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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 ?

Answer» <html><body><p></p>Solution :(i) For the given reaction, `K= [NO]^(4)/ ([N_(2)O]^(2) [O_(2)])` <br/> When volume of the vessel increases, number of moles <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> <a href="https://interviewquestions.tuteehub.com/tag/unit-1438166" style="font-weight:bold;" target="_blank" title="Click to know more about UNIT">UNIT</a> volume (i.e. molar concentration) of eachreactant 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 <a href="https://interviewquestions.tuteehub.com/tag/shift-1205367" style="font-weight:bold;" target="_blank" title="Click to know more about SHIFT">SHIFT</a> in the forward direction. <br/> Alternatively ,increases in volume of the vessel <a href="https://interviewquestions.tuteehub.com/tag/means-1091780" style="font-weight:bold;" target="_blank" title="Click to know more about MEANS">MEANS</a> decrease in pressure. As forward reaction is accmpained by <a href="https://interviewquestions.tuteehub.com/tag/increase-1040383" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASE">INCREASE</a> in the number of moes ( i.e. increase of pressure ) , decrease in pressure will favour forward reaction ( according to Le Chatelier's principle ). <br/> (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.</body></html>


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