(a) The solubility of the Group 2 sulfates decreases down the group. Explain this trend. (b) Describe what is observed when magnesium and barium are reacted separately with an excess of dilute sulfuric acid. magnesium barium (c) The solubility product, $K_{\text{sp}}$, of $\text{BaSO}_4$ is $1.08 \times 10^{-10} \text{ mol}^2 \text{ dm}^{-6}$ at $298\text{ K}$. Calculate the solubility of $\text{BaSO}_4$ in g per $100\text{ cm}^3$ of solution. (i) The equation for the…

(a) Both lattice and hydration enthalpies become less exothermic; hydration decreases more, so enthalpy of solution becomes less exothermic and solubility decreases. · (b) Mg: fizzing. Ba: fizzing and white precipitate. · (c) 2.43 x 10^-4 g per 100 cm^3 · (d)(i) -652 kJ mol^-1 · (d)(ii) BaSO4 has a more negative lattice energy because Ba2+ is smaller and has a higher charge than Cs+, giving stronger attraction. · (e)(i) 214 K (-58.7 °C) · (e)(ii) -24.9 kJ mol^-1

Chemistry 9701/41 — May/June 2022

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

questions

100

marks

120

minutes

Topics Introduction to A Level Organic Chemistry · Equilibria · Transition Elements · Reaction Kinetics · Electrochemistry · Hydrocarbons · +6 more

Q1Group 2Chemical EnergeticsEquilibriaFree sample

(a)

The solubility of the Group 2 sulfates decreases down the group.

Explain this trend.

DifficultyMedium

Worked solution

Answer

Down the group, both the lattice enthalpy and the hydration enthalpy of the sulfates become less exothermic (less negative).
The hydration enthalpy decreases more than the lattice enthalpy.
Therefore $\Delta H_{\text{sol}} = \Delta H_{\text{hyd}} - \Delta H_{\text{latt}}$ becomes less exothermic/more endothermic, so solubility decreases.

Final answer

Both lattice and hydration enthalpies become less exothermic; hydration decreases more, so enthalpy of solution becomes less exothermic and solubility decreases.

Detailed explanation

Background Concept

The solubility of an ionic solid depends on the balance between the energy needed to break up its lattice and the energy released when the separated ions are hydrated by water. For a sulfate $\text{MSO}_4$ , the enthalpy of solution can be thought of as:

\Delta H_{\text{sol}} = \Delta H_{\text{lattice(breaking)}} + \Delta H_{\text{hydration}}

Down Group 2 the cation $\text{M}^{2+}$ gets larger, so its charge density decreases. Both the lattice enthalpy and the hydration enthalpy become less exothermic (less negative). The key is which changes more.

Understanding the Question

This is an 'explain' question. You are told the trend — solubility of Group 2 sulfates decreases down the group — and you must account for it using lattice and hydration enthalpies. The command word 'explain' means you need to give the reason, not just restate the trend.

Approach

State the two enthalpy terms that change, identify which one changes more, and link that to the enthalpy of solution. Finally connect a less exothermic enthalpy of solution to lower solubility.

Step-by-Step Reasoning

Down the group, $\text{Mg}^{2+}$ , $\text{Ca}^{2+}$ , $\text{Sr}^{2+}$ , $\text{Ba}^{2+}$ all carry the same $2+$ charge but the ionic radius increases. Larger ions have lower charge density.
Lower charge density means water molecules are attracted less strongly to the cation, so the hydration enthalpy becomes less exothermic (less negative).
The lattice enthalpy also becomes less exothermic because the larger ions are held less strongly in the crystal.
The crucial point is that the hydration enthalpy decreases by a greater amount than the lattice enthalpy. This is because hydration is more sensitive to ionic radius and charge density than lattice energy.
Hence $\Delta H_{\text{sol}}$ becomes less exothermic (more endothermic). Dissolving becomes energetically less favourable, so solubility decreases.

Key Takeaways

Solubility trends in Group 2 are explained by the competition between lattice and hydration enthalpy.
Both terms become less exothermic down the group, but hydration changes more.
The sign and size of $\Delta H_{\text{sol}}$ decide whether dissolving is energetically favourable.

Common Mistakes

Saying 'lattice energy decreases' without specifying 'becomes less exothermic/less negative'.
Saying hydration and lattice change by the same amount; the mark is awarded for identifying hydration as the dominant term.
Ignoring the enthalpy of solution link and just saying 'ions get bigger'.

Things to Be Careful About

Use precise wording: 'less exothermic' or 'less negative' rather than just 'decreases'.
The mark scheme accepts either 'hydration decreases more' or 'lattice decreases less' as the dominant-factor point.
Do not discuss entropy here; the syllabus explanation for this trend is enthalpy-based.

Techniques used

compare lattice and hydration enthalpy trendsapply enthalpy of solution relationshiprelate ionic radius to charge density

(b)

Describe what is observed when magnesium and barium are reacted separately with an excess of dilute sulfuric acid.

magnesium

barium

DifficultyEasy

Worked solution

Answer

magnesium: fizzing/effervescence (magnesium dissolves).
barium: fizzing/effervescence and a white precipitate/solid forms.

Final answer

Mg: fizzing. Ba: fizzing and white precipitate.

Detailed explanation

Background Concept

Group 2 metals react with dilute acids to form the salt and hydrogen gas. The sulfate solubility trend matters: $\text{MgSO}_4$ is soluble, while $\text{BaSO}_4$ is very insoluble. With excess dilute sulfuric acid, the sulfate ion is present in large amount, so any insoluble sulfate precipitates immediately.

Understanding the Question

This is a 'describe what is observed' question. You need to state the visible signs for each metal separately. The word 'excess' is important: there is plenty of sulfate available to precipitate $\text{BaSO}_4$ .

Approach

For each metal, think about (i) reaction with acid producing $\text{H}_2$ gas, and (ii) whether the sulfate formed is soluble or insoluble.

Step-by-Step Reasoning

Magnesium reacts with sulfuric acid: $\text{Mg(s)} + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{MgSO}_4(\text{aq}) + \text{H}_2(\text{g})$ $\text{MgSO}_4$ is soluble, so the only observation is fizzing/effervescence from hydrogen gas, and the magnesium disappears.
Barium also reacts with acid to produce $\text{H}_2$ , so there is fizzing. But $\text{BaSO}_4$ is insoluble, so a white precipitate/solid of $\text{BaSO}_4$ forms immediately.

Key Takeaways

All Group 2 metals fizz in acid due to $\text{H}_2$ evolution.
Insoluble sulfates appear as white precipitates when sulfate ions are added to their cations.

Common Mistakes

Writing 'bubbles' instead of 'fizzing/effervescence' is usually accepted, but avoid saying 'barium dissolves' because the metal is coated by insoluble $\text{BaSO}_4$ .
Forgetting the white precipitate for barium.

Things to Be Careful About

The mark scheme wants 'fizzing' for magnesium and 'fizzing and white solid/precipitate' for barium.
'White precipitate' is better than just 'precipitate'.

Techniques used

predict observations from solubility of sulfatesdescribe gas evolutionidentify precipitate formation

(c)

The solubility product, $K_{\text{sp}}$ , of $\text{BaSO}_4$ is $1.08 \times 10^{-10} \text{ mol}^2 \text{ dm}^{-6}$ at $298\text{ K}$ .

Calculate the solubility of $\text{BaSO}_4$ in g per $100\text{ cm}^3$ of solution.

DifficultyMedium-Easy

Worked solution

Working

$\text{BaSO}_4(\text{s}) \rightleftharpoons \text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq})$

$K_{\text{sp}} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = s^2$

$s = \sqrt{1.08 \times 10^{-10}} = 1.04 \times 10^{-5}\ \text{mol dm}^{-3}$

$M_r(\text{BaSO}_4) = 233.4$

mass in $100\text{ cm}^3 = 1.04 \times 10^{-5} \times 233.4 \times \frac{100}{1000} = 2.43 \times 10^{-4}\ \text{g}$

Answer

$2.43 \times 10^{-4}\ \text{g per }100\text{ cm}^3$

Final answer

2.43 x 10^-4 g per 100 cm^3

Detailed explanation

Background Concept

For a sparingly soluble salt such as $\text{BaSO}_4$ , the solubility product is the equilibrium constant for dissolution:

\text{BaSO}_4(\text{s}) \rightleftharpoons \text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq})

Because the solid is not included in the equilibrium expression,

K_{\text{sp}} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]

If the molar solubility is $s$ , then both ion concentrations equal $s$ , so $K_{\text{sp}} = s^2$ .

Understanding the Question

You are given $K_{\text{sp}}$ and asked for solubility in g per $100\text{ cm}^3$ . This has two stages: find molar solubility in $\text{mol dm}^{-3}$ , then convert to mass in $100\text{ cm}^3$ using molar mass.

Approach

Write the dissolution equilibrium and $K_{\text{sp}}$ expression.
Let molar solubility $= s$ , so $K_{\text{sp}} = s^2$ .
Take the square root.
Convert $\text{mol dm}^{-3}$ to $\text{g dm}^{-3}$ using $M_r$ , then scale to $100\text{ cm}^3$ $(0.100\text{ dm}^3)$ .

Step-by-Step Reasoning

$s = \sqrt{1.08 \times 10^{-10}} = 1.04 \times 10^{-5}\ \text{mol dm}^{-3}$ .
$M_r(\text{BaSO}_4) = 137.3 + 32.1 + 4(16.0) = 233.4\ \text{g mol}^{-1}$ .
Solubility in $\text{g dm}^{-3}$ : $1.04 \times 10^{-5} \times 233.4 = 2.43 \times 10^{-3}\ \text{g dm}^{-3}$ .
In $100\text{ cm}^3 = 0.100\text{ dm}^3$ : $2.43 \times 10^{-3} \times 0.100 = 2.43 \times 10^{-4}\ \text{g}$ .

Key Takeaways

$K_{\text{sp}}$ of a 1:1 salt gives molar solubility as $\sqrt{K_{\text{sp}}}$ .
Always check the volume unit requested.

Common Mistakes

Forgetting to divide by 10 when converting $\text{dm}^3$ to $100\text{ cm}^3$ .
Using 233 instead of 233.4; acceptable but 233.4 is better.
Giving the answer in $\text{mol dm}^{-3}$ instead of g per $100\text{ cm}^3$ .

Things to Be Careful About

The final answer needs at least 2 significant figures.
Include units: g per $100\text{ cm}^3$ .

Techniques used

write solubility equilibrium expressiontake square root of Kspconvert molar solubility to mass per volume

(d)

(i)

The equation for the formation of a gaseous sulfate ion is shown.

\text{S(s)} + 2\text{O}_2\text{(g)} + 2\text{e}^- \rightarrow \text{SO}_4^{2-}\text{(g)} \quad \Delta H = \Delta H_{\text{f}}^{\ominus} \text{ of } \text{SO}_4^{2-}\text{(g)}

Calculate the standard enthalpy change of formation, $\Delta H_{\text{f}}^{\ominus}$ , of $\text{SO}_4^{2-}\text{(g)}$ . It may be helpful to draw a labelled energy cycle. Use relevant data from Table 1.1 in your calculations.

Table 1.1

energy change	value / $\text{kJ mol}^{-1}$
lattice energy of barium sulfate, $\text{BaSO}_4\text{(s)}$	$-2469$
standard enthalpy change of formation of barium sulfate	$-1473$
standard enthalpy change of atomisation of barium	$+180$
first ionisation energy of barium	$+503$
second ionisation energy of barium	$+965$
standard enthalpy change of atomisation of sulfur	$+279$
standard enthalpy change for $\text{S(g)} \rightarrow \text{S}^{2-}\text{(g)}$	$+440$
standard enthalpy change for $\text{O(g)} \rightarrow \text{O}^{2-}\text{(g)}$	$+657$
$\text{O=O}$ bond energy	$+496$

DifficultyMedium

Worked solution

Working

Born–Haber cycle for $\text{BaSO}_4$ :

$\Delta H_{\text{f}}^{\ominus}(\text{BaSO}_4) = \Delta H_{\text{at}}(\text{Ba}) + IE_1 + IE_2 + \Delta H_{\text{f}}^{\ominus}(\text{SO}_4^{2-}(\text{g})) + \Delta H_{\text{latt}}$

$-1473 = 180 + 503 + 965 + \Delta H_{\text{f}}^{\ominus}(\text{SO}_4^{2-}(\text{g})) - 2469$

$\Delta H_{\text{f}}^{\ominus}(\text{SO}_4^{2-}(\text{g})) = -1473 - 180 - 503 - 965 + 2469 = -652\ \text{kJ mol}^{-1}$

Answer

$-652\ \text{kJ mol}^{-1}$

Final answer

-652 kJ mol^-1

Detailed explanation

Background Concept

A Born–Haber cycle is an application of Hess's law to ionic compounds. The enthalpy of formation of an ionic solid can be reached either directly from its elements, or indirectly by atomising and ionising the elements, forming the anion, and then allowing the gaseous ions to come together to form the lattice.

For $\text{BaSO}_4$ :

\Delta H_{\text{f}}^{\ominus}(\text{BaSO}_4) = \Delta H_{\text{at}}(\text{Ba}) + IE_1(\text{Ba}) + IE_2(\text{Ba}) + \Delta H_{\text{f}}^{\ominus}(\text{SO}_4^{2-}(\text{g})) + \Delta H_{\text{latt}}

Here $\Delta H_{\text{latt}}$ is negative because it is the enthalpy change for forming the solid from gaseous ions.

Understanding the Question

You need to find the formation enthalpy of the gaseous sulfate ion, $\Delta H_{\text{f}}^{\ominus}(\text{SO}_4^{2-}(\text{g}))$ , using the data in Table 1.1. The equation in the question defines this quantity directly, so you can treat it as one step in the cycle.

Approach

Set up the Born–Haber cycle for $\text{BaSO}_4$ . Substitute the five relevant values into the equation above and solve for the unknown. The sulfur atomisation, $\text{S(g)} \rightarrow \text{S}^{2-}\text{(g)}$ , $\text{O(g)} \rightarrow \text{O}^{2-}\text{(g)}$ and $\text{O=O}$ bond energy data are not needed because the formation of $\text{SO}_4^{2-}\text{(g)}$ is already a single given/unknown step.

Step-by-Step Reasoning

Route from elements to $\text{BaSO}_4\text{(s)}$ : atomise Ba $(+180)$ , ionise to $\text{Ba}^{2+}$ $(+503+965)$ , form $\text{SO}_4^{2-}\text{(g)}$ (unknown), combine gaseous ions (lattice, $-2469$ ).
This must equal the direct formation enthalpy, $-1473$ .
Equation: $-1473 = 180 + 503 + 965 + \Delta H_{\text{f}}^{\ominus}(\text{SO}_4^{2-}\text{(g)}) - 2469$
Solve: $\Delta H_{\text{f}}^{\ominus}(\text{SO}_4^{2-}\text{(g)}) = -1473 - 180 - 503 - 965 + 2469 = -652\ \text{kJ mol}^{-1}$

Key Takeaways

Born–Haber cycles are just Hess cycles: the direct route equals the sum of the indirect route.
Lattice energy is negative when defined as formation of the solid from gaseous ions.
Not every piece of data in a table is needed; identify the route first.

Common Mistakes

Using the sulfur atomisation or $\text{O=O}$ bond energy values unnecessarily.
Getting the sign of the lattice energy wrong; it must be added as a negative value.
Forgetting to include both ionisation energies of barium.

Things to Be Careful About

The mark scheme requires exactly the five values: $+180$ , $+503$ , $+965$ , $-2469$ and $-1473$ .
Check the arithmetic signs: the final answer is negative, $-652\ \text{kJ mol}^{-1}$ .
Include units and sign in the final answer.

Techniques used

construct Born-Haber cyclesum enthalpy terms with correct signssolve for unknown formation enthalpy

(ii)

Suggest how the lattice energy of $\text{BaSO}_4\text{(s)}$ differs from the lattice energy of $\text{Cs}_2\text{SO}_4\text{(s)}$ .

Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

$\text{BaSO}_4$ has a more negative (larger) lattice energy than $\text{Cs}_2\text{SO}_4$ .

$\text{Ba}^{2+}$ is smaller and has a higher charge than $\text{Cs}^+$ , so the electrostatic attraction between $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$ is stronger.

Final answer

BaSO4 has a more negative lattice energy because Ba2+ is smaller and has a higher charge than Cs+, giving stronger attraction.

Detailed explanation

Background Concept

Lattice energy is the enthalpy change when one mole of an ionic solid is formed from its gaseous ions. Its magnitude depends on the charges on the ions and the distance between them. Higher charge and smaller ionic radius give stronger electrostatic attraction and a more negative (more exothermic) lattice energy.

Understanding the Question

Compare $\text{BaSO}_4$ with $\text{Cs}_2\text{SO}_4$ . Both contain the sulfate ion, but the cations differ: $\text{Ba}^{2+}$ versus two $\text{Cs}^+$ . You need to say which lattice energy is more negative and justify using charge and radius.

Approach

Compare the charge and size of $\text{Ba}^{2+}$ and $\text{Cs}^+$ . Then connect the stronger attraction to the more negative lattice energy.

Step-by-Step Reasoning

$\text{Ba}^{2+}$ has a $2+$ charge; $\text{Cs}^+$ has a $1+$ charge.
$\text{Ba}^{2+}$ is also smaller than $\text{Cs}^+$ . They are isoelectronic, but the greater nuclear charge of barium pulls the electrons closer.
Therefore the electrostatic attraction between $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$ is stronger than between $\text{Cs}^+$ and $\text{SO}_4^{2-}$ .
Hence $\text{BaSO}_4$ has a more negative (larger magnitude) lattice energy than $\text{Cs}_2\text{SO}_4$ .

Key Takeaways

Lattice energy increases with ionic charge and decreases with ionic radius.
When comparing ions, consider both charge and size.

Common Mistakes

Saying ' $\text{BaSO}_4$ has a larger lattice energy' without specifying 'more negative'.
Ignoring the charge difference and only mentioning radius.
Thinking two $\text{Cs}^+$ ions somehow compensate for the lower charge; each $\text{Cs}^+$ only has a $1+$ charge.

Things to Be Careful About

The mark scheme wants three ideas: more negative, smaller/higher charge, stronger attraction.
Use 'more negative' rather than 'bigger' alone.

Techniques used

compare ionic charge and radiusrelate electrostatic attraction to lattice energy

(e)

The reaction of solid hydrated barium hydroxide, $\text{Ba(OH)}_2\cdot8\text{H}_2\text{O}$ , with ammonium salts is endothermic.

(i)

Calculate the minimum temperature at which the reaction of $\text{Ba(OH)}_2\cdot8\text{H}_2\text{O}$ with $\text{NH}_4\text{NO}_3$ becomes feasible. Show all your working.

\text{Ba(OH)}_2\cdot8\text{H}_2\text{O(s)} + 2\text{NH}_4\text{NO}_3\text{(s)} \rightarrow 2\text{NH}_3\text{(g)} + \text{Ba(NO}_3)_2\text{(s)} + 10\text{H}_2\text{O(l)} \quad \Delta H_{\text{r}}^{\ominus} = +132 \text{ kJ mol}^{-1}

\Delta S^{\ominus} = +616 \text{ J K}^{-1} \text{ mol}^{-1}

DifficultyMedium-Easy

Worked solution

Working

At the minimum feasible temperature, $\Delta G^{\ominus} = 0$ .

$T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}} = \frac{132}{0.616} = 214.3\ \text{K}$

$T = 214.3 - 273.15 = -58.7\ ^\circ\text{C}$

Answer

$214\ \text{K}$ ( $-58.7\ ^\circ\text{C}$ ); the reaction becomes feasible above this temperature.

Final answer

214 K (-58.7 °C)

Detailed explanation

Background Concept

Gibbs free energy change determines feasibility:

\Delta G = \Delta H - T\Delta S

A reaction is feasible when $\Delta G < 0$ . If $\Delta H$ is positive and $\Delta S$ is positive, the $-T\Delta S$ term becomes more negative as $T$ increases, so there is a minimum temperature above which the reaction becomes feasible. At that threshold $\Delta G = 0$ .

Understanding the Question

You are given $\Delta H$ and $\Delta S$ for an endothermic reaction and asked for the minimum temperature at which it becomes feasible. This is a calculation: set $\Delta G = 0$ and solve for $T$ .

Approach

Use

T = \frac{\Delta H}{\Delta S}

Ensure units match: $\Delta H$ is in $\text{kJ mol}^{-1}$ , $\Delta S$ in $\text{J K}^{-1}\text{mol}^{-1}$ , so either convert $\Delta S$ to $\text{kJ K}^{-1}\text{mol}^{-1}$ or $\Delta H$ to $\text{J mol}^{-1}$ . Then convert K to $^\circ\text{C}$ if required.

Step-by-Step Reasoning

At threshold, $0 = \Delta H - T\Delta S$ , so $T = \Delta H / \Delta S$ .
$\Delta S = +616\ \text{J K}^{-1}\text{mol}^{-1} = +0.616\ \text{kJ K}^{-1}\text{mol}^{-1}$ .
$T = 132 / 0.616 = 214.3\ \text{K}$ .
In Celsius: $214.3 - 273.15 = -58.7\ ^\circ\text{C}$ .
Because $\Delta S$ is positive, the reaction becomes feasible above this temperature (higher $T$ makes $\Delta G$ more negative).

Key Takeaways

For an endothermic reaction with positive $\Delta S$ , feasibility increases with temperature.
The minimum temperature is where $\Delta G = 0$ .
Always match units of $\Delta H$ and $\Delta S$ .

Common Mistakes

Using $\Delta S = 616$ without converting to kJ, giving $T = 0.214\ \text{K}$ .
Forgetting to convert from K to $^\circ\text{C}$ .
Saying feasible below the threshold; it is feasible above.

Things to Be Careful About

The mark scheme accepts $T = 214.3\ \text{K}$ and $T = -58.7\ ^\circ\text{C}$ , minimum 2 sf.
Include both units if asked for temperature; the question just says 'minimum temperature', so giving K is acceptable, but $^\circ\text{C}$ is often expected.
Show the conversion of $\Delta S$ to kJ.

Techniques used

apply Gibbs free energy equationset ΔG = 0 at thresholdconvert units of ΔS

(ii)

Barium hydroxide reacts readily with ammonium chloride on mixing at room temperature.

\text{Ba(OH)}_2\cdot8\text{H}_2\text{O(s)} + 2\text{NH}_4\text{Cl(s)} \rightarrow 2\text{NH}_3\text{(g)} + \text{BaCl}_2\cdot2\text{H}_2\text{O(s)} + 8\text{H}_2\text{O(l)} \quad \Delta H_{\text{r}}^{\ominus} = +133 \text{ kJ mol}^{-1}

Some relevant standard entropies are given in Table 1.2.

Table 1.2

substance	$\text{Ba(OH)}_2\cdot8\text{H}_2\text{O(s)}$	$\text{NH}_4\text{Cl(s)}$	$\text{NH}_3\text{(g)}$	$\text{BaCl}_2\cdot2\text{H}_2\text{O(s)}$	$\text{H}_2\text{O(l)}$
$S^{\ominus} / \text{J K}^{-1} \text{ mol}^{-1}$	$427$	$95$	$192$	$203$	$70$

Calculate the standard Gibbs free energy change, $\Delta G^{\ominus}$ , for this reaction at $25\text{ }^\circ\text{C}$ .

DifficultyMedium

Worked solution

Working

$\Delta S^{\ominus} = [203 + (8 \times 70) + (2 \times 192)] - [427 + (2 \times 95)]$

$= 1147 - 617 = +530\ \text{J K}^{-1}\text{mol}^{-1}$

$\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus} = 133 - 298 \times 0.530$

$= 133 - 157.9 = -24.9\ \text{kJ mol}^{-1}$

Answer

$-24.9\ \text{kJ mol}^{-1}$

Final answer

-24.9 kJ mol^-1

Detailed explanation

Background Concept

Standard entropy change for a reaction is products minus reactants using standard molar entropies:

\Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})

Then the Gibbs free energy change at a given temperature is:

\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}

A negative $\Delta G$ means the reaction is feasible under standard conditions.

Understanding the Question

You are given the balanced equation, $\Delta H^{\ominus}$ , and standard entropies. You need to calculate $\Delta S^{\ominus}$ from the table, then use it to find $\Delta G^{\ominus}$ at $25\ ^\circ\text{C}$ ( $298\ \text{K}$ ).

Approach

Sum product entropies, sum reactant entropies, subtract.
Convert $\Delta S$ from J to kJ.
Substitute into $\Delta G = \Delta H - T\Delta S$ .

Step-by-Step Reasoning

Products: $\text{BaCl}_2\cdot2\text{H}_2\text{O(s)}$ $203$ ; $8\text{H}_2\text{O(l)}$ $8 \times 70 = 560$ ; $2\text{NH}_3\text{(g)}$ $2 \times 192 = 384$ . Total $= 1147\ \text{J K}^{-1}\text{mol}^{-1}$ .
Reactants: $\text{Ba(OH)}_2\cdot8\text{H}_2\text{O(s)}$ $427$ ; $2\text{NH}_4\text{Cl(s)}$ $2 \times 95 = 190$ . Total $= 617\ \text{J K}^{-1}\text{mol}^{-1}$ .
$\Delta S^{\ominus} = 1147 - 617 = +530\ \text{J K}^{-1}\text{mol}^{-1} = +0.530\ \text{kJ K}^{-1}\text{mol}^{-1}$ .
$\Delta G^{\ominus} = 133 - 298 \times 0.530 = 133 - 157.9 = -24.9\ \text{kJ mol}^{-1}$ .
Negative $\Delta G$ means the reaction is feasible at $25\ ^\circ\text{C}$ .

Key Takeaways

Entropy change is a sum of standard entropies with stoichiometric coefficients.
$\Delta G$ combines enthalpy and entropy at a specified temperature.
A negative $\Delta G$ indicates feasibility under standard conditions.

Common Mistakes

Forgetting to multiply by stoichiometric coefficients (e.g. $8\text{H}_2\text{O}$ , $2\text{NH}_3$ , $2\text{NH}_4\text{Cl}$ ).
Leaving $\Delta S$ in J and using $\Delta H$ in kJ without converting, giving the wrong sign or magnitude.
Sign errors in the products-minus-reactants subtraction.

Things to Be Careful About

The mark scheme allows ecf for $\Delta S$ if the arithmetic is correct.
Final answer should be $-24.9\ \text{kJ mol}^{-1}$ (1 dp).
Include units: $\text{kJ mol}^{-1}$ .
Note the reaction is feasible because $\Delta G$ is negative.

Techniques used

calculate entropy change from standard entropiesapply Gibbs free energy equationconvert J to kJ

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