What is wrong direction of Keq when reaction is reversed?

K_reverse = 1 / K_forward. If a question gives Keq for one direction, you must invert it for the reverse.

At 500 K, 2.00 mol of PCl_5 is placed in a 2.00 L vessel and decomposes via PCl_5(g) PCl_3(g) + Cl_2(g). At equilibrium [Cl_2] = 0.30 mol L^-1. Calculate K_eq.

For 2NO_2(g) N_2O_4(g), K_eq = 4.0 at 298 K. A mixture has [NO_2] = 0.50 mol L^-1 and [N_2O_4] = 0.50 mol L^-1. (a) Calculate Q.

§-Syllabus dot point

NSWChemistrySyllabus dot point

Inquiry Question 3: How can the position of equilibrium be described and what does the equilibrium constant represent?

Deduce the equilibrium expression (in terms of Keq) for homogeneous reactions, and perform calculations to find the value of Keq and concentrations of substances within an equilibrium system

A focused answer to the HSC Chemistry Module 5 dot point on the equilibrium constant. Deriving the Keq expression, interpreting its magnitude, the ICE table method, and worked HSC calculations for finding Keq and equilibrium concentrations.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

In plain English

Imagine a busy two-way revolving door between a "Reactants Room" and a "Products Room", with people constantly walking through in both directions. Once the flow settled into a steady back-and-forth, you could survey how many people are standing in each room at any instant. The equilibrium constant $K_{eq}$ is just that fixed head-count ratio, roughly how many people end up in the Products Room compared with the Reactants Room, once the flow has stabilised. If far more people end up in the Products Room, $K_{eq}$ is a big number. If most people are stuck in the Reactants Room, $K_{eq}$ is small. Turning up the "temperature" of the building (literally heating or cooling it) is the only thing that changes this ratio permanently; squeezing more people into one room (changing concentration) or shrinking the building (changing volume/pressure) only temporarily disturbs the flow before it settles right back to the SAME head-count ratio. The ICE table is simply the bookkeeping trick for tracking how many people left the Reactants Room and arrived in the Products Room, using the door's fixed "two people out for every one person in" style ratio (the stoichiometric coefficients) to keep the count honest.

What this dot point is asking

NESA wants you to write the equilibrium expression for a given reaction, calculate the value of Keq from experimental data, and use Keq to find equilibrium concentrations. This is the highest-yielding calculation dot point in Module 5 and appears as a 4-6 mark question almost every year.

The answer

The equilibrium expression

For the general reaction:

aA + bB \rightleftharpoons cC + dD

The equilibrium constant in terms of concentration is:

K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}

Products on top, reactants on bottom, each raised to its stoichiometric coefficient.

Rules:

Pure solids and pure liquids are excluded from the expression. Their "concentration" is essentially constant.
Solvents (water in dilute aqueous reactions) are also excluded.
Aqueous and gaseous species are included.
Keq is temperature-dependent only. Concentration and pressure changes do not change Keq, only the position of equilibrium.

Interpreting the value

The magnitude of Keq tells you where the equilibrium lies.

If $K_{eq} \gg 1$ , equilibrium lies to the right (products favoured).
If $K_{eq} \approx 1$ , there are comparable concentrations of reactants and products.
If $K_{eq} \ll 1$ , equilibrium lies to the left (reactants favoured).

Keq has no fixed unit (the unit depends on the change in moles of gas/aqueous species in the equation). In HSC answers, write the numerical value and, if asked, state whether the equilibrium favours reactants or products.

The reaction quotient Q

For a system not yet at equilibrium, the same expression is evaluated using the current concentrations and is called the reaction quotient Q.

If $Q < K_{eq}$ : too many reactants. Reaction proceeds forward to reach equilibrium.
If $Q > K_{eq}$ : too many products. Reaction proceeds reverse.
If $Q = K_{eq}$ : system is already at equilibrium.

The ICE method

Almost every Keq calculation uses an ICE table (Initial, Change, Equilibrium). The change row uses the stoichiometric coefficients with a variable $x$ for the extent of reaction. The same ICE method underpins the Ksp calculations for sparingly soluble salts and the weak-acid pH calculations you meet later in the module.

The four facts examiners test every year

Expression: $K_{eq} = \dfrac{[\text{products}]^{\text{coeff}}}{[\text{reactants}]^{\text{coeff}}}$ , with pure solids, pure liquids and the solvent excluded.
Temperature is the only lever: concentration, volume/pressure and catalysts change the position of equilibrium (and how fast it is reached) but never the numeric value of $K_{eq}$ .
Reversing the equation inverts the constant: $K_{reverse} = 1 / K_{forward}$ ; scaling the equation by $n$ raises $K_{eq}$ to the power $n$ .
$Q$ versus $K_{eq}$ predicts direction: $Q < K_{eq}$ drives the reaction forward, $Q > K_{eq}$ drives it in reverse, $Q = K_{eq}$ means no further net change.

An owned illustrative graph shows how the reaction quotient $Q$ evolves toward $K_{eq}$ as a system approaches equilibrium from a pure-reactant start:

Worked example

For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ at 700 K, Keq = 49. If 1.00 mol of $H_2$ and 1.00 mol of $I_2$ are placed in a 1.00 L sealed flask, calculate the equilibrium concentrations.

Step 1: Write the expression.

K_{eq} = \frac{[HI]^2}{[H_2][I_2]} = 49

Step 2: ICE table.

	$H_2$	$I_2$	$HI$
Initial	1.00	1.00	0
Change	$-x$	$-x$	$+2x$
Equilibrium	$1.00 - x$	$1.00 - x$	$2x$

Step 3: Substitute and solve.

\frac{(2x)^2}{(1.00 - x)(1.00 - x)} = 49

Take the square root of both sides (both sides are perfect squares):

\frac{2x}{1.00 - x} = 7

2x = 7 - 7x

9x = 7 \implies x = 0.778

Step 4: Final concentrations.

$[H_2] = [I_2] = 1.00 - 0.778 = 0.222$ mol/L.
$[HI] = 2 \times 0.778 = 1.556$ mol/L.

Check. $\frac{(1.556)^2}{(0.222)(0.222)} = \frac{2.42}{0.0493} \approx 49$ . Correct.

Examples in context

Example 1. Orica Kooragang Island ammonia plant. The Haber synthesis $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ runs continuously at the Orica Kooragang Island ammonia plant near Newcastle, feeding the explosives and fertiliser markets. Plant engineers monitor $K_{eq}$ at the reactor temperature (around 720 K) using gas chromatography sampling at the outlet. Because $K_{eq}$ at that temperature is roughly $1.6 \times 10^{-4}$ , only about 15 percent of the feed converts on a single pass. The unconverted $N_2$ and $H_2$ are separated from $NH_3$ by refrigeration and recycled. The ICE-table arithmetic that students do by hand on the HSC is exactly the calculation the plant control system performs every minute.

Example 2. Macquarie River dissolved carbonate equilibrium. Limestone gorges along the Macquarie River downstream of Wellington release calcium carbonate that dissolves in the slightly acidic stream water. The relevant equilibrium $CaCO_{3(s)} + CO_{2(aq)} + H_2O_{(l)} \rightleftharpoons Ca^{2+}_{(aq)} + 2HCO_{3(aq)}^{-}$ has $K_{eq}$ that depends only on temperature. WaterNSW field officers measuring $[Ca^{2+}]$ and $[HCO_3^-]$ during summer high-temperature periods compare $Q$ against $K_{eq}$ to predict whether calcite will continue dissolving (when $Q < K_{eq}$ ) or precipitate out as scale in irrigation infrastructure (when $Q > K_{eq}$ ). The HSC ICE-table treatment maps directly onto this real monitoring workflow.

Try this

Q1. Define the equilibrium constant $K_{eq}$ and state two factors that do not change its value at constant temperature. [2 marks]

Cue. Ratio of product to reactant concentrations raised to stoichiometric coefficients at equilibrium; concentration changes and pressure or volume changes do not alter $K_{eq}$ .

Q2. At 500 K, $2.00 \text{ mol}$ of $PCl_5$ is placed in a $2.00 \text{ L}$ vessel and decomposes via $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ . At equilibrium $[Cl_2] = 0.30 \text{ mol L}^{-1}$ . Calculate $K_{eq}$ . [3 marks]

Cue. Set up the ICE table with initial $[PCl_5] = 1.00$ , change $-x$ and $+x$ on each product; $x = 0.30$ , then substitute into $K_{eq} = \frac{[PCl_3][Cl_2]}{[PCl_5]}$ .

Q3. For $2NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$ , $K_{eq} = 4.0$ at 298 K. A mixture has $[NO_2] = 0.50 \text{ mol L}^{-1}$ and $[N_2O_4] = 0.50 \text{ mol L}^{-1}$ . (a) Calculate $Q$ . (b) State the direction of net reaction. (c) Explain what would happen to $K_{eq}$ if the volume were halved. [2+1+2 marks]

Cue. (a) $Q = 0.50 / 0.25 = 2.0$ . (b) $Q < K_{eq}$ , so net forward reaction. (c) Volume changes shift position but leave $K_{eq}$ unchanged.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC5 marksAt 700 K, 2.00 mol of H₂ and 2.00 mol of I₂ are placed in a 1.00 L sealed flask. At equilibrium, [HI] = 3.11 mol/L. Calculate the value of Keq for the reaction H₂(g) + I₂(g) ⇌ 2HI(g).

Show worked answer →

A 5 mark answer needs the ICE table, substitution, and a final value with units (or "dimensionless" stated).

ICE table (concentrations in mol/L):

	$H_2$	$I_2$	$HI$
Initial	2.00	2.00	0
Change	$-x$	$-x$	$+2x$
Equilibrium	$2.00 - x$	$2.00 - x$	$2x$

Given $[HI] = 3.11$ mol/L, $2x = 3.11$ , so $x = 1.555$ mol/L.

Equilibrium concentrations:
$[H_2] = [I_2] = 2.00 - 1.555 = 0.445$ mol/L.

Keq expression:

K_{eq} = \frac{[HI]^2}{[H_2][I_2]} = \frac{(3.11)^2}{(0.445)(0.445)} = \frac{9.67}{0.198} \approx 48.9

Markers reward (1) correct ICE setup, (2) the correct Keq expression with stoichiometric exponents, (3) substitution and numerical answer to 3 significant figures, (4) noting Keq is dimensionless for this reaction (equal moles of gas on both sides).

2018 HSC3 marksFor the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium constant Keq at 500 K is 6.0 × 10⁻². Comment on the position of equilibrium and state two ways the value of Keq could be changed.

Show worked answer →

Keq = $6.0 \times 10^{-2}$ is less than 1, so the equilibrium lies to the left (reactants favoured). At 500 K, the mixture contains more $N_2$ and $H_2$ than $NH_3$ .

Keq depends only on temperature. Two ways to change its value:

Increase the temperature. The forward reaction is exothermic, so increasing temperature shifts equilibrium left, decreasing Keq.
Decrease the temperature. This shifts equilibrium right (exothermic direction), increasing Keq.

Markers reward (1) the correct interpretation that Keq < 1 means reactant-favoured, (2) the statement that only temperature changes Keq, (3) two correct examples with the predicted direction of change.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksWrite the equilibrium expression for

2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}

, and state whether the units of

K_{eq}

are needed if concentrations are in mol/L.

Show worked solution →

Expression. Products over reactants, each raised to its stoichiometric coefficient:

K_{eq} = \frac{[SO_3]^2}{[SO_2]^2[O_2]}

Units. There are 2 mol of gas on the product side and 3 mol of gas on the reactant side (2 + 1), a net change of $2 - 3 = -1$ . Since the net change is non-zero, $K_{eq}$ carries units of $\text{mol}^{-1}\text{L}$ (i.e. $\text{L mol}^{-1}$ ) in this case, so the numeric value should be reported WITH units here (unlike the equal-moles case in the main worked example).

Marking criteria: 1 mark for the correct expression with both exponents, 1 mark for correctly reasoning about units from the net change in moles of gas.

foundation3 marksFor

H_{2(g)} + CO_{2(g)} \rightleftharpoons H_2O_{(g)} + CO_{(g)}

, at a certain temperature

K_{eq} = 0.64

. A mixture has

[H_2] = 0.20

[CO_2] = 0.20

[H_2O] = 0.40

[CO] = 0.10

(all mol/L). Calculate

Q

and state the direction the reaction will proceed.

Show worked solution →

Step 1: write the $Q$ expression (same form as $K_{eq}$ , current concentrations).

Q = \frac{[H_2O][CO]}{[H_2][CO_2]}

Step 2: substitute.

Q = \frac{(0.40)(0.10)}{(0.20)(0.20)} = \frac{0.040}{0.040} = 1.0

Step 3: compare with $K_{eq}$ .

$Q = 1.0 > K_{eq} = 0.64$ , so the mixture has too much product relative to equilibrium. The reaction proceeds in reverse (towards reactants) until $Q$ falls to 0.64.

Marking criteria: 1 mark for the correct $Q$ expression, 1 mark for correct substitution and numeric value, 1 mark for the correct direction with the comparison stated explicitly ( $Q > K_{eq}$ ).

core4 marksFor

N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}

K_{eq} = 0.36

at 320 K. Calculate

K_{eq}

for the reverse reaction

2NO_{2(g)} \rightleftharpoons N_2O_{4(g)}

at the same temperature, and for the reaction written as

\tfrac{1}{2}N_2O_{4(g)} \rightleftharpoons NO_{2(g)}

Show worked solution →

Reverse reaction. Reversing a reaction inverts the equilibrium constant:

K_{reverse} = \frac{1}{K_{forward}} = \frac{1}{0.36} = 2.777\ldots \approx 2.8\ (2\ \text{s.f.})

Halved reaction. Multiplying every stoichiometric coefficient of the ORIGINAL forward reaction by $\tfrac{1}{2}$ raises $K_{eq}$ to the power $\tfrac{1}{2}$ (a square root):

K_{new} = (K_{forward})^{1/2} = (0.36)^{1/2} = 0.60

Check by direct substitution. For $\tfrac{1}{2}N_2O_4 \rightleftharpoons NO_2$ , $K_{new} = \dfrac{[NO_2]}{[N_2O_4]^{1/2}}$ , which is exactly the square root of $K_{eq} = \dfrac{[NO_2]^2}{[N_2O_4]}$ , confirming $K_{new} = \sqrt{0.36} = 0.60$ .

Marking criteria: 1 mark for correctly inverting $K_{eq}$ for the reverse direction, 1 mark for the correct reversed numeric value, 1 mark for correctly applying the power rule when the equation is scaled by $\tfrac{1}{2}$ , 1 mark for the correct final value $0.60$ .

core5 marksAt 480 K, 0.500 mol of

PCl_5

is placed in a 2.00 L flask and partially decomposes:

PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}

. At equilibrium, 30.0 percent of the

PCl_5

has decomposed. Calculate

K_{eq}

at 480 K, giving your answer to 3 significant figures with correct units.

Show worked solution →

Step 1: initial concentration of $PCl_5$ .

[PCl_5]_0 = \frac{n}{V} = \frac{0.500\ \text{mol}}{2.00\ \text{L}} = 0.250\ \text{mol L}^{-1}

Step 2: amount decomposed. 30.0 percent of 0.250 mol/L decomposes:

x = 0.300 \times 0.250 = 0.0750\ \text{mol L}^{-1}

Step 3: ICE table (mol/L).

	$PCl_5$	$PCl_3$	$Cl_2$
Initial	0.250	0	0
Change	$-0.0750$	$+0.0750$	$+0.0750$
Equilibrium	0.175	0.0750	0.0750

Step 4: substitute into the $K_{eq}$ expression.

K_{eq} = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(0.0750)(0.0750)}{0.175} = \frac{0.005625}{0.175} = 0.03214\ldots

Step 5: round to 3 significant figures and assign units. The net change in moles of gas is $2 - 1 = +1$ (non-zero), so $K_{eq}$ carries units of mol/L:

K_{eq} = 0.0321\ \text{mol L}^{-1}\ (3\ \text{s.f.})

Marking criteria: 1 mark for the correct initial concentration, 1 mark for correctly finding $x$ from the percentage decomposed, 1 mark for a correctly completed ICE table, 1 mark for correct substitution into the expression, 1 mark for the final value to 3 s.f. with correct units.

core5 marksThe graph below is an owned illustrative plot of the reaction quotient

Q

against time for a system starting from pure reactants at

t = 0

, approaching

K_{eq} = 49

at 700 K. (a) Describe the shape of the curve and explain, in terms of collision theory, why the rate of change of

Q

slows over time. (b) State what the flat portion of the graph represents at the particle level, given the system is a DYNAMIC equilibrium.

Show worked solution →

(a) Shape and explanation. $Q$ starts at 0 (pure reactants, no product yet) and rises steeply at first, then the rate of increase slows, and the curve flattens as it approaches the horizontal line at $K_{eq} = 49$ from below, reaching it at about $t = 6$ minutes. As the forward reaction proceeds, reactant concentrations fall and product concentrations rise, so by collision theory the forward reaction rate (which depends on reactant concentration) decreases while the reverse reaction rate (which depends on product concentration, initially zero) increases. $Q$ keeps rising only as long as the forward rate exceeds the reverse rate; as the two rates converge, the net rate of product formation falls towards zero, flattening the curve.

(b) The flat portion. The flat portion at $Q = K_{eq}$ is dynamic equilibrium: the forward and reverse reactions are still occurring continuously and at equal rates, so the macroscopic concentrations (and hence $Q$ ) stay constant even though molecules are still being converted in both directions at the particle level. The flat line does NOT mean the reaction has stopped.

Marking criteria: (a) 1 mark for describing the shape (steep rise then plateau), 1 mark for linking the slowing rate to falling reactant concentration via collision theory, 1 mark for linking the plateau to the forward and reverse rates becoming equal. (b) 1 mark for stating both reactions continue at the particle level, 1 mark for explicitly rejecting "the reaction has stopped".

exam7 marksAmmonia is manufactured industrially by the Haber process,

N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}

\Delta H = -92\ \text{kJ mol}^{-1}

. Evaluate the choice of an intermediate operating temperature (around 450 degrees C) and moderately high pressure (around 200 atm) as a compromise between the equilibrium constant and the reaction rate, and explain why a catalyst is essential to this compromise.

Show worked solution →

This is a 7-mark EVALUATE: markers reward a reasoned judgement that weighs BOTH the equilibrium constant and the rate, not just a list of factors.

Band 6 plan.

State the conflict: because the forward reaction is exothermic, LOW temperature favours a large $K_{eq}$ (high equilibrium yield of $NH_3$ , equilibrium shifted right), but low temperature gives a very SLOW rate of reaching that equilibrium (fewer successful collisions per unit time).
State the pressure argument: the forward reaction reduces moles of gas (4 mol reactant to 2 mol product), so by Le Chatelier increasing pressure shifts equilibrium right, increasing both the position of equilibrium and the rate (more frequent collisions), with no trade-off in this variable.
Explain the catalyst's role: a catalyst (iron with $Al_2O_3$ / $K_2O$ promoters) provides an alternative pathway with lower activation energy, increasing the RATE of both forward and reverse reactions equally. Crucially, a catalyst does NOT change $K_{eq}$ or the equilibrium position, so it lets the plant reach the SAME equilibrium yield much faster at a lower temperature than would otherwise be economically viable.
Judgement: 450 degrees C is a deliberate compromise, not the temperature that maximises yield (which would be lower) or the temperature that maximises rate alone (which would be higher); combined with the catalyst, it delivers an acceptable yield in an economically practical time. Very high pressure (above about 200 atm) is avoided because of the high capital and safety cost of pressurised plant, even though it would further increase yield and rate.

Model paragraph (excerpt). The 450 degrees C operating temperature at the Haber plant reflects a deliberate trade-off rather than an attempt to maximise either variable alone: because the forward reaction is exothermic, a lower temperature would raise $K_{eq}$ and hence the equilibrium yield of ammonia, but the reaction would proceed too slowly to be economically useful, since fewer particles would possess the activation energy required for successful collision. Raising pressure to around 200 atm avoids this conflict because it improves both yield and rate simultaneously, as the forward reaction reduces the total moles of gas; however, pressures much higher than this are avoided due to the prohibitive engineering cost of containment. The iron catalyst resolves the remaining tension by increasing the rate of both forward and reverse reactions without altering $K_{eq}$ , allowing industrially acceptable amounts of ammonia to be produced per hour at a temperature far lower than a purely rate-optimised process would require.

Marker's note: top-band answers (1) correctly attribute the temperature trade-off to $\Delta H$ and Le Chatelier, not just "increases rate", (2) note pressure improves BOTH factors with no trade-off (distinguishing it from temperature), (3) explicitly state a catalyst does not change $K_{eq}$ , and (4) close with an explicit judgement that 450 degrees C/200 atm is an economic compromise, not a theoretical optimum.