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 gas...
Question 1 Report
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\(_2\)\(_{(l)}\).
(i) For how long did reaction occur?
(ii) Why was there a loss in mass?
(iii) State whether reaction rate was fastest at the beginning, the middle or towards the end of the reaction. Give reason for our answer.
(iv) List three reaction conditions that can affect the slope of the curve
(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.
Answer Details
(a)(i) Graham's law of diffusion states that, at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or relative molecular mass.
\[r \propto \frac{1}{\sqrt{d}}\qquad \text{or}\qquad r \propto \frac{1}{\sqrt{M}}\]
(ii) Since X is triatomic, its formula is \(X_3\).
\[M_r(X_3)=3\times16=48\]
\[\text{Vapour density}=\frac{M_r}{2}=\frac{48}{2}=24\]
Vapour density of X = 24.
(iii) Molecules of Z have the greater average velocity. At the same temperature, molecules of Y and Z have the same average kinetic energy. Since a molecule of Y has twice the mass of a molecule of Z, the lighter Z molecule moves faster.
\[\frac{1}{2}m_Yv_Y^2=\frac{1}{2}m_Zv_Z^2\]
Since \(m_Y=2m_Z\),
\[v_Z=\sqrt{2}\,v_Y\]
(b)(i) The reaction occurred for 12 minutes, after which the curve became horizontal.
(ii) Carbon(IV) oxide, \(\mathrm{CO_2}\), was evolved and escaped from the open vessel. This caused the mass of the flask and its contents to decrease.
(iii) The reaction rate was fastest at the beginning of the reaction because the curve had its greatest gradient at the beginning. The acid concentration was then highest, so effective collisions occurred most frequently.
(iv) Conditions that affect the slope of the curve include:
- concentration of hydrochloric acid;
- temperature of the reaction mixture;
- surface area or particle size of magnesium carbonate.
(c) \[\mathrm{PCl_{5(g)}\rightleftharpoons PCl_{3(g)}+Cl_{2(g)}}\qquad \Delta H=+95\ \mathrm{kJ\,mol^{-1}}\]
(i)
\[K_c=\frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}\]
(ii)
- Increased pressure: equilibrium shifts to the left, favouring \(\mathrm{PCl_5}\), because there is one mole of gas on the left but two moles of gas on the right.
- Increased temperature: equilibrium shifts to the right, because the forward reaction is endothermic.
- Removal of chlorine: equilibrium shifts to the right to replace the removed \(\mathrm{Cl_2}\).
Energy profile diagram for the forward reaction
The products are at a higher energy level than the reactant; hence \(\Delta H\) is positive and the forward reaction is endothermic.
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