(a) Explain in terms of the kinetic theory why a tyre should not be overinflated. (b)The following results were obtained at room temperature in an experimen...
Question 1 Report
(a) Explain in terms of the kinetic theory why a tyre should not be overinflated.
(b)The following results were obtained at room temperature in an experiment to verify one of the gas laws using a glass syringe:
| Pressure (P) of air in syringe (atm) | Volume (V) of air in syringe (\(cm^3\)) | \(\frac{I}{V}\) |
| 0.100 | 10.00 | 0.100 |
| 0.125 | 8.00 | 0.125 |
| 0.150 | 6.60 | 0.150 |
| 0.175 | 5.60 | 0.179 |
| 0.200 | 4.80 | 0.208 |
| 0.225 | 4.40 | 0.227 |
(i) Plot a graph of P against \(\frac{1}{v}\), using 1 cm to represent 0.01 atm on the vertical axis and 1cm to represent 0.02 unit on the horizontal axis.
(ii) Which of the gas laws is in agreement with the results?
(c) The flow chart below represents the stages involved in the manufacture of H\(_2\)SO\(_4\).
| S + O\(_2\) \(\to\) SO\(_2\) | SO\(_2\) +x \(\to\) SO\(_3\) | SO\(_3\) +Conc. H\(_2\)SO\(_4\) \(\to\) Y | Y +H\(_2\)O \(\to\) Conc H\(_2\)SO\(_4\) |
| stage I | stage II | stage III | stage IV |
(i) Name the process represented by the chart.
(ii) Identify reactant X and product Y.
(iii) What are the operating temperature and pressure at stage II?
(iv) Mention the stage which requires a catalyst and state the catalyst used.
(v) Give the reason why the SO\(_3\) produced in stage II is not dissolved directly in water to form the acid
(d) When K\(_4\)Cr\(_2\)C\(_7\) dissolves in water, the following equilibrium is established:
\(\mathrm{Cr_2O_{7(aq)}^{2-} + H_2O_{(l)} \to 2CrO_{4(aq)}^{2-} + 2H_{aq}}\)
(i) State the colour observed on adding a few drops of dilute H\(_2\)SO\(_4\) to the system.
(ii) Explain your answer in (d)(1).
(iii) What principle is applicable to this explanation?
Answer Details
(a) Why a tyre should not be overinflated (kinetic theory)
The air inside a tyre consists of molecules in rapid, random motion that continually collide with the inner walls; these collisions produce the pressure. Overinflating forces more molecules into the fixed volume, so the collisions become more frequent and the pressure rises. If the tyre becomes hot (from friction with the road or from sunshine), the molecules move even faster and strike the walls harder and more often, raising the pressure further. The pressure may then exceed what the tyre can withstand and it bursts.
(b) Verifying a gas law
The third column of the table is \( \dfrac{1}{V} \); it was obtained from the volume, for example \( \dfrac{1}{10.00} = 0.100 \) and \( \dfrac{1}{8.00} = 0.125 \).
| P / atm | V / cm\(^3\) | 1/V / cm\(^{-3}\) |
|---|---|---|
| 0.100 | 10.00 | 0.100 |
| 0.125 | 8.00 | 0.125 |
| 0.150 | 6.60 | 0.150 |
| 0.175 | 5.60 | 0.179 |
| 0.200 | 4.80 | 0.208 |
| 0.225 | 4.40 | 0.227 |
(b)(i) The graph of \( P \) (vertical axis) against \( \dfrac{1}{V} \) (horizontal axis) is a straight line through the origin.
(b)(ii) Since \( P \) is directly proportional to \( \dfrac{1}{V} \), the results agree with Boyle's law (at constant temperature, \( PV = \text{constant} \)).
(c) Manufacture of \( H_2SO_4 \)
(i) The process is the Contact process.
(ii) Reactant \( X = \) oxygen (air), \( O_2 \); product \( Y = \) oleum (fuming sulphuric acid), \( H_2S_2O_7 \).
(iii) At stage II \( (2SO_2 + O_2 \rightleftharpoons 2SO_3) \) the operating temperature is about \( 450^{\circ}C \) and the pressure is about 1 to 2 atmospheres.
(iv) Stage II requires a catalyst; the catalyst is vanadium(V) oxide, \( V_2O_5 \).
(v) The \( SO_3 \) is not dissolved directly in water because the reaction is very exothermic and produces a dense, choking mist (fog) of fine sulphuric acid droplets that is difficult to condense; instead \( SO_3 \) is absorbed in concentrated \( H_2SO_4 \) to give oleum.
(d) Equilibrium: \( Cr_2O_{7(aq)}^{2-} \) (orange) \( + H_2O_{(l)} \rightleftharpoons 2CrO_{4(aq)}^{2-} \) (yellow) \( + 2H^+_{(aq)} \).
(i) On adding dilute \( H_2SO_4 \), the colour observed is orange.
(ii) The acid increases the concentration of \( H^+ \) ions; the equilibrium shifts to the left to remove the added \( H^+ \), forming more of the orange dichromate ion \( Cr_2O_7^{2-} \).
(iii) The principle applied is Le Chatelier's principle.
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