Chemistry

1. 1.

   Explain what is dynamic about equilibrium in a closed reversible reaction.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Particles still interconvert, but every interval has equal conversion in each direction. Closing the system prevents reactants or products escaping and upsetting that balance.

   Marking points

   • Forward and reverse reactions both continue.
   • Their rates are equal.
   • Macroscopic concentrations remain constant, not necessarily equal.

   Examiner tip: Equilibrium is not a stopped reaction.

2. 2.

   Distinguish a weak acid from a dilute acid, and explain whether dilution changes an acid's intrinsic strength at fixed temperature.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Strength describes an equilibrium tendency; concentration describes how much acid is present. A concentrated weak acid and a dilute strong acid are both possible.

   Marking points

   • Weak refers to partial dissociation in water.
   • Dilute refers to low amount per unit volume.
   • Dilution changes concentration and dissociation fraction, but not Ka at fixed temperature.

   Examiner tip: Do not use 'weak' as a synonym for 'small concentration'.

3. 3.

   At equilibrium for H2 + I2 <=> 2HI, [H2] = 0.20, [I2] = 0.10 and [HI] = 0.40 mol dm^-3. Calculate Kc.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Use stoichiometric coefficients as powers, not multipliers. Equal total powers in numerator and denominator cancel the concentration units.

   Marking points

   • Kc = [HI]^2/([H2][I2]).
   • Use equilibrium values: 0.40^2/(0.20 x 0.10).
   • Kc = 8.0, dimensionless in this concentration expression.

   Examiner tip: The coefficient 2 on HI becomes a square.

4. 4.

   Calculate the pH of 0.0200 mol dm^-3 HCl, assuming complete dissociation and negligible water contribution.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. HCl supplies one proton per molecule, so use the acid concentration directly. The negative logarithm of 2.00 x 10^-2 is 2 - log10(2), about 1.699.

   Marking points

   • [H+] = 0.0200 mol dm^-3.
   • Use pH = -log10[H+].
   • pH = 1.699, or 1.70 rounded.

   Examiner tip: Do not use the weak-acid square-root approximation for HCl.

5. 5.

   A buffer contains HA and A^-. Explain separately how it resists added H+ and added OH-, and why a large addition can overwhelm it.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The conjugate pair supplies a sink for both types of perturbation. It redistributes HA and A^- rather than keeping pH mathematically constant; depletion removes that protection.

   Marking points

   • A^- consumes added H+ to form HA.
   • HA reacts with added OH- to form A^- and water.
   • These reactions reduce the change in free H+ concentration.
   • Finite amounts of buffer components can be depleted, so capacity is limited.

   Examiner tip: Write the reaction for each addition rather than claiming the buffer neutralises everything indefinitely.

6. 6.

   N2 + 3H2 <=> 2NH3 is exothermic. Evaluate increasing pressure and decreasing temperature to maximise industrial production, separating equilibrium yield from rate and operating cost.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Maximum equilibrium fraction is not necessarily maximum economical throughput. A catalyst helps approach equilibrium faster but cannot change its position; recycling recovers unreacted gases.

   Marking points

   • Higher pressure favours ammonia, the side with fewer gas moles.
   • Lower temperature favours the exothermic forward equilibrium.
   • Too low a temperature slows reaction, limiting production per time.
   • Compression costs/safety and a rate-yield compromise require moderate conditions with a catalyst and recycling.

   Examiner tip: Do not claim a catalyst raises the equilibrium yield.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.