"Chemical Thermodynamics" — Practice

Question 1

For the reaction $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ at 298 K, the enthalpy change is $-92.4$ kJ. Find the internal energy change.

Accepted Answer

Gas moles go from 4 to 2, so $\Delta n_g = -2$. Then $\Delta U = \Delta H - \Delta n_g RT = -92.4 - (-2)(8.314	imes10^{-3})(298) = -92.4 + 4.95 = -87.4$ kJ. The internal energy change is smaller in magnitude, because the surroundings did work on the system as the gas volume contracted.

Question 2

One mole of an ideal gas expands isothermally and reversibly at 300 K from 2 L to 20 L. Calculate the work done.

Accepted Answer

Use w equals $-2.303$ nRT log of the volume ratio. That is $-2.303$ times 1 times 8.314 times 300 times log 10, and log 10 is exactly 1, so w equals $-5744$ J, that is about $-5.74$ kJ. The negative sign means work is done by the gas on the surroundings. Since the process is isothermal and the gas ideal, the internal energy change is zero and the gas must absorb 5.74 kJ of heat.

Question 3

An ideal gas expands into an evacuated container. Find the work done and the change in internal energy, with reasoning.

Accepted Answer

The external pressure is zero, so $w = -p_{ext}\Delta V = 0$ however large the expansion — the gas pushes against nothing. For an ideal gas the internal energy depends only on temperature, and a free expansion changes no temperature, so $\Delta U = 0$ as well. The first law then forces $q = 0$ too. This is the cleanest demonstration that work depends on the path and not merely on the change of state: the same expansion carried out reversibly would have delivered the maximum possible work.

Question 4

Given standard enthalpies of combustion of carbon, hydrogen and ethanol as $-393.5$, $-285.8$ and $-1367$ kJ mol$^{-1}$, calculate the standard enthalpy of formation of ethanol, $\mathrm{C_2H_5OH}$.

Accepted Answer

The target is $\mathrm{2C(s) + 3H_2(g) + 	frac{1}{2}O_2(g) ightarrow C_2H_5OH(l)}$. By Hess's law this equals $2\Delta_cH(\mathrm{C}) + 3\Delta_cH(\mathrm{H_2}) - \Delta_cH(\mathrm{C_2H_5OH})$, that is $2(-393.5) + 3(-285.8) - (-1367) = -787.0 - 857.4 + 1367 = -277.4$ kJ mol$^{-1}$. Subtracting the ethanol term rather than adding it is the step to get right.

Question 5

Why is the standard enthalpy of formation of oxygen gas zero, while that of ozone is $+142$ kJ mol$^{-1}$?

Accepted Answer

Because $\mathrm{O_2}$ is the standard state of the element oxygen, and every element in its standard state is assigned zero by definition — it is the reference point the whole table is measured from. Ozone is a different allotrope, not the standard state, so forming it from $\mathrm{O_2}$ has a real and measurable enthalpy change. The positive value says ozone is the less stable form.

Question 6

Estimate the enthalpy change for $\mathrm{H_2(g) + Cl_2(g) \rightarrow 2HCl(g)}$ from bond enthalpies of 436 for H-H, 242 for Cl-Cl and 431 for H-Cl, in kJ mol$^{-1}$.

Accepted Answer

Bonds broken: $436 + 242 = 678$. Bonds formed: $2 	imes 431 = 862$. So $\Delta H = 678 - 862 = -184$ kJ mol$^{-1}$, exothermic. Forgetting to double the H-Cl term is the usual slip, and it would give $+247$ — the wrong sign as well as the wrong size.

Question 7

Predict the sign of the entropy change for each of: melting of ice, $\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)}$, and dissolving sugar in water.

Accepted Answer

Melting ice: positive, since a rigid lattice becomes a mobile liquid. The nitrogen dioxide reaction: negative, since three moles of gas become two. Dissolving sugar: positive, since an ordered crystal disperses through the solvent. Counting gas moles settles the middle case at a glance.

Question 8

Water freezes spontaneously at $-10$ °C even though the entropy of the water decreases. Explain how this is consistent with the second law.

Accepted Answer

The system entropy does fall, since a liquid becomes an ordered solid. But freezing is exothermic, and that released heat flows into the surroundings and raises their entropy by $q/T$. Below $0$ °C the surroundings gain more than the system loses, so $\Delta S_{total}$ is positive and the second law is satisfied. The same statement in system-only terms is that $\Delta G = \Delta H - T\Delta S$ is negative there, because $T$ is small enough that the negative $\Delta H$ outweighs the negative $T\Delta S$.

Question 9

For a reaction, $\Delta H$ is $+30.6$ kJ mol$^{-1}$ and $\Delta S$ is $+66.0$ J K$^{-1}$ mol$^{-1}$. At what temperature does it become spontaneous?

Accepted Answer

Both are positive, so this is the high-temperature case. Setting $\Delta G = 0$ gives $T = \Delta H/\Delta S = 30\,600/66.0 = 464$ K, about $191$ °C. Above that temperature $T\Delta S$ exceeds $\Delta H$ and the reaction becomes spontaneous. Converting kJ to J is essential — omitting it gives 0.46 K.

Question 10

A reaction has an equilibrium constant of 100 at 298 K. Calculate its standard Gibbs energy change.

Accepted Answer

$\Delta G^\circ = -2.303RT\log K = -2.303(8.314)(298)\log 100 = -(5706)(2) = -11\,412$ J, that is $-11.4$ kJ mol$^{-1}$. This matches the rule that every 5.7 kJ mol$^{-1}$ at 298 K is one factor of ten in $K$.

Question 11

Explain the difference between the standard Gibbs energy change and the Gibbs energy change of a reaction, and state the value of each at equilibrium.

Accepted Answer

$\Delta G^\circ$ compares pure reactants in their standard states with pure products in theirs. It is a single fixed number at a given temperature and locates the equilibrium position through $\Delta G^\circ = -RT\ln K$; at equilibrium it is generally not zero. $\Delta G$ is the running value at the actual composition — the slope of the free energy curve — and it is exactly zero at equilibrium, which is the minimum of that curve rather than the pure products.

Question 12

Ammonium nitrate dissolves in water and the beaker becomes noticeably cold. Explain why the process is spontaneous.

Accepted Answer

Because entropy, not enthalpy, is driving it. Dissolving ammonium nitrate is endothermic: the lattice enthalpy slightly exceeds the hydration enthalpy, so the solution absorbs heat and cools. But the ordered crystal disperses into freely moving hydrated ions, a large positive $\Delta S$, and at room temperature $T\Delta S$ outweighs the positive $\Delta H$. The result is a negative $\Delta G$ and a spontaneous process that feels cold.

Practice — "Chemical Thermodynamics"

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