Chemistry
posted by Thomas on .
A mixture of .10 mol of NO, .050 mol of H2 and .10 mol of H20 is
placed in a 1 liter vessel at 300 K. The following equilibrium is
established.
2 NO + 2 H2 <> N2 + 2 H20
At equilibrium (N0) = .062 M. What are the equilibrium concentrations
of H2 N2 and H20

2 NO + 2 H2 <> N2 + 2 H20
initial:
NO = .1
H2 = 0.05
N2 = not given
H2O = 0.1
If NO at equilibrium is 0.062 and it was 0.1 to start, it must have changed by 0.038.
You don't know the N2 at equlibrium because the initial amount is not listed but it is the initial amount + 0.019.
Since the coefficient of H2 is 2 and that of NO is 2, H2O must have decreased by 0.038 to leave 0.050.038 = ??
N2 has coefficient of 1; therefore, it has increased by 1/2 NO or H2; thus, it has increased + 0.019.
Same reasoning increase H2O = +0.038 to make equilibrium 0.1 + 0.038 = ??