• # question_answer $2{{N}_{2}}O(g)+{{O}_{2}}(g)\rightleftharpoons 4NO(g);\Delta H>0$ What will be the effect on equilibrium when (i) Volume of the vessel increases? (ii) Temperature decreases?

(i) For the given reaction,                                                            $K=\frac{{{[NO]}^{4}}}{{{[{{N}_{2}}O]}^{2}}[{{O}_{2}}]}$ When volume of the vessel increases, number of moles per unit volume (i.e. molar concentration) of each reactant and product decreases. As there are 4 concentration terms in the numerator but 3 concentration terms in the denominator, to keep K constant, the decrease in $[{{N}_{2}}O]$ and $[{{O}_{2}}]$ should be more, i.e., equilibrium will shift in the forward direction. Alternatively, increase in volume of the vessel means decrease in pressure. As forward reaction is accompanied by increase in the number of moles (i.e. increase of pressure), decrease in pressure will favour forward reaction (according to Le Chatelier's principle). (ii) As $\Delta H$is +ve, i.e., reaction is endothermic, decrease of temperature will favour the direction in which heat is absorbed, i.e., backward direction.