"Chemical Thermodynamics" — Practice

Question 1

Three moles of an ideal gas expand isothermally at $300$ K from $10$ L to $40$ L, reversibly and then against a constant pressure of $0.5$ atm. Find $w$, $q$ and $\Delta U$ in each case.

Accepted Answer

$\Delta U=0$ for both, since the temperature is unchanged, so $q=-w$ in each case.

Reversible: $w=-3\left(8.314ight)\left(300ight)\ln4=-7482	imes1.386=-10370$ J, and $q=+10370$ J.

Irreversible: $w=-\left(0.5	imes1.013	imes10^{5}ight)\left(30	imes10^{-3}ight)=-1520$ J, and $q=+1520$ J.

The same state change, two very different heat and work values — the definition of path functions.

Question 2

For the reaction $	ext{N}_2\left(gight)+3	ext{H}_2\left(gight)ightarrow2	ext{NH}_3\left(gight)$, $\Delta H=-92.4$ kJ mol$^{-1}$ at $298$ K. Find $\Delta U$.

Accepted Answer

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

$\Delta U=\Delta H-\Delta n_gRT=-92.4-\left(-2ight)\left(8.314	imes10^{-3}ight)\left(298ight)$

$=-92.4+4.95=-87.5$ kJ mol$^{-1}$

The internal energy change is less negative because the surroundings do work on the system as the gas volume contracts.

Question 3

Given $\Delta H^{\circ}_f$ values of $-393.5$ for $	ext{CO}_2\left(gight)$, $-285.8$ for $	ext{H}_2	ext{O}\left(light)$ and $-74.8$ kJ mol$^{-1}$ for $	ext{CH}_4\left(gight)$, find the enthalpy of combustion of methane.

Accepted Answer

$	ext{CH}_4+2	ext{O}_2ightarrow	ext{CO}_2+2	ext{H}_2	ext{O}\left(light)$

$\Delta H=\left[-393.5+2\left(-285.8ight)ight]-\left[-74.8+0ight]$

$=-965.1+74.8=-890.3$ kJ mol$^{-1}$

Elemental oxygen has zero formation enthalpy by definition, which is why it never appears in such a calculation.

Question 4

Use a Born-Haber cycle to find the lattice enthalpy of potassium bromide, given sublimation $+89$, ionisation $+419$, half-dissociation of bromine $+96$, electron gain $-325$ and formation $-392$ kJ mol$^{-1}$.

Accepted Answer

$\Delta H_f=\Delta H_{sub}+IE+	frac12D+\Delta H_{eg}+U$

$-392=89+419+96-325+U$

$-392=279+U$

$U=-671$ kJ mol$^{-1}$

The lattice term is again the largest, and formation is exothermic only because it repays the three endothermic steps with interest.

Question 5

The enthalpy of hydrogenation of 1,3-butadiene is $-239$ kJ mol$^{-1}$ and that of 1-butene is $-127$ kJ mol$^{-1}$. Estimate the resonance energy of butadiene.

Accepted Answer

Two isolated double bonds would give $2	imes\left(-127ight)=-254$ kJ mol$^{-1}$.

The measured value is $-239$.

Resonance energy $=254-239=15$ kJ mol$^{-1}$

Far smaller than benzene's $152$ kJ mol$^{-1}$, which is the quantitative statement of the difference between simple conjugation and aromatic delocalisation.

Question 6

The enthalpy of neutralisation of hydrocyanic acid with sodium hydroxide is $-12.1$ kJ mol$^{-1}$. Find the enthalpy of ionisation of the acid and comment on its strength.

Accepted Answer

$\Delta H_{ionisation}=-12.1-\left(-57.1ight)=+45.0$ kJ mol$^{-1}$

This is a very large positive value, meaning a great deal of energy is required to separate the proton from cyanide.

Hydrocyanic acid is therefore a **very weak** acid, consistent with its $pK_a$ of about $9.2$.

The more energy ionisation costs, the less negative the measured neutralisation enthalpy becomes.

Question 7

Find the entropy change when $2$ mol of an ideal gas is heated at constant pressure from $300$ K to $600$ K, with $C_p=29.1$ J K$^{-1}$ mol$^{-1}$, and separately when it expands isothermally to four times its volume.

Accepted Answer

Heating: $\Delta S=nC_p\ln\dfrac{T_2}{T_1}=2\left(29.1ight)\ln2=58.2	imes0.693=40.3$ J K$^{-1}$

Expansion: $\Delta S=nR\ln4=2\left(8.314ight)\left(1.386ight)=23.0$ J K$^{-1}$

Both are positive, since heating and expansion each increase the accessible microstates.

Doubling the temperature produces the larger effect here, because $C_p$ exceeds $R$ substantially.

Question 8

Predict the sign of $\Delta S$ for the dissolution of ammonium nitrate in water, and explain why the salt dissolves despite the process being endothermic.

Accepted Answer

The ordered crystal lattice is destroyed and the ions are dispersed, so $\Delta S$ is strongly **positive**.

Dissolution is endothermic, with $\Delta H\approx+26$ kJ mol$^{-1}$, because hydration does not fully repay the lattice enthalpy.

At room temperature $T\Delta S$ exceeds $\Delta H$, so $\Delta G$ is negative and the salt dissolves — cooling the solution as it does so.

This is exactly how instant cold packs work.

Question 9

For a reaction $\Delta G^{\circ}=+21.4$ kJ mol$^{-1}$ at $298$ K. Find $K$, and find $\Delta G$ when $Q=10^{-6}$.

Accepted Answer

$\ln K=-\dfrac{21400}{8.314	imes298}=-8.64\qquad\Rightarrow\qquad K=1.8	imes10^{-4}$

$\Delta G=21400+8.314\left(298ight)\ln\left(10^{-6}ight)=21400-34240=-12840$ J mol$^{-1}$

Strongly negative, so the reaction proceeds forward at that composition even though $\Delta G^{\circ}$ is positive.

Since $Q$ is well below $K$, there is not yet enough product and the system must make more.

Question 10

A reaction has $\Delta H=+58$ kJ mol$^{-1}$ and $\Delta S=+176$ J K$^{-1}$ mol$^{-1}$. Find the temperature above which it is spontaneous, and $\Delta G$ at $400$ K.

Accepted Answer

$T=\dfrac{58000}{176}=330$ K

Above $330$ K the entropy term wins and the reaction becomes spontaneous.

At $400$ K: $\Delta G=58000-400\left(176ight)=58000-70400=-12400$ J mol$^{-1}$

Negative, confirming spontaneity at that temperature. This is the dinitrogen tetroxide dissociation, which is why the gas becomes browner on warming.

Question 11

The equilibrium constant of a reaction falls from $4.0$ at $300$ K to $1.0$ at $340$ K. Find $\Delta H^{\circ}$ and state whether the reaction is exothermic.

Accepted Answer

$\ln\dfrac{1.0}{4.0}=-\dfrac{\Delta H^{\circ}}{8.314}\left(\dfrac{1}{340}-\dfrac{1}{300}ight)$

$-1.386=-\dfrac{\Delta H^{\circ}}{8.314}\left(-3.92	imes10^{-4}ight)$

$\Delta H^{\circ}=-\dfrac{1.386	imes8.314}{3.92	imes10^{-4}}=-2.94	imes10^{4}$ J mol$^{-1}$

Negative, so the reaction is **exothermic** — consistent with $K$ falling as temperature rises, exactly as Le Chatelier requires.

Question 12

Explain why zinc oxide can be reduced by carbon above about $1200$ K but not below, using Ellingham reasoning.

Accepted Answer

For $2	ext{Zn}+	ext{O}_2ightarrow2	ext{ZnO}$, $\Delta n_g=-1$, so $\Delta S$ is negative and $\Delta G^{\circ}$ becomes **less** negative as temperature rises.

For $2	ext{C}+	ext{O}_2ightarrow2	ext{CO}$, $\Delta n_g=+1$, so $\Delta S$ is positive and $\Delta G^{\circ}$ becomes **more** negative on heating.

The two lines therefore cross. Below the crossing, zinc holds oxygen more strongly and carbon cannot take it; above it, carbon does.

Coupling the two gives a net negative $\Delta G^{\circ}$ only above roughly $1200$ K, which is why zinc smelting requires high temperatures.

Practice — "Chemical Thermodynamics"

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