Module 4 · L12 of 13~35 min⚡ +50 XP in Learn · +25 to complete
Calculating ΔS° & Standard Entropy
In 1906, Walther Nernst at the University of Berlin measured the specific heats of solids near absolute zero and formulated the Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is exactly zero. This gave chemists an absolute reference point, allowing the tabulation of S°(H₂O, l) = 69.9 J K⁻¹ mol⁻¹, S°(CO₂, g) = 213.8 J K⁻¹ mol⁻¹, and S° for every other substance. Unlike ΔH°f, these are not conventions, they are real measurements from 0 K.
Today's hook, In 1906, Walther Nernst at the University of Berlin proved that entropy reaches zero at 0 K, giving every substance an absolute S° value. S°(H₂O, l) = 69.9, S°(CO₂, g) = 213.8 J K⁻¹ mol⁻¹. Unlike ΔH°f of elements (zero by convention), S° of elements is NOT zero, it's a real measurement. Why does that distinction matter?
0/5QUESTS
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You're here
Warm up and recall
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Lesson question: how do we put an actual number on an entropy change, instead of only predicting whether ΔS goes up or down?
1Look up S° values. Every substance has a tabulated standard entropy S° in J K⁻¹ mol⁻¹.
2Apply the formula. ΔS° = ΣS°(products) − ΣS°(reactants), each S° scaled by its coefficient.
3Interpret the sign. Positive ΔS° means entropy increased; check it against your L11 prediction.
Quick prerequisite: from L11 you can already predict whether ΔS is positive or negative from the change in moles of gas; here the S° values are simply tabulated for you to read off. If those sign rules feel shaky, start with the Supported route below. Syllabus reference: predict entropy changes from balanced equations, classifying as increasing or decreasing (ACSCH075).
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Know what matters most
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Know what matters most
Must know
ΔS° = ΣS°(products) − ΣS°(reactants)
Interpreting the sign of ΔS° (positive = entropy increases)
Checking a calculated ΔS° against the qualitative L11 prediction
Should know
The step-by-step arithmetic, scaling each S° by its coefficient
S° units are J K⁻¹ mol⁻¹ (convert ÷1000 for L13)
S° is always positive, even for elements, never set S°(element) = 0
Going deeper
Why S° = 0 only for a perfect crystal at 0 K (Third Law)
Residual entropy in real, imperfect crystals
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Warm up and recall
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Recall, your gut answer first
+5 XP warm-up
Here's a puzzle: you've learned that the standard enthalpy of formation of elements in their standard state is zero by definition, $\Delta H_f°[\text{O}_2\text{(g)}] = 0$. So you might expect the same rule applies to entropy: $S°[\text{O}_2\text{(g)}] = 0$.
It doesn't. The standard entropy of O₂ at 25°C (298 K) is 205 J K⁻¹ mol⁻¹ a large, positive number.
Why would an element in its standard state have non-zero entropy? What's different about entropy that makes an absolute reference possible?
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What you'll master, and the words for it
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What you'll master
Know
Key Facts
The Third Law of Thermodynamics and its significance for entropy
Why S°(elements) ≠ 0 while ΔHf°(elements) = 0
The formula $\Delta S° = \Sigma S°(\text{products}) - \Sigma S°(\text{reactants})$
Understand
Concepts
How absolute entropy accumulates from 0 K to 298 K
Why qualitative ΔS predictions should match quantitative ΔS° calculations
How phase, complexity, molar mass and temperature affect S°
Can Do
Skills
Calculate ΔS° using standard entropy data tables
Verify quantitative ΔS° against qualitative prediction from L11
Convert ΔS° from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ for use in L13
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Key terms
Standard reaction entropy (ΔS°rxn)
ΔS°rxn = Σ S°(products) − Σ S°(reactants); multiply each S° by its stoichiometric coefficient.
Standard entropy (S°)
The absolute entropy of a substance at 298 K and 100 kPa; unit is J K⁻¹ mol⁻¹ (not kJ).
S° and physical state
S°(gas) >> S°(liquid) > S°(solid) for the same substance; gases have far more accessible microstates.
S° and molecular complexity
More atoms or heavier atoms in a molecule generally means higher S°; more ways to distribute energy.
Units mismatch
ΔS° is in J K⁻¹ mol⁻¹ but ΔH° is in kJ mol⁻¹; when calculating ΔG° = ΔH° − TΔS°, convert ΔS° to kJ K⁻¹ mol⁻¹ (÷1000).
ΔS° for phase transitions
Vaporisation has large positive ΔS°; freezing has large negative ΔS°; magnitude determined by latent heat / temperature.
Cross-lesson links: This lesson is the numerical counterpart to L11. Where L11 asked you to predict whether ΔS is positive or negative qualitatively, this lesson gives you the tabulated S° values to calculate ΔS° precisely using ΔS°rxn = ΣS°(products) − ΣS°(reactants). The calculated ΔS° value from this lesson feeds directly into L13's Gibbs free energy equation: ΔG° = ΔH° − TΔS°. Note the critical unit mismatch: S° is in J K⁻¹ mol⁻¹ but ΔH° is in kJ mol⁻¹, you must convert ΔS° to kJ K⁻¹ mol⁻¹ before substituting into the Gibbs equation.
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The Third Law and Absolute Entropy
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The Third Law and Absolute Entropy
Should knowcore concept
Imagine cooling a pure crystal of diamond from room temperature to absolute zero. At −200°C it is still vibrating. At −270°C it barely vibrates. Extrapolating Walther Nernst's specific heat measurements from 1906, the vibrations approach zero, and at exactly 0 K, there is only one possible arrangement of atoms: every particle in its lowest energy state. That single microstate, W = 1, means S = k ln 1 = 0. Entropy has a genuine zero.
Third Law of Thermodynamics: The entropy of a perfect crystal at absolute zero (0 K) is exactly zero. At 0 K, every particle is in its lowest possible energy state and there is only one possible microstate, no disorder whatsoever.
Going deeper: the perfect-crystal assumption
This assumes a perfect, fully ordered crystal with a single ground-state arrangement. Real crystals with frozen-in disorder, such as solid CO, can retain a small residual entropy even at 0 K.
From this reference point, we can measure the entropy accumulated by a substance as it heats from 0 K to 25°C (298 K): every energy input (heating, phase change) adds entropy. The diagram below is a schematic example, not every substance melts and boils below 298 K. The result is the standard entropy (S°) of the substance at 25°C and 100 kPa.
Key implication: S° is always positive for real substances at 298 K. Elements in their standard state have S° > 0, because they have been "warmed up" from 0 K. This is fundamentally different from ΔHf°: we define ΔHf°(elements) = 0 as a human convention; S° of elements = non-zero values because it is an absolute measurement.
Selected standard entropy values S° at 25°C, 100 kPa
Substance
S° (J K⁻¹ mol⁻¹)
Note
H₂(g)
130.7
Element, not zero!
O₂(g)
205.2
Element, not zero!
N₂(g)
191.6
Element, not zero!
C(graphite)
5.7
Very low, ordered solid
H₂O(l)
69.9
Liquid, medium entropy
H₂O(g)
188.7
Gas, much higher
NH₃(g)
192.4
Polyatomic gas
NaCl(s)
72.1
Ionic solid
Na⁺(aq)
59.0
Aqueous ion
Cl⁻(aq)
56.5
Aqueous ion
Going deeper: how single aqueous-ion S° values are referenced
Single aqueous-ion values such as S°(Na⁺,aq) and S°(Cl⁻,aq) are tabulated relative to the convention S°(H⁺,aq) = 0, so unlike neutral substances they depend on this convention. Use one consistent (NESA-compatible) data table; for whole reactions the convention cancels.
Critical error to avoid: In any ΔS° calculation, never set S° of an element to zero. Look up its value from the data table. Setting S°[O₂] = 0 or S°[H₂] = 0 is a serious and common error. It is only ΔHf°(element) = 0, a different quantity entirely.
Third Law of Thermodynamics: entropy of a perfect crystal at 0 K = 0 (S = k ln 1 = 0). Standard entropy S° values (J mol⁻¹ K⁻¹) are therefore absolute, not relative. Unlike ΔH°f, S° for elements in standard state is NOT zero.
Pause, copy the highlighted definition into your book before moving on.
Quick check: Which statement correctly distinguishes S°(elements) from ΔHf°(elements)?
Correct. The Third Law establishes a physical absolute zero for entropy (S = 0 at 0 K). Elements at 298 K have accumulated real entropy from 0 K. ΔHf° = 0 for elements is purely a convention chosen to simplify calculations.
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Calculating ΔS° Using Standard Entropy Values
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Calculating ΔS° Using Standard Entropy Values
Must knowcore concept
We just saw that S° values are absolute because entropy has a genuine zero at 0 K (Third Law). That raises a question: how do we calculate ΔS° for a reaction from tabulated S° values? This card answers it → with the same products-minus-reactants formula as ΔH°, applied to S° values in J mol⁻¹ K⁻¹.
ΔS° is calculated the same way as ΔH° using the formation method, products minus reactants, scaled by stoichiometric coefficients.
Multiply each S° by its stoichiometric coefficient
ΔS° = ΣS°(products) − ΣS°(reactants)
State answer in J K⁻¹ mol⁻¹; check sign against L11 prediction
Common errors
Using unbalanced equation
Setting S°(element) = 0
Forgetting to scale by coefficient
Reversing order (reactants − products)
Reporting in kJ K⁻¹ mol⁻¹ without converting
Compare with qualitative prediction: if your quantitative ΔS° is positive, it should be consistent with your qualitative prediction from L11 (e.g., increase in moles of gas → positive ΔS°). If they conflict, check your arithmetic or your qualitative reasoning.
Unit conversion reminder: ΔS° is in J K⁻¹ mol⁻¹. When you use ΔS° in the Gibbs formula (Lesson 13), you must convert to kJ K⁻¹ mol⁻¹ by dividing by 1000. Failing to convert is the single most common error in L13 calculations.
Sign error: The formula is $\Delta S° = \Sigma S°(\text{products}) - \Sigma S°(\text{reactants})$, products first, same as ΔHf°. Reversing to reactants − products gives the wrong sign.
ENTROPY SCENARIOS, INTERACTIVEInteractive
Explore each scenario, observe how particle freedom and ΔS sign relate to the qualitative rules
ΔS° = Σ[S°(products)] − Σ[S°(reactants)], each S° multiplied by stoichiometric coefficient. Units are J mol⁻¹ K⁻¹ (not kJ), critical: divide by 1000 before using in ΔG = ΔH − TΔS. Never assume S°(element) = 0; always look up tabulated values.
Add the highlighted point to your notes before the check below.
Explain it: A student calculates ΔS° for a reaction and gets −87 J K⁻¹ mol⁻¹. They then claim they need to "convert" to −0.087 kJ K⁻¹ mol⁻¹ before recording their answer. Are they right? In one or two sentences, explain when the conversion is necessary and when it is not.
The primary answer for ΔS° should be reported in J K⁻¹ mol⁻¹ (−87 J K⁻¹ mol⁻¹). The conversion to kJ K⁻¹ mol⁻¹ (−0.087) is only necessary when substituting into ΔG = ΔH − TΔS, because ΔH is in kJ mol⁻¹, units must match.
Qualitative Prediction vs Quantitative Calculation
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Qualitative Prediction vs Quantitative Calculation
Must knowcore concept
We just saw how to calculate ΔS° from tabulated S° values using ΣS°(products) − ΣS°(reactants). That raises a question: how do the qualitative rules from L11 and the quantitative calculation relate, can they serve as a mutual check? This card answers it → yes, they must agree in sign; disagreement signals an error in stoichiometry or phase assignment.
Qualitative rules from L11 give you the sign of ΔS; quantitative calculation from L12 gives you the magnitude, using both reinforces understanding and catches errors.
Qualitative prediction vs quantitative ΔS°, three reactions
In every case, the sign of qualitative ΔS matches the sign of quantitative ΔS°, confirming the qualitative rules are reliable predictors. The magnitude also shows that vaporisation (liquid → gas) produces an exceptionally large entropy increase (+118.8 J K⁻¹ mol⁻¹) compared to reactions at constant phase.
HSC exam advice: Even when a qualitative prediction is obvious, show the full quantitative calculation if a data table is provided, HSC marks are awarded for the calculation, not just the correct sign. Always show your working.
Deeper insight: Notice that ΔS° for vaporisation of water (+118.8 J K⁻¹ mol⁻¹) is much larger than the ΔS° of many bond-forming reactions. The gas phase really does have dramatically more entropy than the liquid phase, this is why boiling points involve significant entropy changes.
Qualitative ΔS prediction (Δn(gas) rules from L11) and quantitative ΔS° calculation (ΣS°(products) − ΣS°(reactants)) must agree in sign. When they disagree, recheck stoichiometry and phase assignments. Quantitative values (J mol⁻¹ K⁻¹) are required for ΔG = ΔH − TΔS calculations.
Pause, write the highlighted rule into your book.
Fill in the blank: For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), the qualitative prediction is ΔS is [negative/positive] because Δn(gas) = ___. The quantitative calculation gives ΔS° = ___ J K⁻¹ mol⁻¹. The two results [agree/disagree].
Negative (Δn(gas) = 2 − 4 = −2, gas moles decrease); ΔS° = −198.9 J K⁻¹ mol⁻¹; agree, both give a negative value.
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Same science, choose your support
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Same science, choose your support
pick one
All three routes practise the same Must-Know skill: calculate ΔS° from tabulated S° values and classify the change as increasing or decreasing entropy. Do one; try another for more practice.
Supported word bank + sentence frame start here if unsure
For H₂O(l) → H₂O(g), S°[H₂O(l)] = 69.9 and S°[H₂O(g)] = 188.7. Complete the frame:
"ΔS° = ______ = 188.7 − 69.9 = +118.8 ______. Because ΔS° is ______, entropy ______ (liquid → gas)."
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Show a hint
Always do products minus reactants. A gas has far more microstates than a liquid, so going l → g must give a positive ΔS°, entropy increases.
Core independent HSC-standard the exam standard
Calculate ΔS° for 2CO(g) + O₂(g) → 2CO₂(g). Use S°: CO(g) = 197.7, O₂(g) = 205.2, CO₂(g) = 213.8 J K⁻¹ mol⁻¹. State whether entropy increases or decreases, and confirm the sign matches the change in moles of gas.
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Stretch reasoning in a new context push yourself
A reaction has Δn(gas) = 0, yet its calculated ΔS° comes out clearly positive. Explain how a positive entropy change is possible with no change in the number of gas moles, and describe one type of reaction where this could happen.
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Worked examples
Worked examples · reveal as you go
Worked example 1, Haber Process ΔS°+5 XP on full reveal
Products minus reactants. Negative ΔS° is consistent with qualitative prediction: $\Delta n_\text{gas} = 2 - 4 = -2$ (decrease in moles of gas → ΔS < 0). ✓
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Try it now: Calculate ΔS° for the combustion of hydrogen: $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\text{H}_2\text{O(l)}$. S°[H₂(g)] = 130.7, S°[O₂(g)] = 205.2, S°[H₂O(l)] = 69.9 J K⁻¹ mol⁻¹. Then convert to kJ K⁻¹ mol⁻¹.
Positive ΔS°, consistent with qualitative prediction: ionic solid dissolving → ions disperse into solution → more microstates → ΔS > 0. ✓
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Final answer: ΔS° = +43.4 J K⁻¹ mol⁻¹. Qualitative and quantitative results agree for this case: NaCl dissolution gives a positive ΔS°. This is a relatively modest positive value because the ions are constrained by hydration shells; for some other salts hydration orders the water enough to give a negative ΔS°, so this agreement is case-specific, not a universal pattern.
Always state units and verify sign against qualitative reasoning in your HSC answer.
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Key idea
Fill the blanks+4 XP
Complete this standard entropy calculation for 2H₂(g) + O₂(g) → 2H₂O(l). S°: H₂ = 131, O₂ = 205, H₂O(l) = 70 J mol⁻¹ K⁻¹.
Multiply each S° by stoichiometric coefficient; products first; units J K⁻¹ mol⁻¹
Third Law: $S = 0$ for perfect crystal at 0 K, absolute reference enables tabulation of S° for all substances
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Critical: $S°[\text{element}] \neq 0$, contrast with $\Delta H_f°[\text{element}] = 0$
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Unit conversion (for L13): $\Delta S°(\text{kJ}) = \Delta S°(\text{J}) \div 1000$
Setting S°(elements) = 0
Common errors · the 3 traps that cost marks
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Setting S°(elements) = 0
Students write S°[H₂(g)] = 0 or S°[O₂(g)] = 0 because "they are elements in their standard state."
Fix: Only ΔHf°(elements) = 0, that is a human convention for enthalpy. Entropy has an absolute reference (Third Law: S = 0 only at 0 K). At 298 K all elements have S° > 0. Always look up S° from the data table for every species, including elements.
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Forgetting to convert J → kJ before the Gibbs equation
Students substitute ΔS° = −198.9 (J K⁻¹ mol⁻¹) directly into ΔG = ΔH − TΔS where ΔH is in kJ mol⁻¹.
Fix: Always convert: ΔS°(kJ) = ΔS°(J) ÷ 1000. Using the J value inflates the TΔS term by 1000 and gives a completely wrong ΔG. Check units at every step of the Gibbs calculation (Lesson 13).
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Reversing the formula (reactants − products)
Students subtract products from reactants: ΔS° = ΣS°(reactants) − ΣS°(products), giving the wrong sign.
Fix: The formula is always products minus reactants same convention as ΔHf°. A memory cue: "products of the reaction, products come first in the formula." If you get ΔS° = +198.9 for the Haber process, you've used the wrong order.
Student A calculates ΔS° for $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\text{H}_2\text{O(l)}$ and writes:
"S°[H₂(g)] = 0 and S°[O₂(g)] = 0 because they are elements in their standard state."
Identify the error, explain why it is wrong, and state what values should be used.
Error: The student has confused ΔHf°(elements) = 0 with S°(elements) = 0. These are completely different quantities. ΔHf°(elements) = 0 is an arbitrary human convention (a reference point). S°(elements) ≠ 0 because entropy has an absolute reference point (Third Law: S = 0 only for a perfect crystal at 0 K). At 298 K, elements have accumulated entropy. Correct values: S°[H₂(g)] = 130.7 J K⁻¹ mol⁻¹ and S°[O₂(g)] = 205.2 J K⁻¹ mol⁻¹.
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Student B calculates ΔS° for the combustion of hydrogen — drill 4's reaction, $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\text{H}_2\text{O(l)}$, taking S°[H₂O(l)] = 69.9 J K⁻¹ mol⁻¹ — and writes ΔS° = −0.3268 kJ K⁻¹ mol⁻¹. They are confused, their teacher told them ΔS° must be in J K⁻¹ mol⁻¹. Who is right?
Both are "right", they express the same value in different units. ΣS°(products) = 2(69.9) = 139.8; ΣS°(reactants) = 2(130.7) + 205.2 = 466.6; ΔS° = 139.8 − 466.6 = −326.8 J K⁻¹ mol⁻¹, and −326.8 J K⁻¹ mol⁻¹ ÷ 1000 = −0.3268 kJ K⁻¹ mol⁻¹. The primary unit for standard entropy tabulation is J K⁻¹ mol⁻¹, this is what data tables use, and what the answer should be reported in. However, when using ΔS° in ΔG = ΔH − TΔS (Lesson 13), you must convert to kJ K⁻¹ mol⁻¹ to match ΔH (in kJ mol⁻¹). Student B has the right idea for Lesson 13 but should report the primary answer in J K⁻¹ mol⁻¹.
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Revisit your thinking
Go back to your Think First response. Now that you've studied Nernst's Third Law and absolute entropy from his 1906 University of Berlin measurements:
The key insight: Nernst proved entropy has a genuine absolute zero (S = 0 at 0 K for a perfect crystal), unlike enthalpy, which uses an arbitrary reference point. ΔH°f(elements) = 0 is a human convention; S°(elements) ≠ 0 is a physical measurement counting real microstates.
At 298 K, every real substance has gone through heating and phase changes since 0 K, and accumulated real entropy. S°(graphite) = 5.7 J K⁻¹ mol⁻¹; S°(diamond) = 2.4 J K⁻¹ mol⁻¹. Both non-zero, both real.
When you calculate ΔS°rxn = ΣS°(products) − ΣS°(reactants), remember to convert to kJ K⁻¹ mol⁻¹ before substituting into ΔG° = ΔH° − TΔS° in L13. The units mismatch is the most common calculation error in this topic.
According to the Hess’s Law tool, the total enthalpy change of a reaction is independent of the pathway taken.
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Multiple choice
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That tells the system what to drill next.
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Misconceptions to fix before short answer
Wrong: Entropy always increases in every chemical reaction.
Right: The Second Law states that total entropy of the universe increases for spontaneous processes, but the system alone can have ΔS < 0. Reactions with negative ΔS can still be spontaneous if ΔH < 0 and the enthalpic drive outweighs the entropic penalty (as you'll see in L13 Gibbs).
Wrong: S°(elements) = 0, just like ΔHf°(elements) = 0.
Right: ΔHf°(elements) = 0 is a convention; S°(elements) > 0 at 298 K is a physical measurement anchored to the Third Law. Never set S° of any element to zero in a calculation.
Wrong: ΔS° can be used directly in kJ mol⁻¹ units in the Gibbs equation.
Right: ΔS° is tabulated in J K⁻¹ mol⁻¹. Before substituting into ΔG = ΔH − TΔS, divide ΔS° by 1000 to convert to kJ K⁻¹ mol⁻¹.
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Short answer
UnderstandBand 4
Q6. Explain the Third Law of Thermodynamics and why it allows us to tabulate absolute entropy values (S°) for all substances. In your answer, explain why S°(elements) ≠ 0, unlike ΔHf°(elements) = 0. 4 MARKS
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ApplyBand 5
Q7. Calculate ΔS° for the formation of ammonia: $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \to 2\text{NH}_3\text{(g)}$. Then convert ΔS° to kJ K⁻¹ mol⁻¹ and explain when this conversion is necessary.
Use: S°[N₂(g)] = 191.6, S°[H₂(g)] = 130.7, S°[NH₃(g)] = 192.4 J K⁻¹ mol⁻¹. 5 MARKS
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EvaluateBand 6
Q8 (preview of L13, Gibbs free energy). The combustion of methane is: $\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \to \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$. ΔH° = −890 kJ mol⁻¹.
Using S°: CH₄(g) = 186.3, O₂(g) = 205.2, CO₂(g) = 213.8, H₂O(l) = 69.9 J K⁻¹ mol⁻¹:
(a) Calculate ΔS° for this reaction. (b) Determine ΔG° at 25°C (298 K) and comment on whether the reaction is spontaneous. Show all working and unit conversions. 6 MARKS
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Comprehensive Answers
Show comprehensive answers ▼
Q6, Third Law & Absolute Entropy (4 marks)
Third Law: The entropy of a perfect crystal at absolute zero (0 K) is exactly zero. At 0 K, all particles are in their lowest energy state, there is only one possible microstate, so S = 0. This provides an absolute reference point for entropy measurement. [1 mark]
Absolute S° values: Since entropy starts at zero at 0 K, we can measure the total entropy accumulated by any substance as it is heated from 0 K to 298 K (through every heating step and phase transition). The result, the standard entropy S°, is therefore an absolute value, not a relative one. This is why S° can be tabulated directly for every substance. [1 mark]
Why S°(elements) ≠ 0: The value ΔHf°(elements) = 0 is a human convention, we arbitrarily choose elements in their standard state as the reference point for enthalpy (a relative scale). Entropy, by contrast, has an absolute zero defined by physics (the Third Law). At 298 K, elements have absorbed thermal energy from 0 K and accumulated real, measurable entropy, e.g. S°[O₂(g)] = 205.2 J K⁻¹ mol⁻¹. [2 marks]
$\Sigma S°(\text{reactants}) = 1(191.6) + 3(130.7) = 191.6 + 392.1 = 583.7 \text{ J K}^{-1}\text{ mol}^{-1}$ [1 mark, award only if S°[N₂] and S°[H₂] are not set to zero]
When necessary: This conversion is required when substituting ΔS° into ΔG = ΔH − TΔS. Since ΔH is in kJ mol⁻¹ and T is in K, the product TΔS must also be in kJ mol⁻¹, which requires ΔS in kJ K⁻¹ mol⁻¹. Using ΔS = −198.9 instead of −0.1989 would inflate the TΔS term by a factor of 1000. [1 mark]
Q8 (preview of L13), Methane Combustion ΔS° and ΔG° (6 marks)
Spontaneity: ΔG° = −817.6 kJ mol⁻¹ < 0 → the reaction is spontaneous at 25°C. The large negative ΔH° (−890 kJ mol⁻¹) dominates the negative ΔS° term. Both ΔH < 0 and the overall ΔG < 0 confirm thermodynamic favourability. (Note: kinetically hindered without ignition, spontaneous does not mean instantaneous.) [1 mark]
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Check what actually stuck
Check what actually stuck
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Boss battle
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Five timed questions on Calculating ΔS° & Standard Entropy. Beat the boss to bank a tier, gold (perfect + fast), silver (80%+), or bronze (cleared).
