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.
Explanation
(a)(i) Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its density.
Rate a \(\frac{1}{\sqrt{d}}\) or \(\frac{1}{\sqrt{m}}\) where m is the molecular mass and d is the density.
(ii) 1 molecule = X\(_3\). Relative molecular mass of X = 16 x 3 = 48
Vapour density = 1/2 x Relative molecular mass. Vapour density of x = \(\frac{48}{2}\) = 24
(iii) If equal volumes of Y and Z are at the same temperature and pressure and the mass of Y is twice that of 7, molecules of Z will have the greater average velocity. Since the temperature is the same, molecules of Y and Z have the same average kinetic energy. Hence the smaller molecule will have the greater velocity.
(b) Given tha rate curve for the following reaction: MgCO\(_3\) + 2HCI -> MgCl\(_2\) + CO\(_2\) + H\(_2\)O.
(i) The reaction occurred for 12 minutes
(ii) There was a loss in mass because one of the products was a gay (CO\(_2\)) which escaped to the atmosphere.
(iii) The rate was fastest at the beginning of the reaction. The slope of the curve was highest or steepest at the point.
(iv) The slope of the curve can be affected by change in the concentration of HCI, change in temperature, change in particle size/surface area of the MgCO\(_3\)
(c) PCl\(_5\) -> PCl\(_3\) + Cl\(_2\)
\(\Delta\)H = +95 KJmol\(^{-1}\) K\(_{eq}\) = \(\frac{[PCl_3][Cl_2]}{[PCl_5]}\)
or K\(_p\) = \(\frac{P^{PCl}_3P^{Cl}_2}{P^{Cl}_5}\)
(i) Increased pressure will shift the equilibrium position to the left
(ii) increased temperature will shift the equilibrium position to the right
(iii) Removal of chlorine will shift the equilibrium position to the right.