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 :
\[Ba{{O}_{2}}(s)\rightleftharpoons BaO(s)+{{O}_{2}}(g);\Delta H=+ve\] According to the 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 \[{{k}_{1}}={{k}_{2}}.{{p}_{{{O}_{2}}}}\] 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 the concentration of reactants. The given equation is an endothenmic reaction. It 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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