A) increased mass of \[Ba{{O}_{2}}\]
B) increased mass of\[BaO\]
C) increased temperature of equilibrium.
D) increased mass of\[Ba{{O}_{2}}\]and\[BaO\]both
Correct Answer: C
Solution :
\[Ba{{O}_{2}}(s)BaO(s)+{{O}_{2}}(g).\Delta H=+ve\] According to law of mass action. The rate of forward reaction\[={{R}_{1}}\] \[{{R}_{1}}\propto [Ba{{O}_{2}}]\] or \[{{R}_{1}}={{k}_{1}}[Ba{{O}_{2}}]\] But concentration of solid\[=1\]then,\[{{R}_{1}}={{k}_{1}}\] Similarly the rate of backward reaction\[={{R}_{2}}\] \[{{R}_{2}}\propto [BaO][{{O}_{2}}]\] or \[{{R}_{2}}={{k}_{2}}[BaO][{{O}_{2}}]\] \[\because \]Conc. of\[[BaO]=1\] or \[{{R}_{2}}={{k}_{2}}[{{O}_{2}}]\] At equilibrium, \[{{R}_{1}}={{R}_{2}}\] \[{{k}_{1}}={{k}_{2}}[{{O}_{2}}]\]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 \[Ba{{O}_{2}}\] would increase; and more \[{{O}_{2}}\] is produced.
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