Explain the effect of each of the following stresses on the position of the foll
ID: 957945 • Letter: E
Question
Explain the effect of each of the following stresses on the position of the following equilibrium: 3 NO(g) N2O(g) + NO2(g) The reaction as written is exothermic.
(a) N2O(g) is added to the equilibrium mixture without change of volume or temperature.
(b) The volume of the equilibrium mixture is reduced at constant temperature.
(c) The equilibrium mixture is cooled.
(d) Gaseous argon (which does not react) is added to the equilibrium mixture while both the total gas pressure and the temperature are kept constant.
(e) Gaseous argon is added to the equilibrium mixture without changing the volume.
Explanation / Answer
The effect of change in pressure, temperature, volume, and composition of species on the reaction rate of chemical transformation in equilibrium is well defined using Le Chatelier’s principle stated as,
Le Chatelier’s principle: “If system in equilibrium is subjected to a change of temperature, pressure, or concentration then the equilibrium shift in such way that the effect of the changed condition is nullified”.
Given chemical transformation,
3 NO(g) < ------ > N2O(g) + NO2(g).
3 moles of reactant react to form 2 moles of the product. i.e. reaction proceeds with decrease in volume of the system.
(a)If N2O(g) i.e. product is added externally to the system in equilibrium. i.e. constrain imposed on system is increase in volume of product hence according to Le Chatelier’s principle system will nullify this effect on equilibrium by decreasing concentration of products i.e. by increasing concentration of reactant i.e. Backward reaction will be favored.
(b) The system already proceeds with decrease in volume and if we reduce the volume at constant temperature system will try to increase it. Increase in volume is possible only when more concentration of reactant present in the system and hence again Backward reaction will be favored if we reduce volume of equilibrium mixture at constant temperature.
(c) Lowering of temperature (cooling) favors exothermic reaction in forward direction and endothermic reaction in backward direction.
(d) Gaseous Ar do not react with any of species in given transformation. Addition of inert Ar gas keeping total gas pressure and temperature constant will shift equilibrium in such a direction in which there is increase in number of moles of gas. In given transformation backward reaction is accompanied with increase in moles hence backward reaction will be favored under this situation.
(d) Addition of an inert gas without volume change i.e. at constant volume will have no effect on equilibrium.
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