iGCSE Chemistry 0620 Extended - 6.2 Rate of reaction- Exam Style Questions Paper 4- New Syllabus
Question
$SF_{4}(g) + 2H_{2}O(g) \rightleftharpoons SO_{2}(g) + 4HF(g)$ $\Delta H = -54\text{ kJ/mol}$
(a) Complete a reaction pathway diagram for this reaction (plotting energy vs progress of reaction). Include in your diagram:
- the position and the formulae of the products
- an arrow, labelled $E_{a}$, to show the activation energy
- an arrow, labelled $\Delta H$, to show the enthalpy change of the reaction.
- Calculate the energy needed to break the bonds in the reactants.
- Calculate the energy released when the bonds in the products form.
- Calculate the S=O bond energy.
(c) The equation for the reaction is shown.
State the effect, if any, on the position of equilibrium when the following changes are made. Give a reason for each of your answers.
- The temperature is increased.
- The pressure is increased.
- A catalyst is added.
Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):
• Topic $5.1$ — Exothermic and endothermic reactions (Parts $\mathrm{(a)}$, $\mathrm{(b)}$)
• Topic $6.3$ — Reversible reactions and equilibrium (Part $\mathrm{(c)}$)
• Topic $6.2$ — Rate of reaction (Part $\mathrm{(d)}$)
▶️ Answer/Explanation
(a)
– The energy level of the products ($SO_{2}(g) + 4HF(g)$) must be drawn as a horizontal line below the energy level of the reactants.
– An upward-pointing arrow labelled $E_{a}$ should start from the reactant energy level and go up to the highest point (peak) of the energy curve.
– A downward-pointing arrow labelled $\Delta H$ should start from the reactant energy level and end exactly at the product energy level.
(b) $S=O$ bond energy = $467\text{ kJ/mol}$
Calculation steps:
Reactant bonds broken: $(4 \times S-F) + (4 \times O-H) = (4 \times 330) + (4 \times 460) = 1320 + 1840 = 3160\text{ kJ/mol}$.
Product bonds formed release energy. Let total energy released = $E_{released}$.
$\Delta H = \text{Bonds Broken} – \text{Bonds Formed}$
$-54 = 3160 – E_{released}$
$E_{released} = 3160 + 54 = 3214\text{ kJ/mol}$.
The products contain $2 \times S=O$ bonds and $4 \times H-F$ bonds:
$(2 \times S=O) + (4 \times 570) = 3214$
$2 \times S=O = 3214 – 2280 = 934$
$S=O = \frac{934}{2} = 467\text{ kJ/mol}$.
(c)
Temperature increased: The position of equilibrium moves to the left-hand side. Reason: The forward reaction is exothermic (indicated by $\Delta H = -54\text{ kJ/mol}$), so increasing temperature favours the endothermic reverse reaction to absorb the excess heat.
Pressure increased: The position of equilibrium moves to the left-hand side. Reason: There are $3\text{ moles}$ of gas on the left and $5\text{ moles}$ of gas on the right. Increasing pressure shifts equilibrium to the side with fewer gaseous moles.
Catalyst added: No change in the position of equilibrium. Reason: A catalyst speeds up both the forward and reverse reactions equally.
(d) Reducing the temperature decreases the kinetic energy of the particles. Consequently, the frequency of collisions between particles decreases. Furthermore, a lower proportion (or fraction) of particles will have energy greater than or equal to the activation energy ($E_a$), leading to fewer successful collisions per second.