(a)(i) State Graham's law of diffusion.
State what would happen to the vapour density of \(\mathrm{N_2O_4}\) as the temperature of the system is increased. If the system is cooled, would the gases become lighter or darker in colour? Explain your answer in each case.
(i) an inflated balloon that was left in the sun. burst after some time;
(ii) a pure sample of a liquid did not have a constant boiling point at the top and at the base of a high mountain
(ii) Write one equation each to illustrate the reducing property of the gases you listed in (c)(i) above.
(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 molar mass).
\[ r \propto \frac{1}{\sqrt{\rho}} \qquad\text{or}\qquad \frac{r_1}{r_2} = \sqrt{\frac{\rho_2}{\rho_1}} = \sqrt{\frac{M_2}{M_1}} \]
(a)(ii) The equilibrium.
The diagram shows a reversible reaction, heat driving it forward and cooling driving it back:
\[ \text{N}_2\text{O}_{4(g)} \;\underset{\text{cool}}{\overset{\text{heat}}{\rightleftharpoons}}\; 2\text{NO}_{2(g)} \]
(pale yellow \(\rightleftharpoons\) dark brown)
Effect on vapour density as temperature rises: The forward dissociation is endothermic, so raising the temperature shifts the position of equilibrium to the right. One mole of N2O4 (molar mass 92) breaks into two moles of NO2 (molar mass 46 each), so the number of gas particles increases while the total mass is unchanged. The average molar mass of the mixture therefore falls. Since vapour density \( = \dfrac{M}{2} \), the vapour density of N2O4 decreases as the temperature is increased.
Effect of cooling on colour: On cooling, the exothermic reverse reaction is favoured, so the equilibrium shifts back towards N2O4. The dark brown NO2 is converted into pale (almost colourless) N2O4, so the gases become lighter in colour.
(b)(i) Balloon bursting in the sun. Heat from the sun raises the temperature of the trapped air. The gas molecules gain kinetic energy and move faster, striking the walls more often and more forcefully; the gas also tends to expand. At fixed volume this raises the internal pressure. When the pressure exceeds the elastic limit of the rubber, the balloon bursts.
(b)(ii) No constant boiling point up a mountain. A liquid boils when its saturated vapour pressure equals the external atmospheric pressure. Atmospheric pressure falls with altitude, so at the top of the high mountain the pressure is lower and the liquid boils at a lower temperature, whereas at the base the pressure is higher and the boiling point is higher. The boiling point of a pure liquid is therefore fixed only for a fixed pressure, not for a changing one.
(c)(i) Two gaseous reducing agents: hydrogen (H2) and carbon monoxide (CO).
(c)(ii) Equations showing the reducing property:
Hydrogen reduces copper(II) oxide to copper:
\[ \text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O} \]
Carbon monoxide reduces copper(II) oxide (or iron(III) oxide) to the metal:
\[ \text{CuO} + \text{CO} \rightarrow \text{Cu} + \text{CO}_2 \]
\(2\mathrm{NO_{2(g)}}\)