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(D) The given reaction is: $4NH_{3(g)} + 5O_{2(g)} ightleftharpoons 4NO_{(g)} + 6H_2O_{(g)} + 	ext{Heat}$.This is an exothermic reaction because he

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Decreasing the temperature

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(D) The given reaction is: $4NH_{3(g)} + 5O_{2(g)} ightleftharpoons 4NO_{(g)} + 6H_2O_{(g)} + 	ext{Heat}$.This is an exothermic reaction because heat is released (product side).According to Le Chatelier's principle,for an exothermic reaction,decreasing the temperature will shift the equilibrium to the right to produce more heat.Regarding pressure,the change in moles of gas is $\Delta n_g = (4 + 6) - (4 + 5) = 10 - 9 = 1$.Since $\Delta n_g > 0$,increasing the pressure will shift the equilibrium to the left (towards fewer moles).Adding a catalyst only increases the rate of both forward and backward reactions equally and does not shift the equilibrium position.Therefore,decreasing the temperature is the correct condition to shift the equilibrium to the right.

Consider the following reaction and determine which of the conditions will shift the equilibrium position to the right?
$4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_2O_{(g)} + \text{Heat}$

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Consider the following reaction and determine which of the conditions will shift the equilibrium position to the right?$4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_2O_{(g)} + \text{Heat}$