WAEC Chemistry Question and Solution 4

waec chemistry question and solution

WAEC Chemistry Question and Solution 4

(a) State Le Chatelier’s principle

(b) Use Le Chatelier’s principle to deduce the conditions that favour a high yield of ammonia in the Haber process

(C) Give the chemical test for ammonia.

(d) State what would be observed when aqueous ammonia solution is added to:

(i) zinc chloride solution,

(ii) copper (II) tetraoxosulphate (V) solution

(e) Explain why the H – N – H bond angle in ammonia is less than that of H – C – H in methane (t) Give two uses of ammonia.


Question from no 11 SSCE November 1989 chemistry 2 theory examination



(a) Le Chatelier’s principles states that a system at equilibrium will adjust in such a way as to annul any alteration about any factors affecting the equilibrium like temperature. This means that any action that changes the temperature, pressure, or concentrations of reactants in a system at equilibrium would stimulate a response that will to some extent offsets the change whereas a new equilibrium condition will be recognized.


That’s why, Le Châtelier’s principle states that any alteration to a system at equilibrium will adjust to pay off for that change. It is significant to know that Le Châtelier’s principle is only an essential guide to recognize what occurs when the conditions are altered in any chemical reaction that is in dynamic equilibrium; it does not offer reasons for the changes at the molecular level like timescale of change and underlying reaction mechanism.


b) Le Chatelier’s principle states that if the system is altered in a way that increases the concentration of one of the reacting species, it must favor the reaction in which that species is consumed. In other words, if there is an increase in products, the reaction quotient, , is increased, making it greater than the equilibrium constant, K. Consider an equilibrium reached between four substances, ABC and D:

A + 2B ⇌ C + D

When this is applied to haber process with the equation

N2 (g)+3H2 (g)⇌2NH3 (g)(ΔH=-92.4 kJ/mol)

You need to shift the position of the equilibrium as far as possible to the right in order to produce the maximum possible amount of ammonia in the equilibrium mixture. The forward reaction that results in the production of ammonia is exothermic.


According to Le Chatelier’s Principle, high yield of ammonia will be favored if the temperature is lowered. The system will respond by moving the position of equilibrium to offset this – in other words by increasing the temperature. To get as much ammonia as possible in the equilibrium mixture, you would require a very low temperature.


On the other hand, the lower the temperature the slower the rate of the reaction. A manufacturer is hoping to produce as much ammonia as possible in a day. It is senseless to try to reach an equilibrium mixture which contains a very high proportion of ammonia if it takes a number of years for the reaction to arrive at that equilibrium. The gases need to reach equilibrium within the very short possible. In the haber process a compromise temperature of 400 – 450°C is reached to favour a reasonably high yield of ammonia in the equilibrium mixture (even if it is only 15%), but in a very short time.


Pressure considerations

N2 (g) + 3H3 (g) ⇌ 2NH3 (g) ΔH=−92kJ mol-1

In the above equation, there are 4 molecules on the left-hand side of the equation, but only 2 on the right. According to Le Chatelier’s Principle, if you increase the pressure the system will respond by favoring the reaction which produces fewer molecules. That will lead to a fall in the pressure again.


In order to obtain as much ammonia as possible in the equilibrium mixture, you require a very high pressure as possible. Increase in pressure, therefore, favours high yield of ammonia due to the fact that the reaction flows in the direction of the less number of moles. Increase in the number of moles of the reactants will as well favour the production of ammonia.


(c) The chemical test for ammonia is that ammonia form white fumes of ammonium chloride, when it comes in contact with the rod dipped into hydrogen chloride solution.


(d) (i) When aqueous ammonia solution is added to a clear solution of zinc chloride, white gelatinous precipitate which turns soluble in excess of the ammonia solution is formed.


(ii) When ammonia solution is added to a clear solution of copper(Il) tetraoxosulphate (VI) solution, it forms a precipitate which is soluble and forming a deep blue solution.


(e) The H-N-H bond angle in ammonia is less than the H-C-H bond angle in methane because the electron cloud formed by the unshared pair in NH bond disperses over a greater volume than that of the three pairs connected to hydrogen atoms.


This would likely force the bonding pairs closer to one another and in that way reducing the bond angle making it less than the bond angle in the tetrahedral methane.


(f)Two uses of ammonia are:

i) Ammonia can be used in the production of trioxonitrate (V) acid.

ii) Ammonia solution is made use of in laundry work where it helps to remove temporary hardness by precipitating the Calcium ion as CaCO.

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