Question

Reaction $BaO _{2}( s ) \rightleftharpoons BaO ( s )+ O _{2}(g)$ . In equilibrium condition, pressure of $O_{2}$ depends on

• A. increased mass of $BaO_{2}$
• B. increased mass of $BaO$
• C. increased temperature of equilibrium.
• D. increased mass of $BaO_{2}$ and $BaO$ both

Answer 1

Answer: (C)

increased temperature of equilibrium.

Answer 2

$BaO _{2}(s) \rightleftharpoons BaO (s)+ O _{2}(g) . \Delta H=+ ve$ According to law of mass action. The rate of forward reaction $=R_{1}$ $R_{1} \propto\left[ BaO _{2}\right]$ or $R_{1}=k_{1}\left[ BaO _{2}\right]$ But concentration of solid $=1$ then, $R_{1}=k_{1}$ Similarly the rate of backward reaction $=R_{2}$ $R_{2} \propto[ BaO ]\left[ O _{2}\right]$ or $R_{2}=k_{2}[ BaO ]\left[ O _{2}\right]$ $\because$ Conc. of $[ BaO ]=1$ or $R_{2}=k_{2}\left[ O _{2}\right]$ At equilibrium, $R_{1}=R_{2}$ $k_{1}=k_{2}\left[ O _{2}\right]$ or $k_{1}=k_{2} \cdot p_{ O _{2}}$ where $p_{ O _{2}}=$ Partial pressure of $O _{2}$ or $\frac{k_{1}}{k_{2}}=p_{ O _{2}}$ (Equilibrium constant) $\because \frac{k_{1}}{k_{2}}=k$ or $k=p_{ O _{2}}$ So, from the above it is clear that pressure of $O _{2}$ does not depend upon conc. of reactants. The given equation is an endothermic reaction. If the temperature of such reaction is increased then dissociation of $BaO _{2}$ would increase; and more $O _{2}$ is produced.