Chemical Equilibria

Which statement best describes a system at dynamic equilibrium in a closed container?

• AThe forward reaction has stopped and only the reverse reaction (products to reactants) continues
• BThe forward and reverse reactions occur at equal rates, so concentrations remain constant✓
• CThe concentrations of reactants and products have become exactly equal; that is why they no longer change
• DNo reactions are occurring any longer, because the system has reached its most stable state

Answer: At dynamic equilibrium both forward and reverse reactions continue but at equal rates, so the concentrations of all species stay constant (not necessarily equal). The reactions never stop, which is what 'dynamic' indicates.

For the exothermic equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), which change would shift the position of equilibrium to the right, increasing the yield of SO₃?

• AIncreasing the total pressure✓
• BIncreasing the temperature
• CRemoving some O₂
• DAdding a catalyst

Answer: There are 3 moles of gas on the left and 2 on the right, so increasing the pressure shifts equilibrium to the side with fewer gas moles (right), favouring SO₃. Raising temperature shifts an exothermic reaction left; removing O₂ shifts left; a catalyst does not move the position of equilibrium.

For the equilibrium N₂O₄(g) ⇌ 2NO₂(g), which is the correct expression for the equilibrium constant Kc?

• AKc = [N₂O₄] / [NO₂]²
• BKc = [NO₂] / [N₂O₄]
• CKc = [NO₂]² / [N₂O₄]✓
• DKc = 2[NO₂] / [N₂O₄]

Answer: Kc = [products]/[reactants] with each concentration raised to the power of its stoichiometric coefficient: Kc = [NO₂]² / [N₂O₄]. The coefficient of 2 for NO₂ becomes an index (squared), not a multiplier.

At equilibrium for H₂(g) + I₂(g) ⇌ 2HI(g), the concentrations are [H₂] = 0.20 mol dm⁻³, [I₂] = 0.20 mol dm⁻³ and [HI] = 1.60 mol dm⁻³. Calculate Kc.

• A8.0
• B40
• C0.016
• D64✓

Answer: Kc = [HI]² / ([H₂][I₂]) = (1.60)² / (0.20 × 0.20) = 2.56 / 0.040 = 64. Forgetting to square [HI] gives the wrong answer; here Kc has no units because the powers cancel.

For the gaseous equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the equilibrium partial pressures are p(SO₂) = 0.20 atm, p(O₂) = 0.10 atm and p(SO₃) = 0.60 atm. Calculate Kp.

• A30 atm⁻¹
• B90 atm⁻¹✓
• C18 atm⁻¹
• D9.0 atm⁻¹

Answer: Kp = p(SO₃)² / (p(SO₂)² × p(O₂)) = (0.60)² / ((0.20)² × 0.10) = 0.36 / (0.040 × 0.10) = 0.36 / 0.0040 = 90. Units: atm² / atm³ = atm⁻¹.

Adding a catalyst to a reversible reaction that has reached equilibrium will:

• Aincrease the value of the equilibrium constant K and so raise the final yield
• Bshift the position of equilibrium towards the products
• Chave no effect on the position of equilibrium or the value of K✓
• Ddecrease the value of K but increase the yield of product

Answer: A catalyst speeds up the forward and reverse reactions equally, so it lets equilibrium be reached faster but does not change the position of equilibrium or the value of K. Only temperature changes the value of K.

For an endothermic forward reaction at equilibrium, what happens to the value of the equilibrium constant Kc when the temperature is increased?

• AKc decreases because the endothermic forward reaction (ΔH positive) is suppressed by heating
• BKc stays the same because only the concentrations or the pressures change
• CKc increases because the forward (endothermic) reaction is favoured✓
• DKc becomes zero at high temperature

Answer: Only temperature alters Kc. Raising the temperature favours the endothermic direction (here, the forward reaction), so more product forms and Kc increases. For an exothermic reaction, raising the temperature would decrease Kc.

In the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is exothermic. Industrially it is run at a compromise temperature of about 450 °C rather than a much lower temperature. Why is this compromise used?

• AA higher temperature increases the yield of ammonia, because the forward reaction takes in heat and so is driven to the right as the temperature rises
• BA lower temperature would give a higher yield but too slow a rate, so 450 °C balances an acceptable yield against an acceptable rate✓
• CA higher temperature is needed to increase the value of Kc for this exothermic reaction (ΔH negative), and a larger Kc means a better yield
• DTemperature has no effect at all on the equilibrium yield of ammonia, only on the rate, so 450 °C is simply what suits the iron catalyst

Answer: Because the forward reaction is exothermic, a lower temperature would increase the equilibrium yield, but the rate would be uneconomically slow. The compromise temperature of ~450 °C sacrifices some yield to achieve a fast enough rate (helped by the catalyst).

What characterises a dynamic equilibrium?

• AForward and reverse rates are equal and concentrations constant✓
• BForward and reverse rates are zero and concentrations constant
• CForward and reverse rates are equal and concentrations changing
• DThe forward rate exceeds the reverse rate at all times

Answer: Both reactions continue at the same rate, so nothing appears to change even though the system is active. A closed system is required, otherwise products escape and equilibrium is never reached.

According to Le Chatelier's principle, what happens if pressure is increased?

• AEquilibrium shifts towards the side with more gas moles
• BEquilibrium shifts towards the exothermic direction
• CEquilibrium position is unaffected by any pressure change
• DEquilibrium shifts towards the side with fewer gas moles✓

Answer: The system opposes the change by reducing the number of gas particles. If both sides have equal moles of gas, pressure has no effect on the position — only on the rate at which equilibrium is reached.

What happens to Kc when temperature is raised for an exothermic reaction?

• AKc increases as equilibrium shifts towards the products
• BKc decreases as equilibrium shifts towards the reactants✓
• CKc is unchanged but the position shifts to the reactants
• DKc is unchanged and the position also remains the same

Answer: Temperature is the only factor that alters Kc. Heating an exothermic reaction pushes it back towards reactants to absorb the extra energy, so Kc falls. Catalysts and pressure change nothing about Kc.

What effect does a catalyst have on an equilibrium?

• AIt is reached faster and shifts towards the products
• BIt is reached slower but its position is unchanged
• CIt is reached faster but its position is unchanged✓
• DIt is reached faster and shifts towards the reactants

Answer: A catalyst lowers the activation energy of forward and reverse reactions equally, so both speed up by the same factor. Equilibrium arrives sooner but the yield at equilibrium is identical.

Why are the conditions used in the Haber process a compromise?

• AHigh temperature raises the rate and also raises the equilibrium yield
• BHigh temperature raises rate but lowers the equilibrium yield✓
• CLow pressure raises both the rate and the equilibrium yield obtained
• DHigh pressure lowers the rate but raises the yield

Answer: The reaction is exothermic, so a lower temperature would give a better yield but far too slowly to be useful. Around 450 °C is chosen as a workable balance, with high pressure favouring the side of fewer gas moles.

What does a very large value of Kc indicate?

• AEquilibrium lies well over towards the reactants
• BEquilibrium lies well over towards the products✓
• CEquilibrium is reached extremely quickly
• DThe reaction is strongly exothermic overall

Answer: Kc puts products on top, so a large value means product concentrations dominate at equilibrium. It says nothing about how fast equilibrium is reached — that is kinetics, a separate matter.

In the expression for Kc, which species are omitted?

• APure gases and aqueous ions
• BAny species present in excess
• CPure solids and pure liquids✓
• DAny species that acts as a catalyst

Answer: Solids and pure liquids have effectively constant concentration, so they are folded into the constant and left out of the expression. Catalysts never appear because they are unchanged overall.

What is Kp expressed in terms of?

• APartial pressures of the gaseous species✓
• BTotal pressure of the whole gas mixture
• CConcentrations of the gaseous species
• DMole fractions of every species present

Answer: Kp uses partial pressures, each being the mole fraction multiplied by the total pressure. Only gases appear, for the same reason solids are excluded from Kc.

Why does removing a product as it forms increase the yield?

• AEquilibrium shifts backward to replace what was removed
• BThe value of Kc increases when product is removed
• CEquilibrium shifts forward to replace what was removed✓
• DThe reaction stops being reversible once product is removed

Answer: The system opposes the change by making more product, so continuously removing it drives the reaction onward. Kc itself is unchanged — only the position moves, which is exactly why the trick works.