Answer:
The moles of dinitrogen tetroxide after equilibrium is reached the second time is 18.78 moles.
Explanation:
Initially 35.0 mol
Eq'm 8.0 mol 13.5 mol
Moles of nitrogen dioxide = 35.0 mol
Moles of nitrogen dioxide at equilibrium = 8.0 mol
According to reaction, 2 moles nitrogen oxide gives 1 mol of dinitrogen tetraoxide. Then 27 mol of nitrogen dioxde will give:
Moles of dinitrogen tetroxide = 13.5 mol
Concentration of nitrogen dioxide at equilibrium,
Concentration of dinitrogen tetradioxide at equilibrium,
Equilibrium constant of reaction:
Now, adds another 12.0 moles of nitrogen dioxide.
Initially 35.0 mol
Eq'm 8.0 mol 13.5 mol
After adding 12 moles of nitrogen dioxide the moles
(8.0 mol+12 mol)
Again after attaining equilibrium second time:
(20.0 -x)mol (13.5+x/2) mol
Concentration of nitrogen dioxide at second equilibrium,
Concentration of dinitrogen tetradioxide at second equilibrium,
on solving for x:
we get x = 10.56
Moles of dinitrogen tetroxide after equilibrium is reached the second time: