Question

In a reversible reaction, carbon monoxide gas and hydrogen gas are mixed to produce methanol gas (CH3OH). The enthalpy change of this reaction is AH = 90 kJ mol A mixture of 0.360 moles of CO and 0.640 moles of H2 were allowed to reach equilibrium at 150°C in a 10 dm' sealed container. At equilibrium the mixture contained 0.120 moles of CH,OH. (a) Calculate the K for this reaction (to 1.d.p). (10 marks) Similarly to Ko, K is an equilibrium constant which is dependent on partial pressures exerted by the gaseous components. (b) Calculate K, for this reaction (to 2.d.p. R=0.0821 dm atm k mol). (4 marks) (c) Using Le Chatelier's principle explain the effect on this reaction of: i. increasing the pressure and keeping the temperature constant il increasing the temperature and keeping the pressure constant. (6 marks) Question 2 (20 marks) (a) The atomic radius varies in each of the following list of elements, identify the largest atom in each list and explain your answer referring to the electronic configurations of sodium, aluminium and rubidium. i. Sodium; Magnesium: Aluminium: ii. Sodium; Potassium; Rubidium (5 marks) (b) There are several principles which determine how electrons are configured around the nucleus of an atom. Choose one element from sodium, magnesium or aluminium to explain how the principles help you with the detailed electron configuration. (5 marks) (c) Copy and complete the following table using your Periodic Table (2 marks) Number of Element Name Element Number of Symbol electrons Mg12 Number of protons S 16 16 oxygen 160 (d) Using Lewis diagram to show the formation of each compound, and explain how the structure of the compound influence their physical properties, including electrical conductivity and boiling point. I the compound that would be formed in a reaction between Mg and o ii. the compound that would be formed in a reaction between S and O (8 marks)

Answer 1

Aven deversible reaction is s. - CO(g) + 2 H2(g) - CH₂OH (2) ; AH = -90 kJ mel- Initially; no. of males of conco) - 0.360 no. of moles of M2 (M/H2) = 0.640 Temperature - 150°C ~ 423 K. - Volume of containes = lo dime or lo litre At eqm; moles of CH₂OH(CH3OH) = 0.12 a) Kc = ? CO(g) + Qua(g) = CH20H () Initially 0.360 0.640 At eum 0.360-x 0.640-2x x later CH₂ou are giren). (x0=0.120 (giren)) Ke=kn ll and kn =/ mchaoh (V7 (Total volume) = lee (giren) (nco) (MHz)? Ang; change in no. of meles; 1 (1+2) 1-33-2) Kc = r a I 1-2 10.360-x)(0.64-22) ( 10 ) (x = 0:120 (given)) Kc = [ 0.120 1 (0.360-0.120) (0.64. – 260.120))?/ Kc 3 3.125 (100) = 312.5

U23K * ? Ko'a ke (Royong Kp = (312:5) [10.082) (423)] -2 (True - (312-5) = 0.26 Enso ((0.0821)(423)2 - () Because OH is negative (gires) je exothermic reaction - This at constant pressure ; if temperature is increased, = then the equilibrium will shift in backward directions - (Because according to Le Chatler principle, exothermic reactions ase favourable at low temperature) (1) At constant T; if pressure is increased then the - equilibrium will shifts in Howard the direction - (According to le chatlier principle high pressure favours those reactions which are accompained by decrease in total no, of moles..) o dos con + 9) kc = 312.5 b) Kp = 0.26 c) (i) Reaction or equilibrium shifts in forward - direction (it) Equilibrium shifts in backward direction

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