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Unit 7: Equilibrium flashcards

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7.1 Introduction to Equilibrium

What is a reversible process? Give physical and chemical examples.
A process that can go forward (reactants → products) and in reverse, written with ⇌. Physical: evaporation and condensation; a gas dissolving and escaping; a salt dissolving and precipitating. Chemical: proton transfer (acid–base) and electron transfer (redox).
What is dynamic equilibrium?
In a closed system at constant temperature, a reversible process reaches a state where the forward and reverse processes continue at equal rates, so there is no net change. The reaction has not stopped; at equilibrium the forward rate is not zero.
How do you recognise equilibrium on concentration–time and rate–time graphs?
Concentration or pressure lines go flat (constant composition) and none falls to zero. Rate lines meet and stay together. Two concentration lines crossing is not equilibrium if they are still changing.
Are concentrations equal at equilibrium?
Usually not. Equilibrium means concentrations are constant, not equal. It is the rates, not the rate constants, that are equal.
How do coefficients help read an equilibrium graph?
Changes are in the ratio of the coefficients. For A ⇌ 2B starting with 1.00 atm A, if P_A levels at 0.70 atm, A fell by 0.30, so P_B = 0.60 atm.
Why must the system be closed to reach equilibrium?
If matter escapes, as when an open beaker loses a gas, the reverse process cannot keep pace and the system never settles. A sealed jar of water has a steady level because evaporation and condensation rates are equal.
What happens to the reverse rate when you start with only reactants?
It starts at zero and grows as product builds up, while the forward rate falls, until the two are equal.

7.2 Direction of Reversible Reactions

How do the forward and reverse rates decide the direction of net change?
Net rate = forward rate − reverse rate. Forward > reverse: net reactants → products. Reverse > forward: net products → reactants. Equal: equilibrium. Forward 2.0 × 10⁻³ and reverse 5.0 × 10⁻³ M s⁻¹: net to the left at 3.0 × 10⁻³ M s⁻¹.
How do the forward and reverse rates change as a reaction approaches equilibrium?
The rate of the faster direction falls and the rate of the slower direction rises until they meet. Neither falls to zero on the way, except a rate that starts at zero because a species is absent.
Is the direction of a reversible reaction fixed by the way the equation is written?
No. It depends on the current amounts. A mixture with too much product undergoes net change right to left. The same rate constants can give either direction depending on concentrations.
How do you write forward and reverse rates for elementary steps such as A ⇌ B and 2A ⇌ A₂?
A ⇌ B: forward = k_f[A], reverse = k_r[B]. 2A ⇌ A₂: forward = k_f[A]², reverse = k_r[A₂]. Powers come from coefficients only for elementary steps.
How do you check whether a mixture of A ⇌ B is at equilibrium using rate constants?
Calculate both rates. With k_f = 0.060 s⁻¹ and k_r = 0.020 s⁻¹, [A] = 0.20 M and [B] = 0.60 M give forward = reverse = 0.012 M s⁻¹, so it is at equilibrium (k_f[A] = k_r[B]).
Can a single particle snapshot show that a system is at equilibrium?
No. One snapshot shows amounts, not direction. Equal counts do not mean equilibrium; counts that stay constant across snapshots over time do.

7.3 Reaction Quotient and Equilibrium Constant

How do you write the reaction quotient Qc?
For aA + bB ⇌ cC + dD: Qc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ), using concentrations in mol L⁻¹ at any moment. Products on top; coefficients become powers (2NO gives [NO]², not 2[NO]); terms are multiplied, not added.
What is the relationship between Q and K?
As a reaction runs, Q changes and moves towards a fixed value, the equilibrium constant K. At equilibrium Qc = Kc and Qp = Kp. Only equilibrium values give K; any other values give Q.
Which species are left out of K and Q expressions?
Pure solids and pure liquids, including water as the solvent, because their "concentration" does not depend on how much is present. Keep H₂O(g); leave out H₂O(l). For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = P_CO₂.
What is the difference between Kc and Kp?
Kc uses concentrations in mol L⁻¹; Kp uses partial pressures of gases, usually in atm. They are usually different numbers. Do not mix pressures into Kc or concentrations into Kp; use the one the question asks for.
How do you calculate Qp for a gas mixture?
Substitute partial pressures. For 2SO₂ + O₂ ⇌ 2SO₃ with P_SO₂ = 0.20, P_O₂ = 0.50, P_SO₃ = 0.40 atm: Qp = 0.40² ÷ (0.20² × 0.50) = 8.0.
Does changing the amounts of substances change K?
No. At a fixed temperature K is constant; changing amounts changes Q. A different temperature gives a different K.
Write Kc for Cu²⁺(aq) + 4NH₃(aq) ⇌ Cu(NH₃)₄²⁺(aq).
Kc = [Cu(NH₃)₄²⁺] / ([Cu²⁺][NH₃]⁴).

7.4 Calculating the Equilibrium Constant

How do you calculate K from equilibrium data?
Write the balanced equation and the expression (no solids or pure liquids), get the equilibrium values in the right units (mol L⁻¹ for Kc, atm for Kp), substitute, and report K with no unit. For A + B ⇌ C with 0.20, 0.50, 0.40 mol L⁻¹: Kc = 0.40 ÷ (0.20 × 0.50) = 4.0.
How do you use an ICE table to find K when only one equilibrium value is known?
Write Initial, Change and Equilibrium rows. Find x from the known equilibrium value; fill the Change row in the ratio of coefficients (reactants down, products up for a forward reaction). Substitute the Equilibrium row into K.
Work through 2SO₃ ⇌ 2SO₂ + O₂ starting with 0.100 M SO₃ and reaching [O₂] = 0.020 M.
O₂ rose by 0.020, so SO₂ rose by 0.040 and SO₃ fell by 0.040 to 0.060 M. Kc = (0.040)² × 0.020 ÷ (0.060)² = 8.9 × 10⁻³.
What must you do with moles before using them in Kc?
Divide by the volume in litres: c = n / V. 0.60 mol in a 3.00 L vessel is 0.20 mol L⁻¹.
How do you get partial pressures for Kp from a total pressure?
P_X = (fraction of gas moles that are X) × P_total. Do not split a total pressure equally between gases made in a 2 : 1 ratio.
Does K depend on the starting amounts?
No. At a fixed temperature, every experiment gives the same K, whatever the starting amounts. Starting values must never be put into K.

7.5 Magnitude of the Equilibrium Constant

What does the size of K tell you about an equilibrium mixture?
Very large K: mostly products; the reaction goes essentially to completion. Very small K: mostly reactants; it barely proceeds. K near 1: both present in significant amounts. Rough guide: above about 10³ product-favoured, below about 10⁻³ reactant-favoured.
Does a large K mean a reaction is fast?
No. K is about position, not rate. A reaction with K = 10³⁰ may show no change in a week if its activation energy is high. A catalyst changes rate, not K.
Does a very small K mean no product forms?
No. A zero product concentration would make K = 0. A small K means only a trace of product; a large K means a trace of reactant always remains ("essentially complete").
How do you turn K into a percentage for A ⇌ B?
[B]/[A] = K, so fraction as B = K / (1 + K). K = 10² gives 99.0% B; K = 1 × 10⁻⁵ gives about 0.001% B.
How do you read K from a particle diagram for X ⇌ Y?
Volume cancels, so K = number of Y ÷ number of X. With K = 4.0 and 25 particles, there are 20 Y and 5 X.
Does K = 1 always mean equal concentrations of reactants and products?
No. That holds only for expressions of the same shape, like A ⇌ B. For P ⇌ 2R, K = 1 does not mean equal concentrations, because [R] is squared.

7.6 Properties of the Equilibrium Constant

What happens to K when you reverse an equation?
The new K is 1/K. K = 100 becomes 0.010; K = 2.0 × 10⁻³ becomes 5.0 × 10². K is never negative.
What happens to K when you multiply an equation's coefficients?
Multiply by c → raise K to the power c. Doubling: K². Halving: √K. Tripling K = 2.0 × 10⁻³ gives 8.0 × 10⁻⁹; halving gives 0.045. Do not double or halve K itself.
How do you find K for an overall reaction made by adding steps?
Multiply the K values of the steps (do not add them). Steps with K = 4.0 and 0.50 give K = 2.0. Remember any power on a step you multiplied, and check intermediates cancel to the exact target.
How do the K rules parallel Hess's law?
Reversing: ΔH changes sign, K inverts. Multiplying by c: ΔH is multiplied by c, K is raised to the power c. Adding: ΔH values add, K values multiply.
What conditions must hold to combine K values?
All K values must be at the same temperature and of the same type (all Kc or all Kp). The same rules apply to Q, which has the same mathematical form.

7.7 Calculating Equilibrium Concentrations

What is the first step in predicting equilibrium concentrations?
Calculate Q from the starting values and compare with K. Q below K: net forward. Q > K: net reverse. Q = K: already at equilibrium. For A ⇌ 2B with Kc = 0.10, [A] = 0.50 and [B] = 0.30 gives Q = 0.18 > K, so net reverse.
How do you solve for equilibrium concentrations with an ICE table?
Write each change in terms of one unknown in the ratio of coefficients (−x, −2x, +3x), substitute the Equilibrium row into K, and solve. Z ⇌ Y with Kc = 4.0 from [Z] = 1.00 M: x ÷ (1.00 − x) = 4.0, x = 0.80, so [Y] = 0.80 M and [Z] = 0.20 M.
When can you use the small-x approximation, and how do you check it?
When K is small and very little reacts, assume c − x ≈ c. Afterwards check that x is less than about 5% of the starting value; if not, solve the quadratic.
How do coefficients appear in an ICE calculation?
A coefficient of 2 gives a change of 2x, which appears squared in K as (2x)². For 2G ⇌ H going forwards, if [G] falls by 0.30 M, [H] rises by 0.15 M.
Which root of a quadratic do you keep in an equilibrium problem?
The one that makes physical sense: no negative concentrations and no change larger than the starting amount allows.
What does it mean when an ICE calculation reaches Q = K?
The forward and reverse rates are equal. Both reactions continue, but concentrations stay constant. Equilibrium is not equal amounts and not "no reaction".

7.8 Representations of Equilibrium

What does a particle diagram of an equilibrium mixture represent?
A snapshot of a small, fixed volume; the count of each species is proportional to its concentration or partial pressure. In a sequence, counts change while Q moves towards K; when counts stop changing, Q = K.
Why do particles sit in different places in successive equilibrium boxes?
Reactions continue in both directions at equal rates, so individual particles keep reacting even though the counts stay the same. Equilibrium is dynamic.
When can you put particle counts straight into K?
Only when both sides have the same total number of particles, so volume cancels. X₂ + Y₂ ⇌ 2XY with 2 X₂, 1 Y₂ and 4 XY: K = 4² ÷ (2 × 1) = 8.0.
How do you get Kc from a box when particle numbers differ on each side?
Convert counts to concentrations: c = count × mol per particle ÷ box volume. For 2N ⇌ N₂ in 1.0 L with 2 N and 2 N₂ at 0.050 mol each: both 0.10 M, Kc = 0.10 ÷ 0.10² = 10. Raw counts would give 0.5.
Can the same equilibrium be reached from different starting mixtures?
Yes. The same equilibrium box can be reached from mostly reactants or mostly products, if the atoms, volume and temperature are the same.
What rules should a sequence of equilibrium particle diagrams follow?
Same volume and temperature in every box; the same number of atoms of each element; a key using shapes or shading, not colour alone; a joined pair such as J₂ counted as one particle.
What does a box with mostly products or mostly reactants at equilibrium tell you about K?
Mostly products: K is very large. Mostly reactants: K is very small. At the lab scale colour, pressure and concentration stay constant because particle changes cancel.

7.9 Introduction to Le Châtelier’s Principle

What is Le Châtelier's principle?
When a stress upsets an equilibrium (forward and reverse rates no longer equal), the system shifts in the direction that partly counteracts the stress, then settles at a new equilibrium. The shift is partial; old concentrations do not return.
How does adding or removing a reactant or product shift equilibrium?
Adding a species shifts away from it (uses it up). Removing one shifts towards it (replaces it). Precipitating or reacting a species away counts as removing it. K does not change.
How does temperature shift an equilibrium?
Write heat into the equation. Heating shifts in the endothermic direction; cooling shifts in the exothermic direction. Temperature is the only stress that changes K. For 2SO₂ + O₂ ⇌ 2SO₃ (ΔH negative), heating shifts left.
How does changing gas volume (pressure) shift an equilibrium?
Smaller volume (higher pressure) shifts to the side with fewer gas molecules; larger volume to the side with more. No shift if the numbers are equal. 2SO₂ + O₂ ⇌ 2SO₃ compressed shifts right (3 → 2).
Which changes cause no shift in equilibrium?
Adding a catalyst, extra solid or pure liquid, or an unreactive gas at constant volume. Adding solid CaCO₃ to CaCO₃(s) ⇌ CaO(s) + CO₂(g) does not change P_CO₂.
How does diluting a solution shift an equilibrium?
Towards the side with more dissolved particles. Do not count water as the solvent.
Why does compressing a coloured gas mixture darken it at once?
The instant effect: the same particles are squeezed into a smaller volume, so concentration rises before any shift. The colour then changes partly back as the system shifts. Describe the observation if asked.

7.10 Reaction Quotient and Le Châtelier’s Principle

How do you use Q and K to predict the response to a stress?
A stress makes Q ≠ K. Q below K (too few products): net forward. Q > K (too many products): net reverse. The system reacts until Q = K again, with different concentrations. Q = 0.80 and K = 0.25: reverse.
How do you calculate Q after a volume change or dilution?
Every concentration is scaled by s, so new Q = s^Δn × K, where Δn = product coefficients − reactant coefficients (gases or dissolved species). Halving V: s = 2. For N₂O₄ ⇌ 2NO₂, Q = 2K, so reverse. Diluting A(aq) ⇌ 2B(aq) to twice the volume gives Q = ½K, so forward.
How does temperature change K for exothermic and endothermic reactions?
Exothermic: T up → K down (heating favours reverse). Endothermic: T up → K up (heating favours forward).
When you heat a rigid flask at equilibrium, does Q or K change?
K changes. At that instant the concentrations, and so Q, are the same; now Q ≠ K, so the system reacts towards the new K.
Does adding a species change K?
No. Adding or removing a species changes Q; only temperature changes K. Solids, pure liquids and solvent water are left out of Q as they are from K.

7.11 Introduction to Solubility Equilibria

What is K_sp, and what is the equilibrium it describes?
The solubility-product constant: the equilibrium constant for an ionic solid dissolving, e.g. MₓAᵧ(s) ⇌ xMᵃ⁺(aq) + yAᵇ⁻(aq), K_sp = [Mᵃ⁺]ˣ[Aᵇ⁻]ʸ. In a saturated solution with solid present, dissolving and precipitating occur at equal rates. K_sp changes only with temperature.
What is molar solubility, s?
The moles of salt that dissolve per litre of saturated solution (mol L⁻¹). Multiply by molar mass for g L⁻¹. s has a unit; K_sp does not.
How do you relate K_sp and s for different salt formulas?
[Mᵃ⁺] = xs, [Aᵇ⁻] = ys. 1:1: K_sp = s². 1:2 or 2:1: 4s³. 1:3: 27s⁴. 2:3 or 3:2: 108s⁵. MX with K_sp = 4.0 × 10⁻¹⁰ has s = 2.0 × 10⁻⁵ M; MX₂ with s = 1.0 × 10⁻³ M has K_sp = 4.0 × 10⁻⁹.
Can you rank solubility by K_sp?
Only for salts with the same ion ratio. Otherwise calculate s. A 2:1 salt with K_sp = 4.0 × 10⁻¹² (s = 1.0 × 10⁻⁴ M) is more soluble than a 1:1 salt with K_sp = 4.0 × 10⁻¹⁰ (s = 2.0 × 10⁻⁵ M).
How do you predict whether a precipitate forms on mixing two solutions?
Account for dilution on mixing, calculate Q from the ion concentrations, and compare with K_sp. Q > K_sp: precipitate forms.
What are common errors in K_sp calculations?
Forgetting the coefficient inside the bracket (for ZF₂, [F⁻] = 2s); forgetting the power ((2s)² = 4s²); including the solid; using one ion's concentration as s when the ratio is not 1:1 ([Ag⁺] = 2.0 × 10⁻⁴ M for Ag₂X means s = 1.0 × 10⁻⁴ M).
What does K_sp > 1 indicate?
A soluble salt, such as all Na⁺, K⁺, NH₄⁺ and NO₃⁻ salts. Sparingly soluble salts have K_sp much less than 1.

7.12 Common-Ion Effect

What is the common-ion effect?
A salt is less soluble in a solution that already contains one of its ions. Added product ion makes Q > K_sp, so ions precipitate until Q = K_sp again; by Le Châtelier the equilibrium shifts towards the solid. K_sp does not change.
What counts as a common ion for AgCl?
An ion in the K_sp expression: Cl⁻ (from NaCl) or Ag⁺ (from AgNO₃). Na⁺ or NO₃⁻ are not in the expression and do not cause the effect.
How do you calculate molar solubility of MX in a solution of a common ion X⁻?
K_sp = s(c + s) ≈ s × c when s ≪ c, so s ≈ K_sp ÷ c. K_sp = 1.0 × 10⁻¹² in 0.010 M common ion gives s = 1.0 × 10⁻¹⁰ M. Check the approximation afterwards.
How do you calculate molar solubility of MX₂ with a common ion?
In c M X⁻: K_sp = s(c + 2s)² ≈ sc², so s ≈ K_sp ÷ c² (2.0 × 10⁻¹⁰ in 0.020 M X⁻ gives 5.0 × 10⁻⁷ M). In c M M²⁺: K_sp = (c + s)(2s)² ≈ 4cs², so s ≈ √(K_sp ÷ 4c) (gives 5.0 × 10⁻⁵ M in 0.020 M M²⁺).
What happens to [Ag⁺] and K_sp when solid NaCl is added to saturated AgCl?
[Cl⁻] rises, AgCl precipitates, and [Ag⁺] falls. K_sp is unchanged at constant temperature; the molar solubility falls.
What is the effect of removing one of a salt's ions from solution?
The opposite of the common-ion effect: Q falls below K_sp, so more solid dissolves.
How do you find K_sp from data measured in a common-ion solution?
K_sp = [measured ion]ᵐ × [common ion]ⁿ with powers from the formula, using the common-ion concentration (not s) for that ion. Using the pure-water formula (K_sp = s²) gives a K_sp far too small.

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