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 :
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According to law or mass action. |
The rate of forward reaction ![]() |
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or ![]() |
But concentration of solid = 1 |
then, ![]() |
Similarly the rate of backward reaction ![]() |
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or ![]() |
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or ![]() |
At equilibrium |
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where ![]() ![]() |
or ![]() |
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So, from the above it is clear that pressure of ![]() ![]() ![]() |
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