a)(i) State Graham's law of diffusion.
(ii) Calculate the vapour density of a triatomic gas X if its relative: atomic mass is 16.
(iii) Equal volumes of gases Y and Z are maintained at the same temperature and pressure. If the mass of a molecule of Y is twice that of Z state and explain which of the molecules has the, greater average velocity.
(b) The graph below is the ratio curve for the following reaction carried out in an open vessel.
MgCO\(_{3(s)}\) + 2HCI\(_{(aq)}\) \(\to\) MgCl\(_{2(aq)}\) + CO\(_{2(g)}\) + H\(_2\)\(_{(l)}\).
(iii) State whether reaction rate was fastest at the beginning, the middle or towards the end of the reaction. Give reason for our answer.
(c) Consider the following reaction at equilibrium: PCI\(_{5(g)}\) \(\rightleftharpoons\) PCI\(_{3(g)}\)); \(\Delta\)H = +95 kJmol\(^{-}\)
(i) Write an expression for the equilibrium constant K.
(ii) Predict the effect of the following on the equilibrium position.
I. Increased pressure
II. Increased temperature
III. Removal of chlorine Sketch an energy profile diagram for the forward reaction.
(a)(i) Graham's law of diffusion.
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its relative molecular mass).
\[ R \propto \frac{1}{\sqrt{\rho}} \quad\text{or}\quad \frac{R_1}{R_2} = \sqrt{\frac{\rho_2}{\rho_1}} = \sqrt{\frac{M_2}{M_1}} \]
(a)(ii) Vapour density of triatomic gas X (relative atomic mass = 16).
A triatomic molecule contains 3 atoms, so its relative molecular mass is:
\[ M_X = 3 \times 16 = 48 \]
Vapour density is half the relative molecular mass:
\[ \text{V.D.} = \frac{M_X}{2} = \frac{48}{2} = 24 \]
Vapour density = 24.
(a)(iii) Greater average velocity of Y or Z.
Equal volumes of Y and Z at the same temperature and pressure contain equal numbers of molecules (Avogadro's law). At the same temperature the molecules have equal average kinetic energy:
\[ \tfrac{1}{2} m_Y v_Y^2 = \tfrac{1}{2} m_Z v_Z^2 \]
Given \( m_Y = 2 m_Z \):
\[ 2 m_Z v_Y^2 = m_Z v_Z^2 \implies v_Z^2 = 2 v_Y^2 \implies v_Z = \sqrt{2}\,v_Y \approx 1.41\,v_Y \]
Z has the greater average velocity because it is the lighter molecule; for the same kinetic energy the lighter molecule must move faster (velocity is inversely proportional to the square root of the molecular mass).
(b) Reaction curve for \( \text{MgCO}_{3(s)} + 2\text{HCl}_{(aq)} \to \text{MgCl}_{2(aq)} + \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)} \)
Reading the graph: the y-axis is Loss in mass (g) of flask + contents (marked 0.5, 1.0, 1.5) and the x-axis is Reaction time (mins.) (marked 5, 10, 15, 20, 25). The line rises steadily from the origin and then levels off (becomes horizontal) at a loss in mass of about 0.9 g, reaching this plateau at about 12 minutes.
(b)(i) For how long did reaction occur?
The reaction occurred for about 12 minutes. This is the time from the start until the curve becomes flat (constant mass), showing that no more gas was being lost.
(b)(ii) Why was there a loss in mass?
Carbon dioxide gas, \( \text{CO}_2 \), is produced. Because the vessel is open, the \( \text{CO}_2 \) escapes into the air, so the total mass of the flask and its contents decreases.
(b)(iii) When was the reaction rate fastest?
The rate was fastest at the beginning. The slope (gradient) of the curve is steepest at the start, meaning mass was lost most rapidly then. This is because the concentrations of the reactants (\( \text{HCl} \)) and the amount of solid \( \text{MgCO}_3 \) were greatest at the start, giving the highest frequency of effective collisions. As reactants are used up the slope decreases, and at the plateau the rate is zero.
(b)(iv) Three conditions that can affect the slope of the curve.
- Temperature of the reaction mixture.
- Concentration of the hydrochloric acid.
- Surface area (particle size) of the magnesium carbonate.
(A catalyst would also affect the slope.)
(c) \( \text{PCl}_{5(g)} \rightleftharpoons \text{PCl}_{3(g)} + \text{Cl}_{2(g)} \); \( \Delta H = +95\ \text{kJ mol}^{-1} \)
(c)(i) Equilibrium constant expression.
\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]
(c)(ii) Effect on the equilibrium position (Le Chatelier's principle).
- I. Increased pressure: the forward direction produces more gas molecules (1 mole \( \to \) 2 moles). Increasing pressure shifts the equilibrium to the side with fewer moles, i.e. backward (to the left), favouring \( \text{PCl}_5 \).
- II. Increased temperature: the forward reaction is endothermic (\( \Delta H = +95\ \text{kJ mol}^{-1} \)). Raising the temperature favours the endothermic direction, so the equilibrium shifts forward (to the right), producing more \( \text{PCl}_3 \) and \( \text{Cl}_2 \).
- III. Removal of chlorine: removing a product causes the system to replace it, so the equilibrium shifts forward (to the right) to make more \( \text{Cl}_2 \).
Energy profile diagram for the forward reaction (endothermic):
The reactant \( \text{PCl}_5 \) starts at a lower energy. The curve rises to a peak (the activated complex) and then falls to the products \( \text{PCl}_3 + \text{Cl}_2 \), which sit at a higher energy than the reactant. \( E_a \) is the height from reactant to peak; \( \Delta H = +95\ \text{kJ mol}^{-1} \) is the height from reactant level up to product level.
Energy
^
| ,-. (activated complex)
| / \
| / \______ PCl3 + Cl2 (products, higher)
| / } dH = +95 kJ/mol
|________/ _ _ _ _ _ _ _ _ _ _ _ _ _
| PCl5 (reactant, lower)
|___________________________________> Reaction progress
