Reversible Reactions and Equilibrium Worked ExamplesIGCSE Chemistry 0620

(a) Heating removes the water of crystallisation. The pink hydrated salt becomes blue anhydrous cobalt(II) chloride.

(b) The water leaves the boiling tube as a gas (steam). The ice and water cool the vapour so it condenses to a liquid.

(c) Add anhydrous copper(II) sulfate: it turns from white to blue. This shows that water is present, not that it is pure.

(d) Measure its boiling point: pure water boils at exactly $100\,^{\circ}\text{C}$ (or freezes at $0\,^{\circ}\text{C}$).

(e) The blue solid turns pink and the mixture gets warm. The hydrated salt has been re-formed, so the reaction is reversible — shown by the symbol $\rightleftharpoons$. Heat is taken in to remove the water, and given out when the water is added back (the reverse change is exothermic).

(a) The symbol $\rightleftharpoons$ shows the reaction is reversible. $\Delta H$ is negative, so the forward reaction is exothermic.

(b) The reverse reaction has the same size of enthalpy change with the opposite sign: $\Delta H = +41\text{ kJ/mol}$ (endothermic).

(c) A closed system is one in which no substances (reactants or products) can enter or leave. Energy can still be transferred to or from the surroundings.

(d)(i) Count the moles of gas: $1 + 1 = 2$ on the left and $1 + 1 = 2$ on the right. Pressure favours the side with fewer gas molecules, but here they are equal, so there is no change in the position of equilibrium.

(ii) The rate still increases: there are more gas particles per unit volume, so the collision frequency increases. Position and rate are separate ideas.

(a) Equilibrium is reached when the concentrations stop changing — both lines become horizontal at about $8$ minutes.

(b) At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products are no longer changing.

(c) At $1.3$ minutes the HI line is still rising and the $\text{H}_2$ line is still falling — the concentrations are still changing, so the forward reaction is still faster than the reverse. Equal concentrations are not a condition for equilibrium; at equilibrium here they are very different ($0.016$ and $0.002\text{ mol/dm}^3$).

(d) $\text{H}_2$ falls from $0.010$ to $0.002$, a decrease of $0.008\text{ mol/dm}^3$. The equation shows $1$ mol $\text{H}_2$ gives $2$ mol HI, so HI should rise by $2 \times 0.008 = 0.016\text{ mol/dm}^3$ — which matches the graph.

(e) There are $2$ moles of gas on each side, so increasing the pressure causes no change in the position of equilibrium.

Temperature: the forward reaction is exothermic ($\Delta H$ negative). Lowering the temperature moves the equilibrium in the exothermic direction — to the right. So a low temperature is needed.

Pressure: the left has $2 + 1 = 3$ moles of gas; the right has $2$. Increasing the pressure moves the equilibrium to the side with fewer gas molecules — the right. So a high pressure is needed. Answer: C.

A and B use a high temperature, which favours the endothermic reverse reaction. D uses low pressure, which favours the side with more gas molecules (the left).

Temperature increased: the equilibrium moves in the endothermic direction, the reverse reaction, so the amount of $\text{PCl}_5$ decreases. The particles have more energy, so the rate increases.

Pressure decreased: there are $2$ moles of gas on the left and $1$ on the right. Lower pressure favours the side with more gas molecules (the left), so the amount of $\text{PCl}_5$ decreases. Fewer particles per unit volume means fewer collisions per second, so the rate decreases.

Catalyst: it speeds up the forward and reverse reactions equally, so there is no change in the amount of $\text{PCl}_5$ at equilibrium, but the rate increases.

(a) Nitrogen comes from the air. Hydrogen comes from methane (natural gas).

(b) Nitrogen and hydrogen are diatomic molecules, so write $\text{N}_2$ and $\text{H}_2$, then balance.

(c) About $450\,^{\circ}\text{C}$; about $20\,000\text{ kPa}$ ($200$ atm); an iron catalyst.

(d) A catalyst increases the rates of the forward and reverse reactions by the same amount, so equilibrium is reached faster but its position is unchanged.

(a) Sulfur dioxide: burning sulfur in air, or roasting sulfide ores. Oxygen: from the air.

(b) Oxygen is $\text{O}_2$; balance the sulfur and oxygen atoms.

(c) About $450\,^{\circ}\text{C}$; about $200\text{ kPa}$ ($2$ atm); vanadium(V) oxide catalyst.

(d) The position of equilibrium already lies far to the right at $2$ atm, so the yield is already high. A higher pressure would add little extra product but would cost more (stronger equipment, more energy to compress the gases) and would be less safe.

The forward reaction is exothermic, so a lower temperature would actually give a higher percentage of ammonia at equilibrium. But at $250\,^{\circ}\text{C}$ the reaction would be too slow to be economic. $450\,^{\circ}\text{C}$ is a compromise between rate and yield. Answer: B.

A is the wrong way round — lowering the temperature increases the equilibrium yield of an exothermic reaction. C is also wrong: raising the temperature moves the equilibrium to the left. D is wrong because temperature does not change the activation energy.

(a) At every pressure the percentage of L falls as the temperature rises. Increasing the temperature favours the endothermic direction, which must be the reverse reaction here, so the forward reaction is exothermic.

(b) At any temperature the percentage of L is higher at higher pressure. Increasing the pressure moves the equilibrium towards the side with fewer gas molecules, so the right-hand (product) side has fewer gas molecules.

(c) On the $200$ atm curve at $400\,^{\circ}\text{C}$: $45\,\%$.

(d)(i) At $250\,^{\circ}\text{C}$ the rate would be too slow, so less product would be made per hour even though the equilibrium percentage is higher. (ii) A higher pressure needs stronger, more expensive equipment and more energy to compress the gases, and is less safe.

(a) Shape: the curve is steeper at the start and levels off sooner (before $10$ min), at the same concentration, $0.60\text{ mol/dm}^3$.

Reason: the catalyst lowers the activation energy, so equilibrium is reached faster; it speeds up the forward and reverse reactions equally, so the position of equilibrium does not change.

(b) Shape: the curve is steeper at the start and levels off sooner, but at a lower concentration (below $0.60\text{ mol/dm}^3$).

Reason (rate): particles have more kinetic energy, collide more often and a greater proportion of collisions have energy ≥ $E_a$, so equilibrium is reached sooner. Reason (position): the forward reaction is exothermic ($\Delta H$ negative), so a higher temperature moves the equilibrium to the left — less $\text{SO}_3$ at equilibrium.

(a) Position: the forward reaction is endothermic ($\Delta H$ positive), so a high temperature moves the equilibrium to the right, increasing the yield of hydrogen.

Rate: at a high temperature the particles have more kinetic energy, collide more frequently and a greater proportion of collisions have energy ≥ $E_a$, so the rate is faster. Unlike the Haber process, rate and yield both favour a high temperature — there is no need to compromise on temperature.

(b) Count moles of gas: left $1 + 1 = 2$, right $1 + 3 = 4$. A lower pressure moves the equilibrium towards the side with more gas molecules — the right — so more hydrogen is formed.

(c) At a very low pressure there are fewer particles per unit volume, so the rate is lower and less hydrogen is produced per hour.

(d) No effect on the yield: the catalyst increases the rates of the forward and reverse reactions equally, so equilibrium is reached faster but its position is unchanged.

(a) Hydroxide ions react with hydrogen ions, removing them from the equilibrium.

The concentration of $\text{H}^+$ falls, so the equilibrium moves to the right to replace it. Bromine is used up, and the solution turns from orange to colourless (or paler).

(b) Excess acid adds $\text{H}^+$ ions. The equilibrium moves to the left to use them up, bromine is re-formed and the orange colour returns.

(c) Silver ions remove bromide ions as a cream precipitate of silver bromide. The equilibrium moves to the right to replace the bromide, so the orange colour fades.

(d) Pressure only affects equilibria that involve gases. All the substances here are aqueous or liquid, so pressure has no effect.