Equilibrium - Result Question 29
####30. For the reaction
$2 BaO_2(s) \rightarrow 2 BaO(s)+O_2(g)$;
$\Delta H=+ve$. In equilibrium condition, pressure of
$O_2$ is dependent on
[2002]
(a) mass of $BaO_2$
(b) mass of $BaO$
(c) temperature of equilibrium
(d) mass of $BaO_2$ and $BaO$ both
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Solution:
- (c) For the reaction $BaO_2(s) \rightarrow BaO(s)+O_2(g) ; \Delta H=+ve$.
At equilibrium $K_p=P _{O_2}$
[For solid and liquids concentration term is taken as unity]
Hence, the value of equilibrium constant depends only upon partial pressure of $O_2$. Further on increasing temperature, formation of $O_2$ increases as this is an endothermic reaction.
since the steps 1 and 2 are exothermic hence low temprature will favour both the reactions. In step -1 , moles are increasing hence low pressure will favour it. In step 2, moles are decreasing, hence high pressure will favour it.
