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 :
According to law or mass action. The rate of forward reaction \[={{R}_{1}}\] But concentration of solid = 1 then, \[{{R}_{1}}={{k}_{1}}\] Similarly the rate of backward reaction \[={{R}_{2}}\] or At equilibrium \[{{R}_{1}}={{R}_{2}}\] \[{{k}_{1}}={{k}_{2}}[{{O}_{2}}]\,\,or\,{{k}_{1}}={{k}_{2}}\]. where Partial pressure of \[{{O}_{2}}\] So, from the above it is clear that pressure of \[{{O}_{2}}\] does not depend upon Cone, 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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