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 :
\[Ba{{O}_{2}}(s)BaO(s)+{{O}_{2}}(g);\Delta H=+ve\] \[Rat{{e}_{1}}={{k}_{1}}[Ba{{O}_{2}}]\] (forward reaction) \[Rat{{e}_{1}}={{k}_{1}}\] \[(\because \,[Ba{{O}_{2}}]=1)\] Similarly (for backward reaction) \[{{r}_{2}}={{k}_{2}}[BaO]\,\,.\,\,[{{O}_{2}}]\] \[{{r}_{2}}={{k}_{2}}[{{O}_{2}}]\] At equilibrium, \[{{r}_{1}}={{r}_{2}}\] \[{{k}_{1}}={{k}_{2}}[{{O}_{2}}]\] \[\frac{{{k}_{1}}}{{{k}_{2}}}={{p}_{{{o}_{2}}}}\] or \[K={{p}_{{{o}_{2}}}}\] 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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