Answer:
(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.
You need to login to perform this action.
You will be redirected in
3 sec