Why can an endothermic reaction sometimes be spontaneous, and why does an exothermic reaction not necessarily proceed rapidly? In JC2 H2 Chemistry, the expression ΔG° = ΔH° − TΔS° brings together enthalpy, entropy and temperature. Students often remember it yet convert units inconsistently, treat Celsius as absolute temperature, or assume that a negative Gibbs free energy means an observable reaction will happen immediately.
H2 Chemistry 9476 explicitly includes the signs of entropy change, standard Gibbs free energy, spontaneity and temperature effects. It does not require students to calculate a reaction’s standard entropy change from tabulated absolute entropies. The useful skill is to understand what ΔH and ΔS contribute, predict which temperature regime favours a process and recognise the limitations of the standard-state model.
This article gives original worked examples and a method for handling all four sign combinations. It is designed for parents and JC2 students in Hougang who want meaningful A-Level revision, whether learning independently or considering tuition. eduKateSG’s small-group centre is near Sixth Avenue MRT at 8 Fourth Avenue, not in Hougang itself.
Begin with the exact syllabus boundary
In the 2027 SEAB 9476 syllabus, chemical energetics includes explaining entropy, predicting the sign of entropy change for changes of temperature, phase and number of particles, applying ΔG° = ΔH° − TΔS°, evaluating whether a process is spontaneous using the sign of ΔG°, and predicting how temperature affects spontaneity. Candidates are also expected to understand limitations of that prediction.
The syllabus explicitly says quantitative calculations of a reaction’s standard entropy change using standard molar entropies are not required. This means the student should prioritise interpreting a given ΔS° and calculating ΔG° with the correct units, rather than spending the final weeks memorising an unrequired method. Correct scope leaves more time for conceptual transfer questions.
What does spontaneous mean?
In thermodynamics, ‘spontaneous’ describes a process that is thermodynamically favoured in the stated direction under the conditions being considered. It does not mean the process has to be fast, explosive or visibly immediate. Some thermodynamically favoured changes are slow because a large activation barrier limits their rate. The equilibrium state and the speed of approaching it are different kinds of information.
This difference is one of the most valuable cross-topic ideas in JC Chemistry. A negative ΔG° can indicate favourable standard-state thermodynamics, but cannot supply a rate constant or time scale. An answer that says a reaction will happen rapidly ‘because ΔG is negative’ substitutes kinetics without evidence. Ask whether a question concerns energetic feasibility, equilibrium or rate before deciding which model to apply.
The meaning of each term in ΔG°
Write ΔG° = ΔH° − TΔS°. Here ΔG° is the standard Gibbs free-energy change of reaction, ΔH° the standard enthalpy change, ΔS° the standard entropy change, and T the absolute temperature in kelvin. All terms must use compatible energy units; if ΔH° is in kJ mol−1 and ΔS° in J K−1 mol−1, convert one before subtracting.
A student who uses 100 J K−1 mol−1 as though it were 100 kJ K−1 mol−1 makes the entropy contribution one thousand times too large. Another common wrong move is entering 25 for room temperature instead of approximately 298 K. These are not merely calculator slips; they are failures to interpret physical quantities. A unit checklist is the most reliable repair.
Entropy is not a synonym for messiness
Entropy is associated with the distribution of energy and possible microscopic arrangements. At A-Level, students can often make useful qualitative predictions from the number of gaseous particles, whether a phase changes from solid to liquid or gas, and temperature changes. Describing entropy only as ‘disorder’ can conceal the precise physical features the question asks them to compare.
If a solid melts, the particles gain greater freedom of arrangement and the change of entropy is generally positive. If a gas condenses to liquid, the entropy change is generally negative. When a chemical reaction produces more gas molecules under otherwise comparable conditions, the entropy change may be positive, although the complete system needs to be considered. The chemistry is more informative than an unqualified slogan.
Temperature must be in kelvin
The entropy contribution in ΔG° is TΔS°. Kelvin is the absolute temperature scale used in thermodynamic relationships. To convert an illustrative 25°C, add approximately 273.15 to obtain 298.15 K. This is distinct from a temperature difference in calorimetry, where a rise of one degree Celsius corresponds to a rise of one kelvin. Students should not transfer the wrong temperature rule from one chapter into another.
Use a unit line before each calculation: ΔH° in kJ mol−1, ΔS° in kJ mol−1 K−1, and T in K. The product TΔS° then has kJ mol−1, which can be subtracted from ΔH°. If the units are inconsistent, pause before numerical substitution. Dimensional analysis is both a mark-protection habit and a professional scientific skill.
Worked example: endothermic with positive entropy
Suppose ΔH° = +40.0 kJ mol−1 and ΔS° = +100 J K−1 mol−1. Convert entropy to +0.100 kJ K−1 mol−1. Then ΔG° = 40.0 − 0.100T, with T in kelvin. At 298 K, ΔG° = 40.0 − 29.8 = +10.2 kJ mol−1. Under standard conditions at that temperature, the forward process is not thermodynamically favoured by the ΔG° criterion.
At 500 K, ΔG° = 40.0 − 50.0 = −10.0 kJ mol−1, suggesting the process becomes favourable in the standard-state treatment. The positive entropy change can overcome the endothermic enthalpy cost at sufficiently high temperature. This is why the universal statement ‘endothermic reactions cannot be spontaneous’ is incorrect.
Find the critical temperature
For the illustrative ΔH° = +40.0 kJ mol−1 and ΔS° = +0.100 kJ K−1 mol−1, set ΔG° = 0. Thus 0 = 40.0 − 0.100T and T = 400 K. Above that temperature, the calculated standard Gibbs change is negative; below it, positive. The calculation treats ΔH° and ΔS° as suitably constant over the temperature interval, a simplifying assumption whose limits should be recognised.
A student should interpret the result as a threshold in the illustrative model, not an exact switch guaranteeing an observable transformation at 400.1 K. Kinetic barriers and departures from standard conditions remain relevant. The threshold is useful because it explains the competition between the two thermodynamic terms rather than making the student memorise that all endothermic reactions must be heated.
Worked example: exothermic with negative entropy
Suppose ΔH° = −50.0 kJ mol−1 and ΔS° = −120 J K−1 mol−1, or −0.120 kJ K−1 mol−1. Then ΔG° = −50.0 − T(−0.120) = −50.0 + 0.120T. At 298 K, this equals −14.24 kJ mol−1. The exothermic contribution dominates over the unfavourable entropy term at this lower temperature.
At 500 K, ΔG° = −50.0 + 60.0 = +10.0 kJ mol−1. The entropy penalty grows as T increases, making the forward process unfavourable in the standard-state model at that higher temperature. A student who always claims heating makes reactions more spontaneous will get this case wrong. The sign of ΔS is decisive for the temperature response.
A second critical temperature
For ΔH° = −50.0 kJ mol−1 and ΔS° = −0.120 kJ K−1 mol−1, setting ΔG° to zero gives T ≈ 416.7 K. Under the assumed constant ΔH° and ΔS° values, the standard reaction Gibbs change is negative below that threshold and positive above it. It is the reverse temperature pattern from the positive-positive example.
This comparison is useful because both worked examples have the same mathematical structure. What changes is the meaning of the signs. Train the student to classify ΔH and ΔS before calculating, predict whether low or high temperature should favour the process, and only then evaluate the exact threshold. The qualitative expectation checks the algebra.
Four sign combinations at a glance
When ΔH° is negative and ΔS° is positive, the expression ΔG° = negative − positive is negative for all positive temperatures under the constant-term model. When ΔH° is positive and ΔS° is negative, it is positive for all positive temperatures. These are the two straightforward patterns. The student should know why the terms reinforce one another rather than simply memorise a grid.
When ΔH° and ΔS° have the same sign, temperature can determine which contribution dominates. Positive-positive favours sufficiently high temperature, while negative-negative favours sufficiently low temperature. This four-case framework is a fast diagnostic, but numerical work must still use kelvin and consistent energy units. If the signs predict one outcome but the calculated ΔG° suggests the opposite, inspect the arithmetic.
The difference between ΔG° and ΔG
The superscript circle indicates a standard-state Gibbs free-energy change. Real mixtures need not have standard activities or pressures, and their reaction driving force depends on composition as well as temperature. A negative standard-state value is useful information, but it does not mean the forward reaction must continue indefinitely regardless of the current mixture. A system at equilibrium has no net driving force for change in its overall reaction direction.
The 9476 syllabus asks students to understand limitations in using ΔG° to predict spontaneity. An appropriately cautious answer distinguishes conditions and kinetics, rather than treating a single tabulated standard quantity as a full prediction of a real laboratory process. This is a good place to emphasise scientific models: reliable within stated assumptions, not magic rules valid without qualification.
Why negative ΔG° does not predict speed
A reaction may be thermodynamically favoured yet so slow at room temperature that it appears unchanging during an ordinary observation period. The activation energy and mechanism control the rate. A catalyst may lower an effective activation barrier and accelerate the reaction, but it does not change the intrinsic thermodynamic equilibrium constant at the same temperature.
A common incorrect answer is ‘spontaneous means it occurs instantly’. Another is ‘exothermic means fast’. Both confuse a thermodynamic energy difference with a kinetic barrier. When tutoring, use paired prompts that ask one question about sign of ΔG° and another about reaction rate. Students should select a different model for each.
Entropy and gaseous particle count
Consider a reaction in which one mole of gaseous molecules produces two moles of gaseous molecules, with all else broadly comparable. The increase in gaseous particle count often supports a positive entropy-change prediction. The reverse reaction may have a negative entropy change. The qualitative rule is a useful first check, but students should still consider the actual physical states and assumptions.
Do not include solid coefficients in a simple gaseous particle count as though every particle has identical freedom of movement. The 9476 syllabus requires qualitative treatment of entropy changes from phases and particle numbers; it does not ask students to derive complex statistical-mechanical formulas. State what changes physically and why that tends to affect the entropy.
Entropy and phase transitions
Melting generally produces a positive entropy change as a solid becomes a liquid; vaporisation tends to increase entropy further because gas particles have much greater freedom of spatial arrangement. Freezing and condensation are the reverse directions and generally reduce entropy. These predictions can often be made without a numerical ΔS value.
An example such as water freezing can still be thermodynamically favoured at a low enough temperature despite the negative entropy change of the water itself. The system’s enthalpy and surroundings must be accounted for. Students should resist the idea that every spontaneous process individually increases the entropy of only the substance they are watching.
How temperature changes the balance
In ΔG° = ΔH° − TΔS°, increasing T increases the magnitude of the entropy term when ΔS° is treated as fixed. If ΔS° is positive, the negative contribution −TΔS° becomes more favourable with heating; if ΔS° is negative, the contribution becomes more positive and can make a reaction less favourable. This is the reason behind the four sign cases.
The explanation should explicitly name the entropy sign and the changing TΔS° term. Writing ‘higher temperature gives higher entropy’ may be too vague to answer the question, especially when it asks about reaction spontaneity rather than the entropy of a single substance. A scientifically clear answer connects the mathematical change to the direction of feasibility.
Why practical temperature thresholds are approximate
The simple threshold T = ΔH°/ΔS° assumes the enthalpy and entropy terms used remain sufficiently constant over the temperature range. In reality, heat capacities and phase transitions can cause those quantities to vary with temperature, and a reaction’s actual driving force depends on non-standard conditions. These limitations are especially important when extrapolating far from the temperature at which values were obtained.
At JC2, students should not turn this into an unnecessarily complicated thermodynamic derivation. They can acknowledge that the model is based on given standard values and identify a relevant limitation when asked to evaluate a prediction. Precise scientific caveats are more helpful than assuming a calculated threshold has unlimited exactness.
A common sign error and its repair
Consider ΔS° = −0.100 kJ K−1 mol−1. In ΔG° = ΔH° − TΔS°, subtracting a negative entropy contribution adds a positive number to ΔH°. Students sometimes treat a negative ΔS as automatically making ΔG more negative because they see a minus sign in the formula. Writing brackets around ΔS and predicting the qualitative effect first prevents this error.
Practise four numerical values with both positive and negative entropies. For each, state whether −TΔS is positive or negative before multiplying. This one habit can protect a substantial part of the calculation without requiring any new Chemistry content. The student should be able to explain why a decrease in entropy can oppose spontaneity.
Industrial and environmental applications
Chemical processes used in manufacturing, energy conversion and environmental control involve thermodynamics and kinetics together. A favourable Gibbs free-energy change can support a reaction’s potential, but the actual process may require a catalyst, controlled conditions and safe engineering. The industrial choice depends on rate, equilibrium, separation costs, feedstock and environmental impact, not just the sign of one energy value.
A JC2 student can appreciate this broader context without treating every exam problem as a full industrial design. State what the supplied ΔH° and ΔS° can predict and where additional information would be needed. This is the beginning of responsible scientific reasoning: distinguishing a useful model from a complete explanation of the real system.
How to revise this topic efficiently
Begin by learning what ΔH°, ΔS° and ΔG° mean. Practise unit conversions and kelvin. Next classify the four sign combinations without calculations. Then solve the positive-positive and negative-negative threshold examples. Finally answer unfamiliar stimulus questions asking whether a reaction is spontaneous, whether a temperature change helps and what the model cannot predict.
A tutor should observe the student’s first wrong move. If the sign pattern is misunderstood, revisit the competition of terms; if the classification is correct but the number wrong, inspect units and algebra; if the calculation is correct but a claim about reaction speed is wrong, repair the kinetics distinction. Targeted revision is more effective than adding identical worksheets.
Frequently asked questions
Does negative ΔG° mean fast? No; kinetics determines rate. Can an endothermic process be spontaneous? Yes, when a favourable entropy term dominates at suitable temperature. Why use kelvin? Thermodynamic temperature must be absolute. Is a negative ΔS always unfavourable? It contributes positively to −TΔS, but the total ΔG° also depends on enthalpy.
What is critical temperature? An illustrative boundary where calculated ΔG°=0 under stated assumptions. Are standard entropy tables required to calculate ΔS° in H2 9476? The syllabus explicitly excludes that quantitative task. Can a catalyst change ΔG°? Not the reaction’s standard thermodynamic change at fixed conditions; it can affect rate.
A temperature–Gibbs energy graph can show the answer immediately
If ΔH° and ΔS° are treated as constant over the relevant interval, ΔG° = ΔH° − TΔS° has the form of a straight-line equation in T. The intercept at T=0 in the simplified extrapolation is ΔH°, while the gradient is −ΔS°. A positive ΔS° gives a negative gradient, so the calculated Gibbs change decreases as temperature rises. A negative ΔS° gives a positive gradient.
The graph’s intersection with ΔG°=0 is the threshold T=ΔH°/ΔS° where that value is positive and meaningful in the model. Students should understand the slope sign before drawing a graph. An incorrectly signed entropy will reverse the gradient and produce a prediction opposite to the chemical evidence.
Read a graph without assuming every line crosses zero
The positive-enthalpy, negative-entropy combination produces a line that is positive at T=0 and rises with temperature, so its calculated ΔG° stays positive for positive temperatures in the fixed-term model. The negative-enthalpy, positive-entropy combination starts negative and decreases further. Neither has a positive-temperature crossing of zero under those assumptions.
This is a simple visual check on the four sign combinations. A student who solves T=ΔH°/ΔS° and obtains a negative Kelvin temperature should not announce it as a physically attainable threshold. Instead they should interpret the sign pattern and explain why the desired change in spontaneity does not occur over positive temperatures in this model.
Why a critical temperature is not a thermometer switch
The threshold at ΔG°=0 describes equality of the enthalpy and entropy contributions under the specified standard-state approximation. A reaction does not necessarily produce a visible burst of product just above the computed boundary. It may be slow, the actual mixture may be non-standard, and the given ΔH° and ΔS° may vary with temperature.
Words matter in an A-Level answer: ‘becomes thermodynamically favourable in the standard-state model’ is more defensible than ‘instantly reacts’. Train students to explain the mathematical sign change and then state an appropriate limitation. This is a good example of answering a prediction question precisely rather than overclaiming.
A worked numerical graph interpretation
Suppose the illustrative relationship is ΔG°/(kJ mol−1)=+30.0−0.0750T, with T measured in K. The intercept is +30.0 kJ mol−1, corresponding to ΔH° in this model. The gradient −0.0750 kJ mol−1 K−1 corresponds to ΔS°=+0.0750 kJ mol−1 K−1, or +75.0 J mol−1 K−1. The boundary occurs at T=30.0/0.0750=400 K.
At 300 K, the calculated ΔG° is +7.50 kJ mol−1; at 500 K, it is −7.50 kJ mol−1. Students can check this result graphically before relying on the calculator. This problem connects mathematical interpretation, units, signs and chemical reasoning in one compact exercise.
Connecting entropy prediction to gaseous molecules
Consider an illustrative reaction that produces fewer moles of gas than it consumes. Other things being comparable, the reduction in available gaseous arrangements can support a negative entropy-change prediction. This is a useful first pass for reactions such as ammonia synthesis. But entropy depends on the actual species and states, so a full comparison may require additional information.
Do not count every solid, liquid and gas particle as though their contribution to entropy were identical. Instead describe the change in freedom of arrangement and in gas particle number. The 9476 syllabus asks for a qualitative direction, not a detailed statistical calculation, so a clear physical explanation is the correct level of response.
Entropy of the system versus spontaneity of the process
A process can decrease the entropy of the system being observed and still be thermodynamically favoured because energy and entropy changes in the wider situation matter. When a liquid freezes under suitable conditions, the molecules become more ordered in the solid, but the process may release heat into the surroundings. A simple ‘entropy must increase inside the sample’ rule therefore fails.
At JC2, the central calculation uses the reaction’s ΔH° and ΔS° through the given Gibbs relation. Students do not need to build a full derivation of total-universe entropy in every problem. They do need to avoid replacing a correct thermodynamic criterion with an incomplete slogan.
Enthalpy alone cannot predict spontaneity
An exothermic enthalpy change means the system releases heat at constant pressure under the defined process. It does not guarantee that the standard Gibbs change is negative at every temperature. If the reaction has a negative entropy change, the term −TΔS° is positive and can oppose the favourable enthalpy, becoming increasingly important as temperature rises.
Conversely, an endothermic reaction with positive entropy change can be favoured at high enough temperature. Give students two reactions with opposite sign combinations and ask them to predict the correct temperature trend without calculating. Once the conceptual prediction is secure, the numerical threshold becomes a verification rather than a guess.
A common arithmetic trap: kJ versus J
Suppose ΔH° is +48.0 kJ mol−1 and ΔS° is +80.0 J K−1 mol−1. The correct entropy value for substitution alongside ΔH° is 0.0800 kJ K−1 mol−1. If a student enters 80.0 as the coefficient in a kJ equation, the predicted threshold is less than one kelvin instead of the correct 600 K. A three-order-of-magnitude error can look like an innocent calculator slip.
Write the conversion before substitution, and verify that TΔS° has the same energy-per-mole unit as ΔH°. This habit also helps in calorimetry, kinetics and electrochemistry. Units are part of the physical model, not a label to attach only after the answer has been computed.
Critical temperature with a negative denominator
Consider ΔH°=+20.0 kJ mol−1 and ΔS°=−50.0 J K−1 mol−1. Converting gives ΔS°=−0.0500 kJ mol−1 K−1. The formal ratio ΔH°/ΔS° is −400 K, which is not a positive-temperature boundary. The underlying ΔG° expression is +20.0+0.0500T, which remains positive for all positive T in the fixed-term model.
A student who simply writes T=−400 K and says the reaction is spontaneous above it has treated an algebraic root as a physically meaningful temperature. The sign grid should already have predicted that positive enthalpy and negative entropy oppose spontaneity across positive temperatures. Always interpret a number chemically.
Another case: exothermic with positive entropy
Let ΔH°=−30.0 kJ mol−1 and ΔS°=+50.0 J K−1 mol−1. After conversion, ΔG°=−30.0−0.0500T. For any positive temperature, both contributions are negative. Thus the process is predicted to be thermodynamically favourable under the stated standard conditions throughout the modelled positive-temperature range.
This case is a useful contrast with negative-negative, where sufficiently high temperature can make ΔG° positive. Students sometimes see a negative ΔH° and conclude that heating can never hurt feasibility, ignoring ΔS°. The two signs must always be read together before judging the temperature response.
A simple case-based study aid
Instead of a large colourful chart, write four phrases: exothermic plus entropy increase, favourable across positive T; endothermic plus entropy decrease, unfavourable across positive T; endothermic plus entropy increase, high T may favour; exothermic plus entropy decrease, low T may favour. Then justify every phrase by the sign of ΔH° and −TΔS°.
Keep the page small so the student can redraw it from memory. The real test is a changed numerical problem without labels. A student who can classify signs and check units independently is far closer to examination readiness than one who recognises an attractive poster.
A catalyst cannot turn an unfavourable ΔG° into a favourable one at fixed conditions
A catalyst can lower the activation barrier of an alternative pathway and accelerate both forward and reverse processes. It does not alter the thermodynamic ΔG° for the given overall reaction at the same temperature and standard state. A student might observe a faster reaction after a catalyst is added, but that does not mean the reaction enthalpy or entropy has been rewritten.
In a structured response, separate the claim about reaction speed from the claim about the direction of thermodynamic favourability. This distinction is especially important when a question combines industrial catalysts with energetic calculations. True statements about catalysis must be connected to the variable actually being asked about.
How ΔG° links to electrochemical cells
For an electrochemical cell, the 9476 syllabus includes ΔG°=−nFE°cell. The number of moles of electrons n, the Faraday constant F and the standard cell potential E° allow the standard Gibbs free-energy change of the balanced overall redox reaction to be calculated. A positive E°cell corresponds to a negative ΔG° in the specified direction.
That relation does not contradict ΔG°=ΔH°−TΔS°; it provides another way to characterise the standard thermodynamic driving force for the same defined overall process. Students should not multiply a voltage by an arbitrary electron count or confuse the potential with reaction speed. See the linked electrochemistry guide for a separate worked cell example.
The limitations of standard-state reasoning
The superscript circle indicates that a value refers to standard conditions and reference states. Actual reaction mixtures can contain very different concentrations and gas pressures. The spontaneous direction in an actual non-standard mixture may therefore differ from the conclusion inferred only from a standard-state value. Stating this limitation is scientifically responsible when a question asks for evaluation.
A student should not use the qualification to avoid answering an ordinary standard-state question. First give the correct result under the supplied assumptions, then identify the appropriate limit. That order produces a clear answer without pretending the model is universally exact.
Temperature dependence can be more complex outside a short range
The textbook linear relationship between ΔG° and T assumes the supplied standard enthalpy and entropy changes are suitably treated as constant. Across large temperature intervals, heat capacity changes or phase transitions may make that assumption poor. A reaction’s true temperature behaviour can therefore be more complex than one straight line drawn through all possible temperatures.
At H2 level, a sensible response identifies that the calculated threshold applies under the model’s stated approximations. Students should not invent advanced thermodynamic corrections unless data or instructions require them. Strong science means using the simplest adequate model and stating where it may cease to be reliable.
Why one worked example is not enough
A learner may solve the positive-positive example correctly yet fail the negative-negative case because the double negative in −TΔS° is mishandled. Assign at least one problem from each sign family, then combine them in random order. The student should predict the temperature trend first and calculate afterward.
Use a delayed retest a few days later with changed numbers. If the result remains correct without notes and the student can explain why an elevated temperature helps or hinders the particular reaction, the skill is transferable. This is a better marker of mastery than repeating the same successful arithmetic.
How to review a weak Chemistry prelim answer
If the student writes ‘endothermic means cannot occur’, identify a conceptual misconception about entropy. If they correctly predict high-temperature favourability but calculate a nonsensical threshold, inspect unit conversion and algebra. If both are correct but they claim the reaction happens instantly, repair the distinction between thermodynamics and kinetics.
This error-first approach lets a tutor select a precise follow-up rather than prescribe another entire chapter of notes. Parents can ask to see a corrected example and an altered independently solved question. The student’s reasoning is more informative than the number of highlighted pages.
A seven-session revision cycle
Begin with the meaning and signs of ΔH°, ΔS° and ΔG°. Then interpret phase and gaseous-particle changes. Next handle J-to-kJ conversion and kelvin, followed by all four sign combinations. Use one positive-positive and one negative-negative threshold calculation. Finish with graphs of ΔG° against T and an unfamiliar question that requires a limitation of the standard model.
The sessions need not occur on seven consecutive days. JC2 students have other subjects and laboratory responsibilities. Adjust the workload and reserve time to mark answers and retest errors. More hours are not automatically useful if the learner is too tired to reason accurately.
An original self-test for the whole topic
Question one: with ΔH°=+36.0 kJ mol−1 and ΔS°=+90.0 J K−1 mol−1, find the temperature where ΔG°=0. Question two: state whether the reaction is favoured at 300 K and 500 K under the constant-term standard model. Question three: explain why the result does not establish reaction speed. Question four: state one reason an actual mixture might differ from a standard-state prediction.
The threshold is 400 K. At 300 K the calculated ΔG° is +9.00 kJ mol−1; at 500 K it is −9.00 kJ mol−1. A kinetic activation barrier still governs speed, and non-standard concentrations or pressures can alter the actual driving force. A complete answer explains the reasoning rather than merely copying those numbers.
Further study and enquiry
Continue with JC2 entropy, Gibbs free energy and feasibility, why exothermic does not always mean spontaneous, Born–Haber sign reasoning and the JC2 revision plan. The official 2027 H2 syllabus sets the exact requirements.
Families can contact eduKate Singapore or enquire about current H2 tutorials. Confirm current fees, availability, location and syllabus level. The numerical cases here are original illustrations, not official SEAB questions.