.7 - 25.0 = 6.7\ \mathrm{K}$$

Heat taken up by the solution:

$$q_{soln} = m\,c\,\Delta T = (100\ \mathrm{g})(4.18\ \mathrm{J\,g^{-1}\,K^{-1}})(6.7\ \mathrm{K}) = 2800.6\ \mathrm{J}$$

The solution gained it, so the reaction released it:

$$q_p = -2800.6\ \mathrm{J}$$

Moles of water formed. HCl supplies $0.0500\ \mathrm{L} \times 1.00\ \mathrm{mol\,L^{-1}} = 0.0500\ \mathrm{mol}$ of $\mathrm{H^+}$, NaOH the same amount of $\mathrm{OH^-}$, so 0.0500 mol of water forms and neither reagent is left over.

$$\Delta_{neut}H = \frac{-2800.6\ \mathrm{J}}{0.0500\ \mathrm{mol}} = -56012\ \mathrm{J\,mol^{-1}} \approx -56.0\ \mathrm{kJ\,mol^{-1}}$$

The accepted value for a strong acid neutralised by a strong base is close to $-57.1\ \mathrm{kJ\,mol^{-1}}$, and it barely changes whichever strong acid and strong base are used. Both are fully dissociated in solution, so the only chemistry happening is $\mathrm{H^+(aq)} + \mathrm{OH^-(aq)} \rightarrow \mathrm{H_2O(l)}$. That constancy is itself an experimental argument for complete dissociation.

Note the contrast with Question 14. The bomb is sealed, so it delivers $\Delta U$. The cup is open to the air at atmospheric pressure, so it delivers $\Delta H$ directly, with no $\Delta n_g RT$ correction needed.

One approximation has been swallowed here and is worth naming. The specific heat and density used are those of water, not of a salt solution, and the heat absorbed by the polystyrene cup and the thermometer has been ignored. Both approximations push the answer slightly low, which is part of why $-56.0$ falls a little short of $-57.1\ \mathrm{kJ\,mol^{-1}}$. A more careful experiment measures the heat capacity of the calorimeter itself first, by mixing hot and cold water in it, and adds $C_{cal}\Delta T$ to the heat account.

Ans: $\Delta_{neut}H \approx -56.0\ \mathrm{kJ\,mol^{-1}}$ Watch out: The heat is absorbed by the whole 100 g of mixed solution, not by 50 g. And the moles used in the last step are the moles of water formed, not the total moles of acid plus base.

Question 16: Heat released in making a fixed mass of carbon dioxide

Enthalpy of combustion of carbon to $\mathrm{CO_2}$ is $-393.5\ \mathrm{kJ\,mol^{-1}}$. Calculate the heat released upon formation of 35.2 g of $\mathrm{CO_2}$ from carbon and dioxygen gas.

Answer:

The thermochemical equation is

$$\mathrm{C(graphite)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}; \qquad \Delta_c H^{\circ} = -393.5\ \mathrm{kJ\,mol^{-1}}$$

so 393.5 kJ comes out for every mole of $\mathrm{CO_2}$ made.

Molar mass of $\mathrm{CO_2} = 12 + 2(16) = 44\ \mathrm{g\,mol^{-1}}$.

$$n = \frac{35.2\ \mathrm{g}}{44\ \mathrm{g\,mol^{-1}}} = 0.800\ \mathrm{mol}$$

$$q = n \times \Delta_c H^{\circ} = (0.800)(-393.5) = -314.8\ \mathrm{kJ}$$

A reaction enthalpy is an extensive quantity once the amount is fixed: double the amount of substance and you double the heat. What stays fixed is the value per mole, which is why thermochemical data are always tabulated with $\mathrm{mol^{-1}}$ attached. Reading the units of the given value tells you at once whether to multiply by moles.

For this reaction $\Delta_c H^{\circ}$ of carbon and $\Delta_f H^{\circ}$ of $\mathrm{CO_2}$ happen to be the same number, because burning graphite in oxygen is also the formation reaction of carbon dioxide. That coincidence holds only when the combustion product is the compound being formed.

The same figure can be reached from the carbon side, which is a useful cross-check when the question gives the mass of the fuel rather than of the product. 0.800 mol of $\mathrm{CO_2}$ requires 0.800 mol of carbon, that is 9.6 g of graphite, and burning 9.6 g of graphite releases the same 314.8 kJ. Both routes must agree because the equation has a one-to-one stoichiometry.

Ans: 314.8 kJ of heat is released, that is $\Delta H = -314.8\ \mathrm{kJ}$. Watch out: The question asks for "heat released", so the number quoted is 314.8 kJ, but if the answer is written as $\Delta H$ it must carry the minus sign. State which one you are giving.

Question 17: Formation enthalpy of ammonia from a reaction enthalpy

Given $\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \rightarrow 2\mathrm{NH_3(g)}$, $\Delta_r H^{\circ} = -92.4\ \mathrm{kJ\,mol^{-1}}$. What is the standard enthalpy of formation of ammonia gas?

Answer:

A formation reaction makes exactly one mole of the compound from its elements in their reference states. The equation given makes two moles of $\mathrm{NH_3}$, so I halve it:

$$\tfrac{1}{2}\mathrm{N_2(g)} + \tfrac{3}{2}\mathrm{H_2(g)} \rightarrow \mathrm{NH_3(g)}$$

Halving the equation halves the enthalpy change:

$$\Delta_f H^{\circ}(\mathrm{NH_3},\ g) = \frac{-92.4}{2} = -46.2\ \mathrm{kJ\,mol^{-1}}$$

Both $\mathrm{N_2(g)}$ and $\mathrm{H_2(g)}$ are elements in their reference states, so their formation enthalpies are zero and nothing else has to be subtracted.

The general statement behind this one-line calculation is worth writing out, because the next two questions lean on it:

Key Point: $\Delta_r H^{\circ} = \sum a_i\,\Delta_f H^{\circ}(\text{products}) - \sum b_i\,\Delta_f H^{\circ}(\text{reactants})$, with each value multiplied by its stoichiometric coefficient and every element in its reference state contributing zero.

Applied here, $\Delta_r H^{\circ} = 2\,\Delta_f H^{\circ}(\mathrm{NH_3}) - 0 - 0 = -92.4$, which rearranges to the same $-46.2\ \mathrm{kJ\,mol^{-1}}$.

Ans: $\Delta_f H^{\circ}(\mathrm{NH_3},\ g) = -46.2\ \mathrm{kJ\,mol^{-1}}$ Watch out: Fractional coefficients are compulsory here. Writing $\mathrm{N_2} + 3\mathrm{H_2} \rightarrow 2\mathrm{NH_3}$ and calling $-92.4$ the formation enthalpy is the standard error.

Question 18: A reaction enthalpy from a table of formation enthalpies

Enthalpies of formation of $\mathrm{CO(g)}$, $\mathrm{CO_2(g)}$, $\mathrm{N_2O(g)}$ and $\mathrm{N_2O_4(g)}$ are $-110.5$, $-393.5$, $81$ and $9.7\ \mathrm{kJ\,mol^{-1}}$ respectively. Find $\Delta_r H$ for

$$\mathrm{N_2O_4(g)} + 3\mathrm{CO(g)} \rightarrow \mathrm{N_2O(g)} + 3\mathrm{CO_2(g)}$$

Answer:

$$\Delta_r H^{\circ} = \sum a_i \Delta_f H^{\circ}(\text{products}) - \sum b_i \Delta_f H^{\circ}(\text{reactants})$$

Products:

$$\Delta_f H^{\circ}(\mathrm{N_2O}) + 3\,\Delta_f H^{\circ}(\mathrm{CO_2}) = 81 + 3(-393.5) = 81 - 1180.5 = -1099.5\ \mathrm{kJ\,mol^{-1}}$$

Reactants:

$$\Delta_f H^{\circ}(\mathrm{N_2O_4}) + 3\,\Delta_f H^{\circ}(\mathrm{CO}) = 9.7 + 3(-110.5) = 9.7 - 331.5 = -321.8\ \mathrm{kJ\,mol^{-1}}$$

$$\Delta_r H^{\circ} = -1099.5 - (-321.8) = -777.7\ \mathrm{kJ\,mol^{-1}}$$

The large negative value makes chemical sense. Carbon monoxide is a good reducing agent with a fairly weak hold on its oxygen, while carbon dioxide is a very stable molecule, so moving three oxygen atoms from a nitrogen oxide onto CO releases a great deal of energy.

Watch the two positive entries in the data. $\Delta_f H^{\circ}$ of $\mathrm{N_2O}$ and of $\mathrm{N_2O_4}$ are both positive, meaning those oxides sit above their elements in enthalpy. Positive formation enthalpies are perfectly legal and must be carried into the sum with their own signs, not turned negative on the way in.

Ans: $\Delta_r H^{\circ} = -777.7\ \mathrm{kJ\,mol^{-1}}$ Watch out: Every formation enthalpy must be multiplied by its stoichiometric coefficient before the sum. Dropping the 3 in front of CO and $\mathrm{CO_2}$ gives $-211.7$, a favourite distractor.

Question 19: Formation enthalpy of methanol by Hess's law

Calculate the standard enthalpy of formation of $\mathrm{CH_3OH(l)}$ from the following data:

$$\mathrm{CH_3OH(l)} + \tfrac{3}{2}\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\mathrm{H_2O(l)}; \qquad \Delta_r H^{\circ} = -726\ \mathrm{kJ\,mol^{-1}} \quad \text{(i)}$$

$$\mathrm{C(graphite)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}; \qquad \Delta_c H^{\circ} = -393.5\ \mathrm{kJ\,mol^{-1}} \quad \text{(ii)}$$

$$\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}; \qquad \Delta_f H^{\circ} = -285.8\ \mathrm{kJ\,mol^{-1}} \quad \text{(iii)}$$

Answer:

The equation I want is the formation of one mole of liquid methanol from its elements:

$$\mathrm{C(graphite)} + 2\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{CH_3OH(l)}$$

I build it from the three given equations.

Carbon appears on the left of (ii), which is where I want it, so I keep (ii) as it is.

Two moles of $\mathrm{H_2}$ are needed on the left, so I take $2 \times$ (iii).

Methanol must end up on the right, but in (i) it is on the left, so I reverse (i) and reverse the sign of its enthalpy.

$$\Delta_f H^{\circ} = \Delta_r H^{\circ}(\mathrm{ii}) + 2\,\Delta_f H^{\circ}(\mathrm{iii}) - \Delta_r H^{\circ}(\mathrm{i})$$

$$= (-393.5) + 2(-285.8) - (-726)$$

$$= -393.5 - 571.6 + 726 = -239.1\ \mathrm{kJ\,mol^{-1}}$$

A quick check on the oxygen bookkeeping: (ii) plus twice (iii) uses $1 + 1 = 2$ mol of $\mathrm{O_2}$, and reversing (i) releases $\tfrac{3}{2}$ mol, leaving $\tfrac{1}{2}$ mol on the left. That is exactly the target equation.

The routine here is worth fixing as a habit, because every Hess's law question yields to it:

1. Write the target equation first, with the correct states and one mole of the substance being formed.
2. Decide for each given equation whether it must be reversed, and whether it must be multiplied.
3. Apply the same operation to the enthalpy value: reversing flips the sign, multiplying by $k$ multiplies the value by $k$.
4. Add the equations and check that every species not in the target has cancelled.

Step 4 is the one students skip, and it is the only step that catches a mistake made in step 2.

Ans: $\Delta_f H^{\circ}(\mathrm{CH_3OH},\ l) = -239.1\ \mathrm{kJ\,mol^{-1}}$

Question 20: Formation enthalpy of methane from combustion data

The enthalpies of combustion of methane, graphite and dihydrogen at 298 K are $-890.3\ \mathrm{kJ\,mol^{-1}}$, $-393.5\ \mathrm{kJ\,mol^{-1}}$ and $-285.8\ \mathrm{kJ\,mol^{-1}}$ respectively. Calculate the enthalpy of formation of $\mathrm{CH_4(g)}$.

Answer:

The three combustion equations are

$$\mathrm{CH_4(g)} + 2\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\mathrm{H_2O(l)}; \qquad \Delta H = -890.3 \quad \text{(i)}$$

$$\mathrm{C(graphite)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}; \qquad \Delta H = -393.5 \quad \text{(ii)}$$

$$\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}; \qquad \Delta H = -285.8 \quad \text{(iii)}$$

Target:

$$\mathrm{C(graphite)} + 2\mathrm{H_2(g)} \rightarrow \mathrm{CH_4(g)}$$

Take (ii) $+\ 2 \times$ (iii) $-$ (i):

$$\Delta_f H^{\circ} = (-393.5) + 2(-285.8) - (-890.3)$$

$$= -393.5 - 571.6 + 890.3 = -74.8\ \mathrm{kJ\,mol^{-1}}$$

Note the shape of the result: for a compound whose elements and whose own combustion all end at the same products,

$$\Delta_f H^{\circ}(\text{compound}) = \sum \Delta_c H^{\circ}(\text{elements}) - \Delta_c H^{\circ}(\text{compound})$$

The reason it works is that all three combustions end at the same place, $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(l)}$. Burning the elements and burning the compound are two routes from the same starting materials to the same products, so the difference between them is the enthalpy of turning the elements into the compound. Drawing that as a triangle, with the elements at the top left, methane at the top right and the combustion products at the bottom, makes the subtraction obvious and takes about ten seconds.

Methane's $\Delta_f H^{\circ}$ is negative, so methane is more stable than the carbon and hydrogen it is made from. That is the usual situation for a stable compound, and it contrasts sharply with the benzene result in the next question.

Ans: $\Delta_f H^{\circ}(\mathrm{CH_4},\ g) = -74.8\ \mathrm{kJ\,mol^{-1}}$ Watch out: The compound's own combustion enthalpy is subtracted, the elements' are added. Reversing that gives $+74.8$, which is offered in every version of this question.

Question 21: A positive formation enthalpy, and what it says about stability

(a) The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(l)}$ are produced and 3267.6 kJ of heat is liberated. Calculate the standard enthalpy of formation of benzene. Standard enthalpies of formation of $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(l)}$ are $-393.5$ and $-285.8\ \mathrm{kJ\,mol^{-1}}$.

(b) Comment on the thermodynamic stability of $\mathrm{NO(g)}$, given

$$\tfrac{1}{2}\mathrm{N_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{NO(g)}; \qquad \Delta_r H^{\circ} = 90\ \mathrm{kJ\,mol^{-1}}$$

$$\mathrm{NO(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{NO_2(g)}; \qquad \Delta_r H^{\circ} = -74\ \mathrm{kJ\,mol^{-1}}$$

Answer:

(a) The combustion equation is

$$\mathrm{C_6H_6(l)} + \tfrac{15}{2}\mathrm{O_2(g)} \rightarrow 6\mathrm{CO_2(g)} + 3\mathrm{H_2O(l)}; \qquad \Delta_c H^{\circ} = -3267.6\ \mathrm{kJ\,mol^{-1}}$$

and the target is

$$6\mathrm{C(graphite)} + 3\mathrm{H_2(g)} \rightarrow \mathrm{C_6H_6(l)}$$

Using $\Delta_c H^{\circ} = \sum \Delta_f H^{\circ}(\text{products}) - \Delta_f H^{\circ}(\mathrm{C_6H_6})$:

$$\Delta_f H^{\circ}(\mathrm{C_6H_6}) = \left[6(-393.5) + 3(-285.8)\right] - (-3267.6)$$

$$= (-2361.0 - 857.4) + 3267.6 = -3218.4 + 3267.6 = +49.2\ \mathrm{kJ\,mol^{-1}}$$

The value is positive: benzene sits above its elements on the enthalpy scale. The tabulated figure is $+49.0\ \mathrm{kJ\,mol^{-1}}$, and the small gap is rounding in the formation data used as input. A minus sign appears in front of this number in some printings, and it is a slip rather than a different convention: the products come to $-3218.4$ and the combustion term contributes $+3267.6$, and adding two numbers of those magnitudes can only give a small positive result.

(b) $\Delta_f H^{\circ}(\mathrm{NO}) = +90\ \mathrm{kJ\,mol^{-1}}$, also positive, so NO is higher in enthalpy than the dinitrogen and dioxygen it came from. On enthalpy grounds it is unstable with respect to its own elements.

The second equation makes it worse. Converting NO to $\mathrm{NO_2}$ releases 74 kJ per mole, so there is a still lower enthalpy state available to it. NO is therefore unstable both with respect to decomposing into $\mathrm{N_2}$ and $\mathrm{O_2}$ and with respect to being oxidised further to $\mathrm{NO_2}$. This is why nitric oxide made in an engine cylinder does not stay as NO in the open air: it browns into $\mathrm{NO_2}$ within minutes.

The general reading of a formation enthalpy is short. A negative $\Delta_f H^{\circ}$ places the compound below its elements, so it will not fall apart into them on enthalpy grounds. A positive $\Delta_f H^{\circ}$ places it above them, and the compound is described as endothermic and thermodynamically unstable with respect to its elements.

Ans: (a) $\Delta_f H^{\circ}(\mathrm{C_6H_6},\ l) = +49.2\ \mathrm{kJ\,mol^{-1}}$, against a tabulated $+49.0\ \mathrm{kJ\,mol^{-1}}$; (b) NO is thermodynamically unstable, both towards its elements and towards oxidation to $\mathrm{NO_2}$. Watch out: "Unstable" here is a thermodynamic statement only. Benzene and NO both survive on the shelf because their decompositions are slow, and thermodynamics says nothing about rate.