Chemical Equilibria

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

• AThe forward reaction has stopped and the reverse reaction continues
• BThe forward and reverse reactions occur at equal rates, so concentrations remain constant✓
• CThe concentrations of reactants and products are equal
• DNo reactions are occurring because the system is stable

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
• 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

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 reaction is endothermic
• BKc stays the same because only concentrations change
• CKc increases because the forward (endothermic) reaction is favoured✓
• DKc becomes zero

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 equilibrium yield of ammonia
• 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
• DTemperature has no effect on the yield, only on the rate

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).