Reaction \[Ba{{O}_{2}}(s)~BaO(s)+{{O}_{2}}(g),~\] |
\[\Delta H=+ve\]. In equilibrium condition, Pressure of \[{{O}_{2}}\] depends on: [AIPMT 2002] |
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 |
or |
But concentration of solid = 1 |
then, |
Similarly the rate of backward reaction |
or |
Conc. Of |
or |
At equilibrium |
or |
where Partial pressure of |
or (Equilibrium constant) |
or |
So, from the above it is clear that pressure of 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 would increase; and more is produced. |
You need to login to perform this action.
You will be redirected in
3 sec