Equilibrium: common doubts

Question 1

Why is chemical equilibrium called dynamic?

Accepted Answer

Because the forward and reverse processes never stop at equilibrium; they simply run at the same rate. In a sealed bottle of water, molecules keep evaporating and condensing, but equal numbers do each every second, so the vapour pressure stays steady. Isotope-labelling experiments confirm this, showing tagged atoms passing between reactants and products long after the overall composition has stopped changing.

Question 2

What happens to K when a reaction is reversed or multiplied by a number?

Accepted Answer

Reversing a reaction inverts its constant, so K' = 1/K. Multiplying the whole equation by n raises the constant to that power, K' = Kⁿ, so halving the equation takes the square root. When two reactions are added, their constants multiply. If N₂ + 3H₂ ⇌ 2NH₃ has constant K, then NH₃ ⇌ ½N₂ + (3/2)H₂ has 1/√K.

Question 3

Why are pure solids and liquids left out of the equilibrium constant?

Accepted Answer

The concentration of a pure solid or liquid is fixed, since its amount per unit volume depends only on its density and not on how much of it is present. That constant value is folded into K, so it does not appear in the expression. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = p(CO₂), whatever the amount of solid in the vessel.

Question 4

How are Kp and Kc related and when are they equal?

Accepted Answer

Use Kp = Kc(RT)^Δn, where Δn is moles of gaseous products minus moles of gaseous reactants and R = 0.0831 bar L mol⁻¹ K⁻¹ for pressures in bar. For N₂ + 3H₂ ⇌ 2NH₃, Δn = 2 − 4 = −2, so Kp = Kc(RT)⁻². When Δn = 0, as in H₂ + I₂ ⇌ 2HI, the two constants are equal. Only gases count towards Δn.

Question 5

What does the magnitude of the equilibrium constant tell us?

Accepted Answer

It shows how far a reaction proceeds before reaching equilibrium. A very large K, above about 10³, means products dominate and the reaction is nearly complete; a very small K, below about 10⁻³, means very little product forms; values in between leave sizeable amounts of both. K says nothing about speed, so a reaction with a huge K can still be very slow.

Question 6

What is the difference between the reaction quotient Q and the equilibrium constant K?

Accepted Answer

Q has the same form as K but uses the concentrations at any chosen moment, while K uses only equilibrium values. Comparing them tells you which way the reaction will go: if Q < K it moves forward, if Q > K it moves backward, and if Q = K it is already at equilibrium. As the reaction proceeds, Q keeps changing until it equals K.

Question 7

Does a catalyst change the position of equilibrium?

Accepted Answer

No. A catalyst lowers the activation energy of the forward and reverse reactions equally, so both speed up by the same factor. Equilibrium is reached sooner, but K and the final composition are exactly what they would be without it. That is why the Haber process uses an iron catalyst for speed, while the yield is controlled by temperature and pressure.

Question 8

Why does only temperature change the value of the equilibrium constant?

Accepted Answer

Changing a concentration or the pressure shifts the equilibrium position, but the system adjusts until the ratio in K is restored, so K stays the same. Temperature changes the forward and backward rate constants by different amounts, which alters their ratio. For an exothermic reaction K falls as temperature rises, and for an endothermic reaction K rises.

Question 9

Why does high pressure favour ammonia formation in the Haber process?

Accepted Answer

Raising the pressure pushes an equilibrium towards the side with fewer moles of gas, since that shift lowers the number of gas molecules and partly relieves the pressure. In N₂ + 3H₂ ⇌ 2NH₃, four moles of gas become two, so high pressure raises the ammonia yield. The reaction is exothermic, so a lower temperature also helps the yield, but too low a temperature makes it very slow.

Question 10

Does adding an inert gas shift an equilibrium at constant volume?

Accepted Answer

No. The inert gas raises the total pressure, but the amounts of the reacting gases and the volume they occupy are unchanged, so their partial pressures and concentrations stay the same. None of the terms in Q change, so Q still equals K and nothing shifts. Treating the higher total pressure as if it were squeezing the reacting gases is a common mistake.

Question 11

Why is BF₃ called an acid when it has no hydrogen?

Accepted Answer

Because it fits the Lewis definition, in which an acid is any species that accepts an electron pair. Arrhenius and Brønsted-Lowry acids must release or donate protons, but boron in BF₃ has only six valence electrons and a vacant orbital, so it readily takes a lone pair from a donor such as NH₃. Metal ions like Mg²⁺ and Co³⁺ act as Lewis acids too.

Question 12

Why does a strong acid have a weak conjugate base?

Accepted Answer

A conjugate acid-base pair differs by one proton, and the two strengths are opposite. A strong acid such as HCl gives up its proton very readily, which means its conjugate base, Cl⁻, has almost no tendency to take a proton back, so it is a very weak base. A weak acid like acetic acid holds its proton tightly, so acetate is a noticeably stronger base.

Question 13

Why is HI a stronger acid than HF even though fluorine is more electronegative?

Accepted Answer

Down group 17 the halogen atom grows larger, so the H–X bond becomes longer and weaker. A weaker bond lets go of H⁺ more easily in water, giving the order HF < HCl < HBr < HI. Here bond strength outweighs electronegativity. Along a period, as in CH₄ < NH₃ < H₂O < HF, bond strengths are similar and electronegativity decides acid strength instead.

Question 14

Is the pH of pure water always 7?

Accepted Answer

No, only at 298 K. The ionisation of water absorbs heat, so its ionic product Kw grows as the temperature rises. In hot water both [H⁺] and [OH⁻] exceed 10⁻⁷ M, so the pH of pure water falls below 7. The water is still neutral, because [H⁺] equals [OH⁻]; pH 7 means neutral only where Kw = 1.0 × 10⁻¹⁴.

Question 15

How do you calculate the pH of a weak acid solution?

Accepted Answer

For a weak acid of concentration c that ionises only slightly, [H⁺] = √(Ka c) and pH = −log[H⁺]. For 0.1 M acetic acid with Ka = 1.8 × 10⁻⁵, [H⁺] = √(1.8 × 10⁻⁶) ≈ 1.34 × 10⁻³ M, so pH ≈ 2.87. The shortcut assumes the degree of ionisation is small; if it is not, solve the full quadratic instead.

Question 16

Is the pH of 10⁻⁸ M HCl equal to 8?

Accepted Answer

No, it is about 6.98, slightly acidic. At such a tiny acid concentration the H⁺ from water's own ionisation cannot be ignored: pure water already holds 10⁻⁷ M, ten times what the acid adds. Allowing for both sources gives [H⁺] of roughly 1.05 × 10⁻⁷ M. A pH of 8 would absurdly describe an acid solution as basic.

Question 17

Why does adding sodium acetate reduce the ionisation of acetic acid?

Accepted Answer

Sodium acetate supplies extra acetate ions, the same ion acetic acid produces when it ionises. By Le Chatelier's principle, CH₃COOH ⇌ H⁺ + CH₃COO⁻ shifts to the left to consume some of the added acetate, so less acid ionises and [H⁺] drops. This suppression of a weak electrolyte by an ion it shares with an added salt is the common ion effect.

Question 18

Why is a sodium acetate solution basic and an ammonium chloride solution acidic?

Accepted Answer

Sodium acetate comes from a strong base and a weak acid, and its acetate ion takes a proton from water to form some acetic acid and OH⁻, so the solution turns basic. Ammonium chloride comes from a weak base and a strong acid, and its NH₄⁺ ion passes a proton to water, forming NH₃ and H₃O⁺, so the solution turns acidic. Na⁺ and Cl⁻ do not hydrolyse.

Question 19

How does a buffer solution resist changes in pH?

Accepted Answer

A buffer contains a weak acid with its salt, or a weak base with its salt, so it can neutralise both added acid and added base. In an acetic acid and sodium acetate buffer, added H⁺ combines with acetate to form acetic acid, and added OH⁻ is neutralised by acetic acid. The salt-to-acid ratio barely shifts, so the pH stays almost constant.

Question 20

How do you calculate the pH of a buffer solution?

Accepted Answer

For an acidic buffer use pH = pKa + log([salt]/[acid]). When the salt and acid concentrations are equal, the log term is zero and pH = pKa, which is why a buffer is chosen with a pKa close to the desired pH. For a basic buffer of a weak base and its salt, use pOH = pKb + log([salt]/[base]) and then pH = 14 − pOH at 298 K.

Question 21

How do you relate Ksp to solubility for salts like AgCl and Ag₂CrO₄?

Accepted Answer

Write the dissolution equation and raise each ion's concentration to its coefficient. For AgCl, Ksp = S². For Ag₂CrO₄, which releases two Ag⁺ per CrO₄²⁻, Ksp = (2S)²(S) = 4S³. Because the formula types differ, a salt with the smaller Ksp can still be the more soluble one, so always calculate S before comparing solubilities.

Question 22

Why does passing HCl gas through brine precipitate pure sodium chloride?

Accepted Answer

HCl gas raises the chloride ion concentration of the saturated solution, so the ionic product [Na⁺][Cl⁻] exceeds the solubility product. Sodium chloride then crystallises out until the product drops back to Ksp. Soluble impurities remain dissolved, so the solid that separates is purer than the original salt. It is the common ion effect working on a solubility equilibrium.

Equilibrium: common doubts, answered

Equilibrium in physical processes