` 2 N_(2) O (g) + O_(2) (g) hArr 4 No (g) , Delta H gt 0 `
What will be the effect on equilibrium when
(i) Volume of the vessel increases ? (ii) Temperature decreases ?

Explanation

(i) For the given reaction, `K= [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 productdecreases. 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 , increases in volume of the vessel means decrease in pressure. As forward reaction is accmpained by increase in the number of moes ( 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 absorted , i.e., backward direction.