(a)(i) State three postulates of the kinetic theory of gases.
(ii) Draw a sketch to illustrate Boyle’s law.
(iii) Explain what is meant by “absolute zero of temperature”.
(b) Equal volumes of a gas X and oxygen are at the same pressure but temperature of 200 K and 800K respectively.
(i) Giving reason for your answer, state which of the gas samples contains the greater number of molecules.
(ii) If the mass of one molecule of the gas X is 2.19 x 10(^{-22}) g, determine the molar mass of X. [Avogadro constant = 6.02 x 10(^{-23})]
(iii) Hence, state with reason which of X and oxygen will diffuse faster under the same conditions. [O= 16].
(c)(i) Mention two reaction conditions that can increase the yield of ammonia in the reaction represented by the following equation: N(_{2(g)}) + 3H(_{2(9)}) (rightleftharpoons) 2NH(_{3(g)}) (Dellta)H = -46kJmol(^{-1}). Draw a labelled energy profile diagram for an endothermic reaction.
(iii) A solid W decomposes on heating according to the following equation: W(_{(s)} to Y_{(s}) + Z_{(g)}) List two factors apart from temperature, which can affect the rate of the reaction.
Explanation
(a)(i) Postulates of the kinetic theory of gases: Gas molecules move randomly in straight lines. They collide with one another and with the walls of the container. The temperature of the gas is a measure of the average kinetic energy of the molecule. The forces of attraction between gas molecules are negligible. The volume occupied by gas molecules is negligible compared to the volume of the container. The collision of the molecules is perfectly elastic (any three).
(ii) Sketches to illustrate Boyle’s Law
(iii) Absolute zero of temperature is the temperature at which the volume of a gas would be (theoretically) reduced to zero. OR It is the temperature at which all motion is assumed to have ceased. OR It is the lowest temperature theoretically possible OR It is temperature at which all gases liquefy or solidify OR the entropy of the system is at rest.
(b) If equal volumes of gas X and oxygen are at the same pressure but at 200 K and 800 K respectively:
(i) X contains the greater number of molecules; Reason: Since X is at a lower temperature, it must contain more molecules to be able to exert the same pressure Alternatively n = (frac{PV}{RT})
n(_x) = (frac{PV}{200R})
nO(_2) = (frac{PV}{800R})
Hence, n(_x) is greater
(ii) To calculate molar mass of X if one molecule of it has a mass of 2.19 x 10(^{-22})g. Calculation: 1 mole of x contains 6.02 x 10(^{23}) molecule.
1 mole of x will have a mass of (2.19 x 10(^{-22}) x 6.02 x 10(^{23}))g = 131.84 g mo1(^{-1}) ; approximately 132g mol(^{-1}).
(iii) Comparison of rate of diffusion of X with that of oxygen: O(_2) will diffuse faster. The lighter the gas, the faster the rate of diffusion. The molar mass of O(_2) is 32g. Hence it is lighter than X and will diffuse faster.
(c)(i) Increasing the pressure or increasing the concentration of reactions of the left hand side.
-lowering the temperature of the system.
-Continuous removal of NH(_3) as it is produced. (any two)
(ii) Energy profile diagram for an endothermic reaction.
(iii) in the reaction represented by the following equation W_(s)) —> Energy Yv + Z(_(s)). Reaction rate can be affected by the following factors apart from temperature.
– Surface area / particle size
– Presence of catalyst.