Grade 12 ChemistryStudy Guide

The mole is chemistry's counting unit: a fixed number of particles large enough to weigh on a bench balance. Dividing a measured mass by the molar mass is the bridge between the two worlds: weigh out 36.0 g of water, divide by its molar mass of 18.02 g/mol, and you know you have 2.00 mol, about 1.2 × 10²⁴ molecules. Every stoichiometry problem starts or ends with this conversion, because balanced equations speak in moles while balances speak in grams.

The word mole was coined by Wilhelm Ostwald in the 1890s, but the idea goes back to Avogadro's 1811 hypothesis that equal gas volumes hold equal numbers of molecules. Since the 2019 SI redefinition, the mole is defined by an exact count, 6.02214076 × 10²³ particles, so molar masses in g/mol are now measured quantities rather than definitions.

A mole is a count and nothing more mysterious: the chemist's dozen, scaled up to something useful. Multiply an amount in moles by the Avogadro constant and you have the literal number of particles in front of you. The only real question is why the constant is that particular size, and the answer is that it was chosen to make the bridge between two worlds land cleanly: $N_A$ is the number of atoms that makes a mole of carbon-12 weigh exactly 12 grams. That choice is what lets a mass in grams read off a balance be converted into a count of atoms, which is the single most useful trick in chemistry.

Scale is the thing worth feeling here. A 250 g glass of water is 250/18.015 = 13.9 mol, so it holds $13.9 \times 6.022\times10^{23} = 8.4\times10^{24}$ molecules. Now count the other way: all the water on Earth, about $1.4\times10^{21}$ litres, divided into 250 g glasses, comes to roughly $5.4\times10^{21}$ glasses. There are about fifteen hundred times more molecules in one glass of water than there are glasses of water in every ocean on the planet. That ratio is why chemists never think about individual molecules and why statistical behaviour is so reliable at this scale.

The constant carries Avogadro's name but not his arithmetic: he proposed in 1811 that equal gas volumes hold equal numbers of particles, and never estimated the number. Jean Perrin did, from Brownian motion, and named it for Avogadro in 1909, work that took him the 1926 Nobel Prize. The 2019 SI redefinition then reversed the logic entirely. $N_A$ is now exact by decree at 6.02214076 × 10²³ per mole, and the mole is defined as that many entities. Carbon-12's role is retired: a mole of it now weighs 12 grams only to within experimental uncertainty, rather than by definition. So 2.00 mol contains exactly 1.204428152 × 10²⁴ particles, with no uncertainty in the constant at all, though your measured 2.00 mol still has its own.

The error that swallows this page whole is leaving "particles of what" unstated. A mole of O₂ is 6.022 × 10²³ molecules but 1.204 × 10²⁴ atoms. A mole of NaCl is 6.022 × 10²³ formula units, which is 6.022 × 10²³ sodium ions plus the same number of chloride ions: 1.204 × 10²⁴ ions in total. A mole of Al₂(SO₄)₃ contains three moles of sulfate. Almost every wrong answer here is a correct calculation attached to the wrong noun, so write the noun down before you multiply: molecules, atoms, ions, or formula units.

Two smaller slips. Dividing a molar mass by $N_A$ gives the mass of one particle in grams, water comes out at $2.99\times10^{-23}$ g, and people routinely lose a factor of 1000 by mixing kilograms into that step, or confuse it with the mass in unified atomic mass units, which is just the molar mass number again with different units. And because $N_A$ is exact, it never limits your significant figures; only the amount you measured does.

Sensible heat is the energy that changes a substance's temperature without changing its phase. The specific heat capacity c is the price of each degree: how many joules one kilogram demands per kelvin of warming. Water's is famously steep at about 4186 J/(kg·K), which is why oceans moderate coastal climates and why a kettle takes its time. Heating 1.5 kg of water from 15 °C to 95 °C costs Q = 1.5 × 4186 × 80 ≈ 502 kJ.

The concept dates to Joseph Black's calorimetry experiments in 1760s Glasgow, which first pried apart the ideas of temperature and heat. Note that ΔT is a temperature difference, so a change of 80 °C equals a change of 80 K exactly: Fahrenheit differences convert by scale alone, with no offset. The formula holds as long as c stays roughly constant over the range and nothing melts or boils along the way.

Latent heat is the energy a phase change absorbs or releases while the temperature holds still. Melting 1 kg of ice at 0 °C soaks up 334 kJ, enough to heat that same water from 0 °C to 80 °C, yet the thermometer never moves until the last crystal is gone. Boiling is costlier still: vaporizing a kilogram of water takes about 2256 kJ, more than five times the energy needed to warm it from ice-cold to boiling.

Joseph Black coined the term in 1762: latent means hidden, because the heat disappears into the phase change instead of the temperature reading. The physics runs everyday life: sweat cools you as it evaporates, steam scalds far worse than boiling water because it dumps its latent heat on condensing against skin, and every refrigerator moves heat by evaporating and condensing a working fluid in an endless loop.

One quantity, three notations, depending on whose book you are holding. Physics writes it L, as here, and splits it into L_f for fusion and L_v for vaporization. Engineering and every steam table print it h_fg, where f is saturated fluid and g is saturated gas, so h_fg = h_g − h_f is the gap between the two columns, which is exactly why it shrinks to nothing at the critical point where those columns meet. Chemistry writes it as an enthalpy of vaporization per MOLE rather than per kilogram, so its numbers look nothing like these until you divide by the molar mass. Same energy, three addresses.

Enthalpy is an extensive quantity: run a reaction twice and you get twice the heat. That is the entire justification for $q = n\Delta H$, and it is why tabulating a single per-mole figure is enough to describe a reaction at any scale from a test tube to a boiler. The sign convention runs from the system's point of view: ΔH negative for an exothermic reaction, because the system's enthalpy falls as heat leaves it. A negative $q$ is not an error message; it is the answer telling you the heat came out.

Something concrete. Heating 200 L of water from 10 °C to 60 °C takes $200 \times 4.186 \times 50 = 41\,860$ kJ. Methane burns at ΔH = −890.3 kJ/mol, so the amount required is $n = 41\,860/890.3 = 47.0$ mol: about 754 g of methane, or 1.05 m³ at STP, before any consideration of how much of that heat actually reaches the water. Chain those three pages together and you have most of a combustion calculation.

The relation's real power is that enthalpy is a state function, which is Hess's law: the heat of a reaction depends only on where it starts and ends, not on the route. So you can add reactions like algebra, and you can build any ΔH you need from tabulated standard enthalpies of formation, products minus reactants, without ever running the reaction. One caution on what is being measured: ΔH is the heat at constant pressure, which is what an open vessel or a flowing burner delivers. A bomb calorimeter holds volume constant and measures ΔU instead, and the two differ by the work done pushing the atmosphere aside, $\Delta H = \Delta U + \Delta n_{gas}RT$.

The question this page most needs you to ask is "per mole of what?" ΔH belongs to the balanced equation as written, not to any one substance in it. $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ has ΔH = −571.6 kJ, but $\text{H}_2 + \tfrac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}$ has ΔH = −285.8 kJ. Same chemistry, same physical world, different bookkeeping, and a factor of two waiting for anyone who reads a table without reading the equation above it. Write the equation first, decide which species your $n$ counts, and make the two agree.

Then the trap that costs more marks than any other in stoichiometry. The limiting reagent is found by moles, never by mass. Given 100 g of hydrogen and 100 g of oxygen, the masses are equal and the amounts are not remotely: 49.6 mol of H₂ against 3.13 mol of O₂, and the reaction needs two hydrogens per oxygen. Oxygen limits by a factor of eight, and 100 g of hydrogen looks generous only because hydrogen is light. Convert everything to moles, divide each by its coefficient in the balanced equation, and the smallest quotient is the limiter. The $n$ that goes into $q = n\Delta H$ is then the extent of reaction that limiter permits, not the amount of whatever you happened to weigh out.

One last note for anyone comparing fuel figures. Combustion enthalpies come in two flavours depending on whether the product water is counted as liquid or as vapour, differing by 44 kJ per mole of water. That is the higher-heating-value and lower-heating-value distinction, and for methane it is about a 10% gap. Two sources can disagree by that much while both being correct.

Molarity answers the practical question "how much stuff is in this bottle?" by counting moles of solute per liter of solution. Dissolve 58.44 g of table salt, exactly one mole of NaCl, in water and top up to the 1.00 L mark of a volumetric flask, and you have a 1.00 M solution. Note the fine print: it is per liter of solution, not per liter of water added, which is why chemists fill to a calibrated mark instead of adding a measured liter of solvent.

Molarity is the workhorse concentration unit because reactions are mole-to-mole affairs: multiplying C by a dispensed volume immediately gives the moles delivered, which is exactly what a titration calculation needs. Physiological saline is about 0.154 M NaCl, ocean water roughly 0.5 M, and concentrated hydrochloric acid around 12 M: a span that dilution calculations cross daily in every lab.

Enthalpy is a state function: it depends only on where the chemistry starts and where it ends, never on the route. So if you can reach the target reaction by adding up a chain of steps whose enthalpies you already know, their sum is the answer. Reverse a step and you flip the sign of its ΔH; double a step and you double its ΔH. Only two steps? Enter 0 for the third.

Germain Henri Hess published this in 1840 in St Petersburg, and the date is the remarkable part: it is before the first law of thermodynamics was stated. Joule's paddle-wheel measurements, Mayer's paper, and Helmholtz's 1847 formulation of energy conservation were all still to come. Hess had no theory of energy conservation to lean on; he simply measured heats of neutralisation with a calorimeter, noticed that the totals were the same whichever order he ran the reactions in, and stated the constancy of heat summation as an empirical law. It later turned out to be a corollary of the first law, which is a fair definition of a good experimentalist.

Worked case: the enthalpy of formation of acetylene cannot be measured directly, you cannot make 2C(s) + H₂(g) → C₂H₂(g) happen in a calorimeter, but its combustion can. Route it: 2C + 2O₂ → 2CO₂ at 2(−393.5) = −787.0 kJ/mol; H₂ + ½O₂ → H₂O at −285.8 kJ/mol; and the reversed combustion 2CO₂ + H₂O → C₂H₂ + 5/2 O₂ at +1299.6 kJ/mol. Sum: −787.0 − 285.8 + 1299.6 = +226.8 kJ/mol, matching the tabulated +227 kJ/mol. Acetylene is one of the few hydrocarbons with a positive formation enthalpy, which is exactly why an oxy-acetylene torch burns hot enough to cut steel.

This is Hess's law with the bookkeeping already done for you. Every compound gets one tabulated number (the enthalpy of forming one mole of it from its elements in their standard states) and every element in its standard state is defined as exactly zero. Build any reaction out of "unmake the reactants back to elements, then make the products", and the whole route collapses to products minus reactants. Multiply each ΔH°f by its coefficient in the balanced equation before you add: the sums that go into this calculator are already coefficient-weighted.

The two classic slips are the minus sign and the physical state. Reactants are subtracted, so a strongly negative reactant enthalpy pushes ΔH°rxn up, not down. And ΔH°f depends on state: liquid water is −285.8 kJ/mol but water vapour is −241.8 kJ/mol, a 44 kJ/mol gap that is exactly the enthalpy of vaporisation and the entire difference between a fuel's higher and lower heating value.

Worked case: burning methane, CH₄ + 2O₂ → CO₂ + 2H₂O(l). Products: −393.5 + 2(−285.8) = −965.1 kJ/mol. Reactants: −74.6 + 2(0) = −74.6 kJ/mol, since O₂ is an element in its standard state. ΔH°rxn = −965.1 − (−74.6) = −890.5 kJ/mol, the number a gas utility is selling you, one mole at a time.

The power-law rate expression, $r = k C_A^{\,n}$, is an empirical fit rather than a law of nature, and treating it as more than that causes most of the trouble it gets into. The order $n$ is a number found by measuring rates at different concentrations and fitting a straight line to $\ln r$ against $\ln C_A$. It is NOT read off the balanced equation. Only an elementary reaction, one that really does happen in a single molecular collision, has orders equal to its stoichiometric coefficients, and most industrially interesting reactions are multi-step sequences that do not.

Fractional orders are therefore ordinary rather than pathological. An order of 1.5 usually signals a chain mechanism; an order of 0.5 often means a dimer dissociating before the rate-determining step; an order that starts near 1 at low concentration and falls toward 0 at high is the signature of a catalyst surface saturating, which the Langmuir–Hinshelwood form describes properly. A NEGATIVE order is real too, and means a species inhibits its own reaction: typically a product competing for the same active sites.

Now the part this site cannot fix for you, and the reason this page carries a warning the others do not. The units of $k$ depend on $n$. At first order $k$ is s⁻¹. At second order it is m³/(mol·s). At an order of 1.5 it is (mol/m³)^−0.5·s⁻¹, which has no name and no entry in any unit converter, because its dimensions are not known until the reader supplies $n$. This calculator therefore takes $k$ as a plain SI number and converts nothing: whatever you type is used exactly as typed.

That makes one specific error very easy and very expensive. Kinetics tables almost universally publish second-order constants in L/(mol·s), and SI wants m³/(mol·s): a factor of 1000. Third-order constants are out by a million. The concentration boxes on this page do convert, so entering 2 mol/L correctly becomes 2000 mol/m³ internally; the $k$ box cannot. Divide a tabulated second-order constant by 1000 before it goes in, and check the resulting rate against something you know (a half-life, a conversion you have measured) before it sizes a vessel. A silent factor of a thousand in a reactor volume is not a rounding error.

Molecules collide constantly, but only the small fraction carrying enough energy to climb the activation barrier actually react. The Boltzmann factor $e^{-E_a/RT}$ is that fraction, and A is roughly the collision frequency with the right geometry, so the rate constant is "how often they meet" times "how often the meeting is violent enough". Because $E_a$ sits in an exponent, the temperature dependence is ferocious: for a typical $E_a$ near 50 kJ/mol, a 10 K rise around room temperature roughly doubles the rate, the old rule of thumb behind refrigerating food and running reactions under reflux.

Svante Arrhenius proposed the form in 1889 after Jacobus van 't Hoff's thermodynamic argument suggested it. He was no stranger to sceptical committees: his 1884 Uppsala doctoral thesis on electrolytic dissociation, the claim that salts split into ions in water, so baffled his examiners that they awarded it the lowest passing grade, nearly ending his career. Nineteen years later the same work won him the 1903 Nobel Prize in Chemistry. A worked case: a reaction with A = 1.0 × 10¹³ s⁻¹ whose rate constant is 1.0 × 10⁻³ s⁻¹ at 300 K has $E_a$ = RT ln(A/k) = 8.314 × 300 × ln(10¹⁶) = 91.9 kJ/mol. Only the ratio A/k matters here, so any consistent pair of rate-constant units works.

Write the Arrhenius equation twice, once at each temperature, and divide: the pre-exponential factor A cancels and only the ratio of rate constants survives. That is a gift to the experimentalist, because A is hard to measure but a rate ratio needs only two runs on the same apparatus. Since only k₂/k₁ appears, the units of the rate constants are irrelevant as long as both are the same: half-lives work just as well, inverted.

The classic result: a reaction whose rate exactly doubles between 300 K and 310 K has $E_a$ = R ln 2 / (1/300 − 1/310) = 8.314 × 0.693 × 9300 = 53.6 kJ/mol. That is where the "rates double every 10 degrees" rule of thumb comes from: it is only true for activation energies near 50 kJ/mol, and it fails badly for very fast or very slow reactions. The two traps are using Celsius instead of kelvin (this calculator converts for you, but a hand calculation will be wildly wrong) and reversing the reciprocal difference, which flips the sign and hands you a negative activation energy.

Take logarithms of the Arrhenius equation and it straightens out: ln k = ln A − $(E_a/R)(1/T)$. Plot ln k on the vertical axis against 1/T on the horizontal and you get a line whose slope is $-E_a/R$ and whose intercept is ln A. Multiplying the slope by −R (8.314 J/(mol·K)) recovers the activation energy from as many data points as you care to collect, which is far more robust than the two-point method because random scatter averages out across the whole set.

A measured slope of −6448 K gives $E_a$ = 8.314 × 6448 = 53.6 kJ/mol, the same value a doubling-per-10-K experiment would yield. The slope has units of kelvin, and it is always negative for a normal reaction: a positive slope means the reaction speeds up on cooling, which happens only in unusual multi-step systems with a negative apparent activation energy. Watch the axis scaling too: 1/T values cluster in a narrow band (1/300 to 1/320 spans only 0.00021), so plotting the reciprocal to too few decimal places wrecks the slope long before the chemistry does.

A zero-order reaction consumes reactant at a fixed rate no matter how much is left, so a plot of concentration against time is a straight line of slope −k. This looks strange until you see the mechanism: it happens when something other than the reactant is the bottleneck (a saturated enzyme, a fully covered catalyst surface, or a photochemical step limited by the lamp). Start at 0.500 M with k = 0.0100 mol/(L·s) and after 20 s exactly 0.200 M has gone, leaving 0.300 M.

Ethanol metabolism is the everyday example: alcohol dehydrogenase saturates at very low blood alcohol levels, so the liver clears roughly 0.015% blood alcohol per hour regardless of how much you drank, which is why "one drink per hour" advice works and why doubling the dose doubles the sobering-up time rather than leaving it unchanged. The unique feature of zero order is that it genuinely runs out: set [A] = 0 and the reaction stops dead at t = [A]₀/k, unlike first-order decay which merely approaches zero forever. Note that k here carries units of mol/(L·s), not the s⁻¹ of a first-order constant.

In a first-order reaction each molecule decomposes independently with a fixed probability per second, so the rate is proportional to how much is left and the concentration falls exponentially. Taking logarithms gives ln[A] = ln[A]₀ − kt, a straight line: the diagnostic test that distinguishes first order from every other order. The defining property is a constant half-life, t½ = ln2/k = 0.693/k, independent of where you start.

The maths is identical to radioactive decay, which is why the same equation covers the elimination of most drugs from the bloodstream, the fading of a chemiluminescent glow stick, and the isomerisation of cyclopropane to propene. Worked case: with k = 0.0231 s⁻¹, a 0.100 M solution falls to 0.100 × $e^{-0.0231 \times 30}$ = 0.0500 M in 30 s, the half-life, since 0.693/0.0231 = 30.0 s. The usual arithmetic slip is mixing k in s⁻¹ with a time in minutes; the calculator converts both to SI first, so enter each with its real unit rather than pre-converting.

Radioactive decay never runs out of sample all at once: it halves, and halves again, forever. Each half-life leaves exactly 50% of what entered it, so after two half-lives a quarter remains, after ten about a thousandth. Carbon-14, with its 5,730-year half-life, is the famous example: a bone whose C-14 content has fallen to one quarter of the living value has been dead roughly 11,460 years. N and $N_0$ can be in any matching unit (grams, atom counts, or becquerels) since only their ratio matters.

Solving for time or half-life inverts the exponential with a base-2 logarithm, which demands 0 < N < $N_0$: you cannot take the log of a zero ratio, and decay never leaves more than it started with. The same mathematics governs drug elimination in pharmacology and the six-hour half-life of technetium-99m, the workhorse isotope of medical imaging.

Half-life and decay constant are the same fact in two dialects. The decay constant λ is the fundamental one: it is the probability per unit time that any individual nucleus, chosen at random and regardless of its history, will decay. That single assumption, a constant per-nucleus probability, produces the exponential law $N = N_0 e^{-\lambda t}$ and nothing else. The half-life is what you get by asking that law when $N/N_0 = \tfrac{1}{2}$: take logarithms and $t_{1/2} = \ln 2/\lambda$, with ln 2 = 0.6931471806. Large λ, short half-life. The two numbers carry identical information, and which one a source quotes is a matter of discipline rather than physics.

Cobalt-60 has a half-life of 5.27 years, so $\lambda = 0.6931/5.27 = 0.1315$ per year, or $4.17 \times 10^{-9}$ per second once the years are converted. That per-second figure is the useful one, because activity is $A = \lambda N$. One gram of Co-60 holds $6.022\times10^{23}/60 = 1.00\times10^{22}$ nuclei, so its activity is $4.17\times10^{-9} \times 1.00\times10^{22} \approx 4.2\times10^{13}$ Bq: about 42 TBq, or 1100 curies, from a single gram. λ is what converts a count of atoms into a radiation hazard.

A third quantity lives in the same family and is the one most often mixed up with the others. The mean lifetime τ, the average time an individual nucleus survives before decaying, is simply $1/\lambda$, which makes it $t_{1/2}/\ln 2 = 1.443\,t_{1/2}$. The mean life is 44% longer than the half-life, and they are not interchangeable. Particle physics tables usually quote τ, nuclear medicine and health physics usually quote $t_{1/2}$, and a value copied across that boundary without conversion is wrong by a factor you will not notice from the magnitude alone. The same structure appears wherever a fixed fractional loss per unit time does: drug clearance, capacitor discharge through a resistor, the attenuation of light through an absorbing medium.

The unit trap is mechanical and unforgiving: λ and $t_{1/2}$ are reciprocals, so their units must be reciprocals too. A λ quoted per year set against a time measured in seconds is off by $3.16\times10^7$. Decide on one time unit at the start and hold it through the whole problem.

One property of λ is worth stating because it is genuinely unusual. It does not depend on temperature, on pressure, on the chemical compound the atom sits in, or on how long the nucleus has already existed. Essentially every chemical rate constant doubles or better for a 10 K rise; λ does not move. Heating a radioactive sample, dissolving it, or bonding it into a molecule changes nothing measurable, because the decay is a nuclear event and chemistry happens far outside the nucleus. In living systems, though, a second clock runs alongside: a radiopharmaceutical is also excreted, and the two combine as $1/T_{eff} = 1/T_{phys} + 1/T_{bio}$, so the effective half-life in a patient is always shorter than the physical one.

When two molecules of the same reactant must find each other, the rate goes as [A]², and integrating gives a straight line only if you plot the reciprocal concentration against time. The slope of that line is k, in L/(mol·s). Starting at 0.100 M with k = 0.500 L/(mol·s), after 10 s the reciprocal has climbed from 10 to 10 + 5 = 15, so [A] = 0.0667 M.

The three integrated laws form the standard diagnostic kit: plot [A] versus t, ln[A] versus t, and 1/[A] versus t, and whichever comes out straight tells you the order. Second-order kinetics has a distinctive personality: it starts fast and then drags, because losing reactant hurts the rate twice over. Gas-phase NO₂ decomposition and many radical recombinations follow it. Be careful with the "pseudo" cases: a reaction that is genuinely second order overall but run with a huge excess of one partner behaves like first order in the other, which is exactly how kineticists tame a two-variable problem into a one-variable measurement.

Set [A] = [A]₀/2 in the second-order integrated law and the reciprocals collapse to t½ = 1/(k[A]₀). The consequence is counterintuitive if you are used to radioactive decay: the half-life is not a constant of the reaction but depends on where you start, and it doubles every time you halve the concentration. With k = 0.200 L/(mol·s) and [A]₀ = 0.500 M the first half-life is 1/(0.200 × 0.500) = 10.0 s; the next half, from 0.250 M down to 0.125 M, takes 20.0 s, then 40.0 s, and so on.

That lengthening tail is the signature of second-order kinetics and a genuinely useful diagnostic at the bench: measure successive half-lives and if they keep doubling, the reaction is second order; if they stay constant, first order; if they keep halving, zero order. It also explains why the last traces of a dimerising impurity are so stubborn to remove: the reaction that cleans it up slows down quadratically as the impurity thins out.

Cato Guldberg and Peter Waage, two brothers-in-law working in Kristiania (now Oslo), published the law of mass action in 1864: at equilibrium the products of the concentrations, each raised to its stoichiometric coefficient, sit in a fixed ratio. For the common one-to-one case A + B ⇌ C + D that ratio is just [C][D]/([A][B]). Their paper appeared in Norwegian and was ignored for a decade until a French translation reached van 't Hoff and Ostwald.

The identical expression evaluated with any set of concentrations, not necessarily equilibrium ones, is the reaction quotient Q, and comparing Q with Kc predicts which way the reaction will run: Q < Kc means it goes forward, Q > Kc means it reverses, Q = Kc means it is already there. Worked example: with [C] = [D] = 0.60 M and [A] = 0.20 M, [B] = 0.30 M, the quotient is 0.36/0.060 = 6.0, so if Kc were 2.0 the mixture would run backwards to consume product. Note that this calculator assumes all four coefficients are 1; with other coefficients each concentration needs its own exponent.

Q is built from exactly the same expression as the equilibrium constant (products over reactants, each raised to its balanced coefficient) but you are allowed to evaluate it with any concentrations, not just equilibrium ones. That is the whole point. Q is a snapshot of where a mixture currently stands; K is where it is heading. Compare them and you have the direction of shift for free: Q < K means too much reactant, so the reaction runs forward; Q > K means too much product, so it reverses; Q = K means the mixture is already at equilibrium and nothing net happens.

Each concentration is divided by the standard state c° = 1 mol/L before it is raised to its power, which is what makes Q a pure number even when the powers do not balance. This page covers the very common aA + bB ⇌ cC shape; for a reaction with two products and matching one-to-one coefficients, use the Kc calculator instead. A species that does not appear on the reactant side can be switched off by entering a coefficient of 0: pure solids and pure liquids are always left out, which is why the water in an aqueous equilibrium never shows up.

Henry Le Chatelier, whose 1884 principle is the qualitative version of this comparison, paid dearly for not having the quantitative one. He ran nitrogen and hydrogen together at high pressure looking for ammonia, an explosion in his laboratory nearly killed an assistant, and he abandoned the work: later calling it "the greatest blunder of my scientific career". The explosion was caused by air left in the apparatus, not by the equilibrium. Fritz Haber took it up two decades later. Worked case, on Le Chatelier's own reaction N₂ + 3H₂ ⇌ 2NH₃: with [N₂] = 0.50 M, [H₂] = 0.20 M and [NH₃] = 0.10 M, Q = (0.10)²/[(0.50)(0.20)³] = 0.010/0.0040 = 2.5. If K at that temperature is 0.060, then Q > K and the mixture will run backwards, decomposing ammonia: the difference between a plant that makes fertiliser and one that does not.

Gas-phase equilibria can be written with partial pressures or with molar concentrations, and the ideal gas law (P = (n/V)RT) converts between them one species at a time. Every mole of gas on the product side contributes a factor of RT and every mole on the reactant side removes one, so only the net change Δn survives: Kp = Kc$(RT)^{\Delta n}$. This calculator uses R = 0.08206 L·atm/(mol·K), the value that pairs partial pressures in atmospheres with concentrations in mol/L.

Δn counts gases only: solids and liquids never appear in either constant. For the Haber synthesis N₂ + 3H₂ ⇌ 2NH₃, Δn = 2 − 4 = −2, so Kp is far smaller than Kc at high temperature. Where Δn = 0, as in H₂ + I₂ ⇌ 2HI, the two constants are numerically identical and the whole conversion evaporates. A worked case: a reaction with Δn = +1 and Kc = 1.00 at 1000 K has Kp = 1.00 × (0.08206 × 1000) = 82.1, a reminder that "the" equilibrium constant is meaningless until you say which basis you meant.

$PV = nRT$ says that for a gas, four quantities are not independent: fix any three and the fourth is decided. Squeeze it and the pressure rises; warm it and it pushes harder or swells; add more of it and both go up. What makes the equation remarkable is not that those things are true, anyone with a bicycle pump knows them, but that one constant serves every gas. Helium, nitrogen and steam all obey it with the same $R = 8.314$ J/(mol·K), which is a strong hint that pressure has nothing to do with what the molecules are and everything to do with how many there are and how fast they are moving.

A worked case in units you would actually read off a gauge. A 20 L cylinder sits at 150 kPa absolute on a 20 °C morning. Rearranged for amount, $n = PV/RT = (150\,000 \times 0.020)/(8.314 \times 293.15) = 3000/2437 \approx 1.23\ \text{mol}$. Note what had to happen before the arithmetic: pascals not kilopascals, cubic metres not litres, and kelvin not Celsius. This page converts your entries for you, but the discipline is worth keeping in your head, because a scrap of paper will not.

The law arrived in pieces. Boyle established $PV$ constant at fixed temperature in 1662; Charles and Gay-Lussac tied volume and pressure to temperature around 1800; Avogadro proposed in 1811 that equal volumes of gases hold equal numbers of particles. Émile Clapeyron folded them into a single expression in 1834. Kinetic theory later derived the whole thing from mechanics: treat molecules as point masses that bounce elastically and never attract one another, average over their collisions with the walls, and $PV = nRT$ falls out, with $RT$ revealed as a measure of the average kinetic energy per mole.

Those two assumptions are also the fine print, and here a common textbook line deserves correcting. The ideal gas law is a limit, not a fact about gases. Real molecules do occupy volume and do attract each other, so the equation is exact only as pressure approaches zero and the gas gets out of its own way. Near condensation it fails plainly: at 100 atm, or anywhere close to the boiling point, the error runs to tens of percent and you want van der Waals or a compressibility factor. Under ordinary room conditions the error is well under 1%, which is why the approximation earns its keep.

Two errors account for most wrong answers on this page, and both are unit errors rather than physics errors. The first is feeding in Celsius. Doubling a gas from 20 °C to 40 °C does not double anything: in kelvin that is 293 to 313, a rise of 7%, and a calculation that used 20 and 40 would be wrong by a factor of nearly two. The second is feeding in a gauge pressure. A tire gauge reading 220 kPa means 321 kPa absolute; $P$ here is absolute pressure, measured from vacuum, because the equation counts molecular impacts and vacuum is where there are none. A third, quieter trap: the familiar 22.4 L per mole belongs to 0 °C and 1 atm. IUPAC redefined standard pressure to 100 kPa in 1982, and at that pressure the molar volume is 22.71 L. Both numbers circulate, and quoting one against the other's conditions is a 1.3% error hiding inside a memorised constant.

Drop silver chloride into water and a little dissolves until the solution is saturated: AgCl(s) ⇌ Ag⁺ + Cl⁻. The solid's concentration never appears in an equilibrium expression, so Ksp is simply [Ag⁺][Cl⁻], and because each formula unit hands over one of each ion, both equal the molar solubility s. Hence Ksp = s². AgCl dissolves to about 1.3 × 10⁻⁵ M, giving Ksp = 1.7 × 10⁻¹⁰: a number small enough that gravimetric chloride analysis can assume essentially all the silver precipitates.

Running it backwards is just as useful: barium sulfate's Ksp of 1.1 × 10⁻¹⁰ gives s = √(1.1 × 10⁻¹⁰) = 1.05 × 10⁻⁵ M, which is why patients can safely swallow a barium meal of a compound whose free Ba²⁺ ion is highly toxic. Almost none of it ever enters solution. The trap is the common-ion effect: this s = √Ksp shortcut assumes pure water. Add 0.10 M NaCl and AgCl's solubility collapses to Ksp/[Cl⁻] ≈ 1.7 × 10⁻⁹ M, a hundredfold drop, which is exactly how analysts force a precipitation to completion.

When a salt dissolves into three ions instead of two the arithmetic changes shape. CaF₂(s) ⇌ Ca²⁺ + 2F⁻ releases one calcium and two fluorides per formula unit, so if s moles dissolve per litre then [Ca²⁺] = s but [F⁻] = 2s, and Ksp = s(2s)² = 4s³. Calcium fluoride's molar solubility of 2.15 × 10⁻⁴ M therefore gives Ksp = 4(2.15 × 10⁻⁴)³ = 4.0 × 10⁻¹¹, matching the handbook value.

Reversed, Mg(OH)₂ with Ksp = 5.6 × 10⁻¹² dissolves to s = ∛(5.6 × 10⁻¹² / 4) = 1.1 × 10⁻⁴ M: enough hydroxide to buffer a stomach at pH about 10 in the flask, which is the entire pharmacology of milk of magnesia. The universal trap is dropping the 4, or forgetting that the doubled ion gets both a coefficient and an exponent. Also note that Ksp alone does not rank solubility across different stoichiometries: AgCl (Ksp 1.7 × 10⁻¹⁰) is actually less soluble than Mg(OH)₂ despite the larger constant, because the exponents differ.

Josiah Willard Gibbs worked out this balance in "On the Equilibrium of Heterogeneous Substances", published in 1876–78 in the Transactions of the Connecticut Academy of Arts and Sciences: a journal so obscure that European chemists only learned of it after Maxwell began championing the work and Ostwald translated it into German in 1892. The idea is a tug of war: enthalpy pulls a reaction toward lower energy, entropy pulls it toward greater disorder, and temperature decides who wins. ΔG negative means the process can run on its own; ΔG positive means it needs driving; ΔG zero is equilibrium.

Limestone decomposition, CaCO₃ → CaO + CO₂, is the textbook illustration: ΔH = +178.3 kJ/mol (strongly endothermic) and ΔS = +160.5 J/(mol·K) (a gas is released). At 25 °C, ΔG = 178.3 − 298.15 × 0.1605 = +130.4 kJ/mol, so nothing happens. Solve ΔG = 0 for T and you get 178300/160.5 = 1111 K, or about 838 °C, which is why lime kilns are fired to roughly that temperature and no lower. The classic unit trap lives right here: ΔH is tabulated in kJ/mol while ΔS is tabulated in J/(mol·K), and forgetting the factor of 1000 makes the entropy term vanish. Enter ΔS in J/(mol·K) and this calculator handles the rest.

Thermodynamic tables and equilibrium tables describe the same chemistry in different currencies, and this equation is the exchange rate. A negative ΔG° means K > 1 and products dominate; a positive ΔG° means K < 1 and reactants win; ΔG° = 0 sits exactly at K = 1. At 25 °C the conversion factor RT is 2.479 kJ/mol, so RT ln 10 = 5.708 kJ/mol: every factor of ten in K is worth 5.7 kJ/mol of free energy. A reaction with K = 1.0 × 10⁵ therefore has ΔG° = −5.708 × 5 = −28.5 kJ/mol.

Because the relationship is exponential, small energy differences produce enormous equilibrium swings: a change of just 11.4 kJ/mol multiplies K by a hundred. That steepness is why enzyme designers and medicinal chemists chase a few kilojoules of binding energy so hard, and why ΔG° values quoted to the nearest kilojoule are already good enough for most predictions. Two cautions: the standard state matters (K must be written with the same reference concentrations and pressures the ΔG° was tabulated for), and this ΔG° is not the ΔG of an actual running mixture. The two differ by RT ln Q, and it is ΔG, not ΔG°, that must reach zero at equilibrium.

Hydrogen ion concentrations in water span more than ten orders of magnitude, so in 1909 the Danish chemist Søren Sørensen, working at the Carlsberg brewery laboratory, compressed them onto a logarithmic scale. Each pH unit is a factor of ten: lemon juice at pH 2 carries a hundred times the hydrogen ion concentration of tomato juice at pH 4. Pure water at 25 °C sits at pH 7, where [H⁺] = 1.0 × 10⁻⁷ mol/L. The logarithm takes the concentration in mol/L, and the inverse direction is exact: [H⁺] = $10^{-\text{pH}}$.

In water treatment pH is the master variable. Boiler water is typically held between 10.5 and 11.5 to suppress corrosion, cooling towers near 7–9 to balance scale against corrosion, and municipal drinking water around 7.0–8.5. A pH swing of a single unit, a tenfold chemistry change, is often the first sign that a chemical feed pump has failed.

Water is never merely water. A small fraction of it is always pulled apart into H⁺ and OH⁻, and the equilibrium between them fixes the product of their concentrations at a constant: $K_w = [\text{H}^+][\text{OH}^-] = 1.0\times10^{-14}$ at 25 °C. Because it is a product that is constant, the two concentrations are locked in a see-saw: push one up and the other must come down by the same factor. Take negative logarithms of both sides and the product becomes a sum: $\mathrm{pH} + \mathrm{pOH} = 14$. So pH and pOH are not two independent measurements. They are one number written two ways, and knowing either gives the other for free.

A worked case. Dissolve enough sodium hydroxide to make a 0.010 M solution: it dissociates completely, so $[\text{OH}^-] = 1.0\times10^{-2}$ and pOH = 2.00. Then pH = 14 − 2.00 = 12.00. Check it the long way, $[\text{H}^+] = K_w/[\text{OH}^-] = 10^{-14}/10^{-2} = 10^{-12}$, giving pH 12.00, and the two routes agree, as they must.

The "14" is not a magic number, it is $\mathrm{p}K_w$, and the exact value at 25 °C is 13.995. The whole logarithmic apparatus came from Søren Sørensen at the Carlsberg laboratory in 1909, who needed a compact way to talk about hydrogen ion concentrations spanning more than ten orders of magnitude in brewing. The p-prefix now attaches to anything worth compressing that way: pOH, pKa, pKw, pCa.

The trap is that 14 belongs to 25 °C and nothing else. Self-ionisation is endothermic, so heating water drives it forward and $K_w$ rises: pKw is about 14.9 at 0 °C, 13.0 at 60 °C, and 12.0 at 100 °C. Which means neutral water at 100 °C has a pH of 6.0, and it is still perfectly neutral, not acidic, because pH still equals pOH. This matters wherever hot water is measured: boiler water sampled at temperature and boiler water sampled after cooling do not read the same, and the difference is the equation moving, not the chemistry changing. Anyone reading pH on a hot sample needs to know whether the meter is compensating and what reference temperature it is compensating to.

Which leads to the deeper misconception. Neutral does not mean pH 7; neutral means pH equals pOH. The two coincide at 25 °C by arithmetic accident, and part company at every other temperature. Two smaller notes: the relation is aqueous only (it says nothing about a non-aqueous solvent, which has its own autoionisation constant if it has one at all) and the pH scale is not fenced between 0 and 14. Those bounds are just where 1 M solutions land. Concentrated hydrochloric acid at 12 M has a negative pH, and concentrated sodium hydroxide runs above 14, with the sum still holding.

Add the dissociation of an acid to the hydrolysis of its conjugate base and the two half-reactions sum to the autoionisation of water, so their equilibrium constants multiply to Kw. That single line means you never need to look up both numbers: acetic acid's Ka of 1.8 × 10⁻⁵ instantly gives acetate a Kb of 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰, an extremely feeble base, exactly as expected from a strong-ish weak acid.

The relation encodes chemistry's most useful see-saw: the stronger the acid, the weaker its conjugate base, and the product is fixed. It also explains why sodium acetate solutions are mildly alkaline while sodium chloride solutions are neutral: chloride's parent HCl is so strong that its conjugate Kb is vanishingly small. Kw itself is temperature-dependent (1.0 × 10⁻¹⁴ at 25 °C but about 5.5 × 10⁻¹⁴ at 50 °C), so the pairing of tabulated Ka and Kb values only balances at the temperature they were measured.

Acid strengths span more than twenty orders of magnitude, so chemists do to Ka exactly what Sørensen did to [H⁺]: take the negative logarithm. Acetic acid's Ka of 1.8 × 10⁻⁵ becomes pKa 4.74, hydrofluoric acid's 6.8 × 10⁻⁴ becomes 3.17, and formic acid's 1.8 × 10⁻⁴ becomes 3.75: three numbers you can hold in your head and compare at a glance. Lower pKa means stronger acid, and every unit is a tenfold jump in dissociation.

The sign is the perennial trip-up: because Ka is smaller than one for every weak acid, its logarithm is negative and the minus sign flips pKa positive. A pKa of 3.75 corresponds to Ka = 10⁻³·⁷⁵ = 1.78 × 10⁻⁴, not 10³·⁷⁵. pKa also does double duty as a practical marker: it is the pH at which an acid is exactly half ionised, which is why pharmacologists read a drug's pKa straight off the label to predict whether it will cross a membrane in the stomach (pH 2) or the small intestine (pH 7).

Kb measures how far a base pulls a proton off water, and taking its negative logarithm produces the same compact scale that pKa gives on the acid side: small pKb means a strong base. Methylamine, with Kb = 4.4 × 10⁻⁴, has pKb = 3.36 and is a distinctly stronger base than ammonia at Kb = 1.8 × 10⁻⁵ and pKb 4.74: the extra electron density from the methyl group is worth more than a full order of magnitude.

Modern practice increasingly skips pKb entirely and quotes the pKa of the conjugate acid instead, since a single acid scale ranks everything from sulfuric acid to hydroxide without ever switching frameworks. The two are locked together by pKa + pKb = 14 at 25 °C, so a base of pKb 9.25 has a conjugate acid of pKa 4.75. Just remember that identity, like pH + pOH = 14, holds only near room temperature: Kw climbs with heat, and at 60 °C the sum is closer to 13.0.

For a weak acid HA the equilibrium expression Ka = x²/(C − x) is a quadratic, but when the acid barely dissociates the x in the denominator is negligible and it collapses to x = [H⁺] = √(Ka·C). Take 0.100 M acetic acid with Ka = 1.8 × 10⁻⁵: [H⁺] = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M, so pH = 2.87. That is a thousand times less acidic than 0.100 M HCl at pH 1.00: the whole difference between a weak acid and a strong one at the same concentration.

The approximation earns its keep but has a known failure mode. It assumes the dissociated fraction is under about 5%, which holds when C is at least a hundred times Ka. Push a fairly strong weak acid to low concentration (0.0010 M chloroacetic acid, Ka = 1.4 × 10⁻³) and the square-root shortcut overshoots badly; you must solve the quadratic. It also ignores water's own contribution to [H⁺], which starts to matter for very dilute or very weak acids where the predicted pH creeps above about 6.5.

Ka tells you how a weak acid behaves in principle; percent ionization tells you what actually happened in the beaker. In 0.100 M acetic acid the equilibrium [H⁺] is 1.34 × 10⁻³ M, so only 1.34% of the molecules have given up a proton and 98.7% are still intact: a vivid picture of what "weak" means. It is also the number that justifies the x-is-small approximation, which is generally trusted below about 5% ionization.

The counterintuitive part is that percent ionization rises as you dilute. Because [H⁺] goes as the square root of concentration while C goes linearly, halving the concentration multiplies the fraction ionized by roughly √2. Dilute that acetic acid to 0.0010 M and it is about 13% ionized even though the solution is far less acidic in absolute terms. Ostwald's dilution law captures the same effect, and it is why a weak acid approaches complete dissociation in the limit of infinite dilution.

A buffer resists pH change because it holds a reservoir of both a weak acid and its conjugate base: add acid and the base mops it up, add base and the acid neutralises it. The Henderson–Hasselbalch equation says the pH depends only on the acid's pKa and the ratio of the two forms, not on how concentrated they are. Equal amounts give pH = pKa exactly, which is why you choose a buffer whose pKa sits within about one unit of your target. Acetic acid has pKa 4.76, so an acetate buffer holding 0.200 M acetate against 0.100 M acetic acid sits at pH = 4.76 + log₁₀(2) = 4.76 + 0.301 = 5.06.

Lawrence Joseph Henderson, a Harvard physiologist, wrote the mass-action version in 1908 while working out how blood keeps its pH steady; Karl Albert Hasselbalch, a Dane, recast it in logarithms in 1917 using Sørensen's brand-new pH scale, and the joint name stuck. Their subject remains the textbook example: blood's bicarbonate system has an effective pKa of 6.10 and runs at a 20:1 ratio of HCO₃⁻ to dissolved CO₂, giving 6.10 + log₁₀(20) = 7.40, the pH your body defends to within about 0.05 units. The classic trap is that the equation is an approximation built on assuming the acid and base concentrations are their formal (added) values; it fails for very dilute buffers, for pH far from pKa, and it never applies to a strong acid, which has no meaningful pKa to work with.

Run Henderson and Hasselbalch's argument on the base side and everything mirrors: pOH is set by the base's pKb plus the logarithm of the conjugate-acid-to-base ratio. An ammonia buffer (pKb 4.75) holding 0.200 M NH₄⁺ against 0.100 M NH₃ has pOH = 4.75 + log₁₀(2) = 5.05, so at 25 °C its pH is 14 − 5.05 = 8.95: comfortably alkaline, which is why ammonia/ammonium buffers are the standard choice for EDTA titrations of calcium and magnesium hardness.

The one thing to keep straight is which species goes on top. In the acid form the base is the numerator; here the conjugate acid is, because adding more BH⁺ makes the solution less basic. Get it upside down and your answer is wrong by twice the log term. Many chemists sidestep the whole issue by converting pKb to pKa (pKa = 14 − pKb at 25 °C) and using the acid form throughout: for ammonia that gives pKa(NH₄⁺) = 9.25, and 9.25 + log₁₀(0.100/0.200) = 8.95, the same answer by a different road.

Adding solvent to a solution spreads the same solute through a larger volume: the moles do not change, only their crowding. Since moles equal concentration times volume, $C_1V_1$ must equal $C_2V_2$ before and after any dilution. To prepare 250 mL of 1.0 M hydrochloric acid from a 12.1 M concentrated stock, solve for $V_1$: (1.0 × 250)/12.1 ≈ 20.7 mL of stock, made up to the 250 mL mark with water. (And always add acid to water, never the reverse.)

The law is not limited to molarity: any concentration measure proportional to moles per volume works, as long as both sides use the same one. Water-treatment operators lean on it daily when dosing inhibitor or biocide from concentrated drums into recirculating loops, and biologists use the identical arithmetic for serial dilutions, where each step divides concentration by a fixed factor.