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On the basis of the Le Chatelier principle, explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction:

N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g) \;\Delta H = -92.38 \; kJ mol^{-1}

What will be the effect of the addition of argon to the above reaction mixture at constant volume?

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As per the information given in the question, N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g) \;\Delta H = -92.38 \; kJ mol^{-1}

Now, we know that as ΔH is negative, it means that the forward reaction is exothermic in nature.

As per the Le Chatelier principle, when the temperature is reduced the reaction will move in forward direction, hence more yield of the product will be obtained.

We know that when the temperature is increased the reaction will be moving in the backward direction that the yield of product will be reduced.

As per the Le Chatelier principle, when the pressure is increased the equilibrium shifts in the direction where there are lesser numbers of gas molecules. Therefore, when the pressure is increased, the equilibrium will be shifting in the forward direction and consequently the yield of the product will increase.

Thus, number of moles of reactants = 1 + 3 = 4

Number of moles of product = 2

Thus, we can conclude that high pressure and low temperature are the desirable conditions for increasing the yield of the products. Addition of an an inert gas at constant volume does not result in a shift. This is due to the fact that the addition of a non-reactive gas does not cause any change in the partial pressures of the other gases in the container.

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