JC2 Chemistry Tutor Hougang

Searching for a JC2 Chemistry tutor in Hougang because Kp partial-pressure calculations keep going wrong? An H2 A-Level Chemistry student may remember the Kp expression yet substitute total pressure for a component’s partial pressure, calculate fractions from starting moles rather than equilibrium moles, or overlook a change in the total number of gas molecules. The remedy is to reason from the balanced equation, a mole ledger and Dalton’s Law before choosing numbers.

H2 Chemistry syllabus 9476 covers Kp expressions, Dalton’s Law, gas equilibria and calculations involving equilibrium quantities. It explicitly does not require the relationship between Kp and Kc to be developed. In this guide I will show how the equilibrium gas mixture supplies partial pressures, how to distinguish concentration changes from temperature changes, and how to check both a calculation and its physical meaning.

The examples work for independent study and small-group correction. eduKateSG’s centre is at 8 Fourth Avenue near Sixth Avenue MRT, not inside Hougang; confirm the actual JC2 class, timing and travel arrangements before making a tuition decision.

Start from the official examination requirement

The 2027 SEAB H2 Chemistry 9476 document requires students to deduce Kc and Kp expressions, calculate equilibrium constants from appropriate data and find quantities at equilibrium without quadratic-equation solving. The gaseous-state section includes Dalton’s Law and ideal gas calculations. This is enough to create demanding integrated questions without needing a derivation of the Kp–Kc relationship, which the syllabus explicitly excludes.

I ask learners to write the syllabus code on their practice set when using older past papers. Older questions can still be useful to teach conservation and gas behaviour, but an unlabelled collection encourages accidental syllabus mixing. It is more efficient to master the current learning outcomes deeply than to collect derivations that are not asked for.

The meaning of partial pressure

For an ideal gas mixture, the partial pressure p_i of gas i is its mole fraction x_i multiplied by the total pressure: p_i=x_i P_total. The mole fraction is n_i/n_total, where n_total counts the moles of every gaseous component in the mixture. It is not the amount of gas initially poured into the container if a reaction has changed the composition.

Suppose a mixture contains 2.00 mol nitrogen and 3.00 mol hydrogen at a total pressure of 5.00 bar. Nitrogen’s mole fraction is 0.400, so its partial pressure is 2.00 bar. Hydrogen’s mole fraction is 0.600, giving 3.00 bar. Their sum is the total pressure. This short check should be performed automatically in more complicated equilibrium questions.

Write the Kp expression correctly

For aA(g)+bB(g)⇌cC(g)+dD(g), Kp=(p_C)^c(p_D)^d/[(p_A)^a(p_B)^b], using equilibrium partial pressures. The exponents are coefficients in the balanced equilibrium equation. They are not empirical reaction orders in a rate law, even though both expressions contain concentrations or pressures raised to powers.

Reading the equation first also protects against reversal errors. If the equation is reversed, the corresponding equilibrium constant is reciprocated. If all coefficients are doubled, its mathematical expression is squared. A numeric value quoted for one reaction direction cannot be copied unchanged to a differently written equation.

Worked example: PCl5 gas dissociation

Consider PCl5(g)⇌PCl3(g)+Cl2(g) at equilibrium with 0.60 mol PCl5 and 0.40 mol each of PCl3 and Cl2. There are 1.40 mol total. At total pressure 7.00 bar, p_PCl5=(0.60/1.40)×7.00=3.00 bar, while each product has p=(0.40/1.40)×7.00=2.00 bar.

The expression is Kp=p_PCl3×p_Cl2/p_PCl5=(2.00×2.00)/3.00=1.33 bar under the pressure-based classroom convention. Notice that the pressure unit emerges from two powers above and one below. Using the raw 0.40 mole amounts rather than 2.00 bar partial pressures would not answer a Kp calculation.

Worked example: Haber equilibrium

For N2(g)+3H2(g)⇌2NH3(g), Kp=(p_NH3)²/[p_N2(p_H2)³]. Suppose equilibrium amounts are 1.00 mol N2, 3.00 mol H2 and 2.00 mol NH3, totalling 6.00 mol. With total pressure 6.00 bar, their partial pressures are 1.00, 3.00 and 2.00 bar respectively. Each number should be labelled by species before substitution.

Kp=2.00²/(1.00×3.00³)=4/27≈0.148 in the bar-based treatment. Here the net pressure exponent is 2−4=−2, so the ordinary school expression carries bar^−2. A different pressure unit gives a different numerical form and corresponding units. The physical reaction has not changed merely because the reporting convention has.

Why initial and equilibrium amounts differ

An equilibrium constant refers to the equilibrium composition. If the question starts with one mole of a gaseous reactant that partially decomposes, inserting the original amount into Kp ignores the reaction. Write a simple initial–change–equilibrium ledger. For A(g)⇌2B(g), if ξ mol A is consumed from n0 mol initially, the final amounts are n_A=n0−ξ and n_B=2ξ, totalling n0+ξ mol.

A sealed system conserves atoms, not necessarily the count of individual gas molecules. When one A becomes two B, the total gas amount increases even though the system is closed. Confusing gas-molecule number with mass conservation is a conceptual error that explains several otherwise mysterious numerical mistakes.

A complete extent-of-reaction calculation

Start with 1.00 mol pure A for A(g)⇌2B(g). If 0.20 mol A reacts, 0.80 mol A and 0.40 mol B remain at equilibrium. The total gas amount is 1.20 mol. At 6.00 bar total pressure, p_A=(0.80/1.20)×6.00=4.00 bar and p_B=(0.40/1.20)×6.00=2.00 bar.

Kp=(p_B)²/p_A=2.00²/4.00=1.00 bar. Two attractive wrong methods are to use 1.00 mol as the final total or to assume B formed is 0.20 rather than 0.40 mol. A clear mole ledger exposes both immediately, while mental calculation conceals their origin.

Dalton’s Law and pV=nRT agree

For ideal gases sharing the same container, P_total V=n_total RT and p_i V=n_i RT. Divide the latter by the former: p_i/P_total=n_i/n_total. The mole-fraction formula is therefore a consequence of the ideal-gas equation, not an isolated mnemonic. It holds within the stated ideal-mixture model and assumes compatible temperature, volume and units.

If pressure is expressed in pascals and volume in cubic metres, use temperature in kelvin and a consistent R such as 8.31 J K−1 mol−1. Mixing bar with pascals or cm³ with m³ without conversion can produce a neat but incorrect answer. The quantities should be recorded with units at each relevant step.

Pressure change versus change in Kp

At fixed temperature, the value of Kp for a specified reaction is unchanged by pressure alterations in the ordinary idealised equilibrium treatment. The composition can shift after a volume change until the same constant is satisfied again. A student who sees more product forming under compression and says ‘Kp increases’ has mistaken the equilibrium position for the equilibrium constant.

Temperature may change Kp, because it changes the relative favourability of the forward and reverse processes for a given reaction. Explain a temperature effect using the forward reaction’s enthalpy direction and the stated conditions. A catalyst, by contrast, changes the rate of approach to equilibrium without changing Kp at the same temperature.

Pressure and gaseous mole coefficients

For N2+3H2⇌2NH3, there are four gaseous reactant moles for two gaseous product moles according to the equation. Reducing volume at constant temperature initially increases pressure and tends to favour the side with fewer gas moles, here the ammonia side. The chemical explanation is about gas-mole count and the resulting change of partial-pressure reaction quotient.

For H2+I2⇌2HI, two gaseous moles appear on each side. A common pressure scaling through volume reduction does not favour either direction purely through gaseous mole count, because the pressure factors cancel. Counting coefficients before announcing a shift avoids the inaccurate rule that greater pressure always favours products.

The inert gas pressure trap

Adding an ideal inert gas to a fixed-volume container at constant temperature increases total pressure without immediately changing the partial pressures of existing reacting gases. Their mole amounts and volume are unchanged. Their reaction quotient therefore remains the same in the ideal model, so the equilibrium does not shift solely because the new total pressure is higher.

At fixed total pressure, adding inert gas expands the volume, diluting reacting-gas partial pressures. That can change the equilibrium composition when gaseous mole counts differ between sides. A sound answer must say which condition is maintained. ‘Add inert gas’ without a volume or pressure qualification is not enough information for a universal shift rule.

Exclude pure solids from the Kp fraction

For CaCO3(s)⇌CaO(s)+CO2(g), the simplified school Kp expression is p_CO2. Pure CaCO3 and CaO do not enter as variable gas partial-pressure terms. Their presence matters to the reaction, but their activities are treated as fixed in the underlying equilibrium expression. The constant is not constructed by putting every formula in the balanced equation into a pressure fraction.

State symbols are therefore part of the chemistry. Students who ignore (g) and (s) can copy a perfectly balanced equation and still produce the wrong mathematical model. Ask which species have measurable equilibrium partial pressures before writing the fraction.

How pressure units work

The conventional H2 pressure expression may carry units according to the net exponent of pressure. For A(g)⇌2B(g), the resulting unit is pressure to the first power. For ammonia synthesis it is pressure to the minus second power. For A(g)⇌B(g), powers cancel and the result is unitless in this pressure-ratio representation.

Formal thermodynamics uses dimensionless activities relative to standard reference states. That more general convention should not be confused with the unit-bearing school calculation when a question explicitly works with partial pressures in bar. Follow the stipulated quantities and be consistent. A dimensional check can identify a missing coefficient or reversed term.

A three-student diagnosis

Imagine Aisha uses starting amounts rather than final amounts, Ben correctly finds equilibrium moles but treats total pressure as every species’ pressure, and Mira reverses products and reactants in her Kp equation. These are different wrong moves despite all three receiving the same low mark. A meaningful correction should address the precise cause rather than assigning more identical worksheets to everyone.

In a small group, each student should independently write the mole ledger before hearing another explanation. The tutor can then supply a different gas reaction as a transfer test. If each student can explain the changed task, their learning is becoming independent instead of dependent on recognising a previous solution.

How to practise under A-Level time pressure

Begin with non-reacting ideal-gas mixtures to master mole fractions. Next practise writing Kp expressions from balanced equations. Then move to complete equilibrium amounts, followed by an ICE table and an unfamiliar gas mixture. Only once this reasoning is reliable should time limits be introduced. Speed built on guessing is not the same as faster correct work.

After marking, record the first wrong step and complete a delayed retest on altered data. Mixed Paper 2 and Paper 3 questions may conceal gas-equilibrium work in an industrial or experimental setting. The ability to select the model without a chapter heading is the real sign of exam readiness.

Frequently asked questions

Does Kp use the total pressure? It uses the partial pressures derived from the equilibrium gas mixture. Does compression change Kp? Not at constant temperature for the same idealised reaction. Is the total number of gas moles constant? Not necessarily; the reaction can change particle count while conserving matter. Do pure solids appear in Kp? Not as variable gas-pressure terms.

Is the Kp–Kc equation examinable in 9476? Its relationship is explicitly excluded from the stated syllabus treatment. Does inert gas always cause a shift? No; distinguish constant volume from constant pressure. How should I check an answer? Confirm the balanced equation, equilibrium amounts, mole-fraction sum, partial-pressure sum, mathematical powers and units.

What Qp tells us before equilibrium

The reaction quotient Qp is written with the same pressure powers as Kp, but it can be evaluated for a gas mixture before equilibrium is reached. At a fixed temperature, if Qp is smaller than Kp for the specified forward reaction, the composition tends to change in the forward direction as equilibrium is approached; if Qp is larger, the reverse direction tends to be favoured. This comparison explains pressure responses more deeply than a memorised shift rule.

To use this idea, students must still calculate partial pressures from the current mixture. A wrong mole fraction leads to a wrong Qp and then a wrong prediction. Ask which species are present immediately after a change and which reaction direction would move the quotient towards Kp. This forces a separation between immediate physical effects and subsequent chemical readjustment.

The moment after compression is not the final equilibrium

Imagine a closed, ideal gas mixture compressed at constant temperature. Immediately after volume reduction, the amounts of individual gases have not yet changed through reaction. Their partial pressures increase because they now occupy less space. The reaction quotient can change at once, depending on the gaseous stoichiometric coefficients; only afterward does chemical reaction alter amounts until the equilibrium condition is restored.

This distinction prevents an important shortcut error. A student may be asked for the initial pressure increase, then for the direction of equilibrium shift, and finally for a new equilibrium amount. Each asks about a different stage. Make a timeline: before compression, immediately after compression, and after chemical equilibration. Do not use the final-state mole quantities while describing the instant of physical compression.

A new dissociation example using a fraction reacted

Suppose pure A(g) dissociates according to A(g)⇌B(g)+C(g). If 1.00 mol A is initially present and 25% reacts, the equilibrium amounts are 0.75 mol A, 0.25 mol B and 0.25 mol C. The total becomes 1.25 mol. At a total pressure of 5.00 bar, p_A=(0.75/1.25)×5.00=3.00 bar, and p_B=p_C=1.00 bar.

Kp is therefore (1.00×1.00)/3.00=0.333 bar under the pressure-based convention. A learner who treats 25% as 0.25 of each gas without a reaction ledger may accidentally assign the wrong starting and final species. Try the same calculation with 40% dissociation, but do not assume the new composition is automatically an equilibrium at the same temperature and pressure as the previous example without corresponding physical consistency.

Why the total number of moles changes

One molecule of A producing one B and one C conserves the constituent atoms, but it increases the number of gaseous molecules from one to two in that reaction event. That is why the total amount for an initial 1.00 mol can become 1.25 mol after 0.25 mol A reacts. Conservation of mass and conservation of gas-particle count are not interchangeable principles.

A useful thought experiment is to count dots representing molecules before and after the reaction while preserving atoms inside each dot. Then ask whether an ideal gas’s partial pressures should be calculated from the original total or the current total. The visual model can resolve a misconception that remains persistent when students see only symbolic mole equations.

Predict how compression affects A⇌B+C

For A(g)⇌B(g)+C(g), the product side contains more gaseous particles. Isothermal compression raises gas pressures immediately, and the equilibrium then tends toward the side with fewer gas moles, here A, under the appropriate idealised assumptions. The shift is a response to the changed reaction quotient, not evidence that the temperature-dependent Kp itself increased or decreased.

When the reaction is written in reverse, B+C⇌A, the pressure-response chemistry is identical but the verbal label ‘forward’ refers to the other direction. A student should always count the gaseous coefficients in the printed equation and identify the actual species formed. The words ‘left’ and ‘right’ depend on how the equation was written.

Reversing the equation changes the named Kp

Using a simplified Kp expression Kp=p_Bp_C/p_A for A⇌B+C, the reverse reaction B+C⇌A has Kp’=p_A/(p_Bp_C)=1/Kp. If the forward numerical value is 0.333 in a given pressure convention, the reverse numerical value is approximately 3.00 with reciprocal pressure dimensions. The physical equilibrium state has not changed; only the named reaction direction has.

Students sometimes use a table Kp for one direction but silently reverse a balanced equation to make it look more familiar. The result is a reciprocal error even when every partial pressure is correct. Write the expression from the actual equation rather than importing a remembered answer. This is a good check in multistep Paper 3 questions.

Changing coefficients changes the mathematical constant

If A⇌B+C is multiplied by two to become 2A⇌2B+2C, the corresponding pressure expression is the square of the original expression. Its numerical Kp value is Kp² under the same convention. The gas mixture and underlying chemical process can be described consistently by both equations, but the mathematical constant belongs to the chosen stoichiometric representation.

This is a reason to include the balanced reaction beside any recorded Kp value. Without the equation, a bare number may be ambiguous. Tutors can teach this with a comparison table showing original, reverse and doubled equations, then ask students to write the appropriate expression and powers without using any new experimental data.

Why an inert gas at fixed volume need not shift equilibrium

For an ideal gas mixture in a rigid container at unchanged temperature, adding an inert gas increases total pressure but does not change the partial pressures of the reacting gases. Each reacting gas has the same number of moles, volume and temperature immediately afterward. Therefore its contribution to Qp remains unchanged, and there need be no shift solely from the extra inert gas.

This can surprise students who know only the phrase ‘higher pressure favours fewer moles’. That phrase assumes a particular way of changing pressure, often by reducing volume. The correct first question is what has changed for the reacting gases. A larger total pressure is not enough information by itself to predict a shift.

Why fixed-pressure inert-gas addition differs

If total pressure is held constant while inert gas is added to an ideal gaseous equilibrium, the volume expands and the reacting-gas partial pressures decrease. For a reaction with unequal total gaseous stoichiometric coefficients, the equilibrium composition may then adjust. A direction prediction requires counting gaseous moles and considering how the partial-pressure quotient changes.

Do not conflate this with adding a reactant gas. Adding more of a reacting gas at fixed volume changes the amount of a species appearing directly in Qp, while adding an inert gas at fixed volume does not. These contrasting situations make excellent paired exam questions because they reveal whether the learner understands partial pressure.

Adding a reactant gas is a different perturbation

For N2+3H2⇌2NH3, adding hydrogen to the equilibrium mixture at fixed temperature and volume immediately increases p_H2 while other reacting gas amounts have not yet changed. Since p_H2 appears in the denominator raised to the third power, Qp initially falls relative to its prior equilibrium value. The forward reaction is then favoured until the equilibrium condition is restored, under the stated model.

An answer should say why hydrogen is directly involved, not simply ‘more reactants shift right’ as an unexplained slogan. If the gas added were ammonia instead, the numerator would change and the immediate response would be different. The structure of Qp makes both predictions traceable to the equation.

The reaction with equal gaseous mole counts

Consider H2(g)+I2(g)⇌2HI(g). Its Kp expression is p_HI²/(p_H2 p_I2). A uniform multiplication of all partial pressures by the same factor leaves the ratio unchanged because the total numerator and denominator pressure powers are equal. This provides a simple mathematical explanation for why pure isothermal compression does not favour one side from the gas-mole-count argument.

That does not mean the reaction has stopped or that concentration changes can never affect it. Adding just one reactant changes its own partial pressure and therefore Qp. Students should distinguish a common scaling of every gaseous partial pressure from a selective change to a particular chemical species.

How a catalyst changes a graph without changing Kp

A catalyst may shorten the time taken to reach an equilibrium composition by increasing the rates of forward and reverse reactions. On a concentration–time graph, an appropriately catalysed process can approach the same final concentrations more quickly under otherwise unchanged conditions. The equilibrium constant at the same temperature remains unchanged.

The distinction is particularly valuable when a question contains both a graph and an industrial description. Do not interpret a steeper initial slope as proof that a different equilibrium yield is reached. One measures rate of change, while the other concerns final composition. Identify the axes and the chemical quantity before interpreting what a catalyst does.

Why industrial conditions involve compromises

The Haber process combines a forward exothermic reaction, fewer gaseous moles on the product side, kinetic limitations and engineering constraints. Higher pressure can favour the equilibrium composition of ammonia, while lower temperature favours the exothermic direction. Yet low temperature may produce an unacceptably slow reaction, and high pressure has financial and safety costs.

A suitable catalyst improves rate, not the equilibrium constant at fixed temperature. A balanced explanation therefore considers pressure, temperature, rate and practical operation separately. Students can connect their Kp calculations to genuine industrial decisions without claiming that a simple equilibrium expression solves the entire manufacturing problem.

The difference between an equilibrium yield and Kp

An equilibrium constant tells us about the ratio of relevant activities or pressures for a balanced reaction at a fixed temperature. The fraction of reactant converted also depends on starting mixture amounts and constraints such as total pressure. Two mixtures can satisfy the same Kp under suitable different compositions, so Kp is not itself a percentage yield.

This is why students should not say ‘Kp is 0.2 so 20% of reactants have formed product’. The units, powers and equation make that shortcut invalid. Compute conversion from the mole ledger if it is requested, and compute Kp from equilibrium partial pressures if that is requested. The quantities answer different questions.

Why pure solids are excluded

In a heterogeneous equilibrium such as CaCO3(s)⇌CaO(s)+CO2(g), the simplified Kp expression depends on the partial pressure of CO2, provided the required solid phases are present. Pure solid activities are treated as fixed in the model. Their absence from the algebra does not imply that the solids are absent from the chemical reaction.

Students who mechanically place every formula in the expression often include a fictitious concentration of CaO(s). State symbols explain why that is wrong. Have learners colour gaseous species one way and pure solids another before composing Kp; the physical form is part of the mathematical definition.

Pressure units as a consistency test

The net power of pressure in the Kp expression is the sum of product gaseous coefficients minus the sum of reactant gaseous coefficients. In a pressure-based classroom convention, that guides the units of the calculated constant. For A⇌B+C the net exponent is +1, and for ammonia synthesis it is −2. If the answer has incompatible units, a coefficient or pressure conversion may be wrong.

Professional thermodynamics may define dimensionless equilibrium constants using activities relative to reference states. The user of a school-level worked problem should still follow the stated pressure convention and not mix numerical values between bar and Pa. Good science explains definitions and units instead of silently changing them.

A quick error ledger for gas equilibria

Keep a small table with four possible first errors: the balanced equation, equilibrium mole amounts, partial-pressure conversion or Kp expression. After a marked question, write exactly which error occurred and the corrected principle. ‘I used starting N2 instead of equilibrium N2’ is a more useful note than ‘careless’.

Within a few days, test the same underlying principle on a different reaction. If the mistake is repeated, return to the corresponding concept rather than increasing the quantity of random worksheets. If the learner can independently solve a changed question, the repaired skill is becoming more durable.

What a sensible tutor does with three students

Each learner should write an initial answer alone. One might correctly construct Kp but misunderstand mole fractions; another might solve the numbers but confuse pressure changes with temperature changes. The tutor compares their reasoning, asks the students to explain their first decisions and provides different follow-up tasks. The peer discussion should reveal distinctions rather than allow one confident student to complete the work for everyone.

A small group is useful when independent practice and feedback remain visible. A parent can ask to see a corrected original question and a changed question completed without prompts. That evidence matters more than a claim that the class ‘covered all gas equilibrium’ in a single lesson.

Exam technique: read the question’s stage

An equilibrium problem may describe an initial mixture, a perturbation, a moment immediately after the change and the eventual final state. Students who use numbers from different stages in one fraction can produce an answer that is mathematically tidy but physically impossible. Label each set of quantities with time or state before calculating.

When the question asks only for a direction, do not invent precise equilibrium amounts without sufficient information. When it asks for a numeric constant, do not substitute a general Le Châtelier explanation instead of the necessary algebra. Following the command word and identifying the correct stage are both part of accurate Chemistry.

Parent questions that do not require teaching Chemistry

Ask the learner to explain the difference between a partial pressure and total pressure using their own example. Then ask how they know the partial pressures are correct. A clear reply should mention mole fractions and that the components add to the total. For a later question, ask why Kp can remain fixed while composition changes after compression.

Parents need not derive a formula to assess whether a study strategy works. The learner’s ability to explain and solve an altered task independently gives a meaningful measure of progress. Protecting study time, sleep and school commitments can be as useful as adding lessons.

A final mini assessment

Question one: for A⇌2B, initial 1.00 mol A produces 0.40 mol B. Find equilibrium moles of both gases and the total. Question two: at 7.00 bar total pressure, calculate each partial pressure. Question three: construct Kp for the stated reaction. Question four: explain the direction of an isothermal volume reduction. These form a compact transfer test.

The correct mole amounts are A=0.80 mol, B=0.40 mol and total=1.20 mol. Partial pressures are approximately 4.67 bar and 2.33 bar, giving Kp≈1.17 bar in the stated pressure convention. Because the product side contains more gaseous particles, compression tends to favour A. The student should show the reasoning rather than memorise these rounded values.

An independent Kp check before moving on

A good final self-test uses new data rather than a memorised PCl5 or ammonia table. Begin with a generic reaction 2A(g)⇌B(g). Suppose the equilibrium mixture contains 1.00 mol A and 0.50 mol B at a total pressure of 6.00 bar. The total gas amount is 1.50 mol, so p_A=4.00 bar and p_B=2.00 bar. For the reaction as written, Kp=p_B/(p_A²)=2.00/16.0=0.125 in the matching bar-based pressure convention, with dimensions bar⁻¹. Work it out before checking the result.

Then ask what would happen if the reaction equation were reversed, without changing the mixture. The new constant is reciprocal, 8.00 in the corresponding bar convention. This deliberately separates the physical gas composition from how the equilibrium equation is defined. If the student can explain the reciprocal, track the units and show each partial pressure without hints, they have mastered more than one worked example.

Finally ask them to state the relevant limitation: these calculations use equilibrium composition and an idealised partial-pressure treatment. An actual non-ideal mixture or a different temperature would require appropriate care. Clear working, not a larger collection of formulas, is the practical goal of the revision.

A worked composition expressed as a percentage

Some equilibrium questions give a gas mixture as percentages rather than mole amounts. If the percentages represent mole fractions or volume fractions of an ideal gas mixture under the same conditions, they can be used to calculate partial pressures. For example, in a mixture at 8.00 bar with 25.0% gas A and 75.0% gas B by mole, p_A = 0.250 × 8.00 = 2.00 bar and p_B = 0.750 × 8.00 = 6.00 bar. The sum checks against the total pressure. Do not mistake percentages by mass for mole fractions.

A useful follow-up is to give the same mixture as 1.00 mol A and 3.00 mol B instead of percentages. The calculated partial pressures should agree. If the problem provides mass data, first convert each gas mass to moles and then form the mole fractions. This is a powerful transfer exercise because it shows that the chemistry does not change when the data representation changes; only the initial arithmetic route differs.

Why a volume fraction can be useful under an ideal model

For an ideal-gas mixture at a common temperature and pressure, the volume fraction corresponding to a gas component can match its mole fraction under the appropriate ideal mixture interpretation. Thus a question reporting a dry gas sample’s composition by volume can sometimes provide the information needed for Dalton’s Law. However, the student must check whether water vapour, excess gas or another component is included in the stated total. An omitted component changes the denominator and hence every partial pressure.

Before calculating Kp from such a composition, ask what the percentages sum to and what sample they describe. This simple reading check prevents a surprising number of errors. It also reflects how scientists handle measurements: understanding whether the sample is wet, dry, complete or selectively analysed is part of interpreting the data, not a footnote to be ignored.

JC2 Chemistry Tutor Hougang | H2 Kp Partial Pressure, Mole Fractions and Gas Equilibrium