For the chemical reaction: BaO_{2(s)} rightle | vedclass

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(C) The given reaction is $BaO_{2(s)} \rightleftharpoons BaO_{(s)} + \frac{1}{2} O_{2(g)}$.Since $BaO_2$ and $BaO$ are in the solid state,their active

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Increasing the temperature

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(C) The given reaction is $BaO_{2(s)} \rightleftharpoons BaO_{(s)} + \frac{1}{2} O_{2(g)}$.Since $BaO_2$ and $BaO$ are in the solid state,their active masses are taken as unity $(1)$.The equilibrium constant expression is $K_p = P_{O_2}^{1/2}$.This shows that the equilibrium pressure of $O_2$ is independent of the amounts of solid reactants or products present.However,since the reaction is endothermic $(\Delta H > 0)$,according to Le Chatelier's principle,an increase in temperature will shift the equilibrium in the forward direction to absorb the heat,thereby increasing the equilibrium pressure of $O_2$.

For the chemical reaction: $BaO_{2(s)} \rightleftharpoons BaO_{(s)} + \frac{1}{2} O_{2(g)}$,$\Delta H = +ve$. At equilibrium,the pressure of $O_2$ depends upon:

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