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
See below.
To determine the equilibrium concentrations of H2, N2, and H2O, we can use the concept of stoichiometry and the given information.
First, let's calculate the initial moles of each substance:
Initial moles of NO = 0.10 mol
Initial moles of H2 = 0.050 mol
Initial moles of H2O = 0.10 mol
Since the equation involves a 2:2 ratio for NO and H2, let's divide the initial moles of NO and H2 by 2:
Initial moles of NO = 0.10 mol / 2 = 0.05 mol
Initial moles of H2 = 0.050 mol / 2 = 0.025 mol
Using the stoichiometry of the balanced equation, we see that for every 2 moles of NO and H2 reacted, 1 mole of N2 and 2 moles of H2O are formed.
So, using the given equilibrium concentration of NO (0.062 M), we can work backward to find the equilibrium concentrations of H2, N2, and H2O.
From the equation, we can see that 2 moles of NO react to form 1 mole of N2. Therefore, if the concentration of NO is 0.062 M, the concentration of N2 is half of that:
[N2] = 0.062 M / 2 = 0.031 M
Since the ratio of H2 to NO is also 2:2, the concentration of H2 is the same as NO:
[H2] = 0.062 M
Finally, for every 2 moles of NO reacted, 2 moles of H2O are formed. Therefore, the concentration of H2O is twice the concentration of NO:
[H2O] = 2 * [NO] = 2 * 0.062 M = 0.124 M
Therefore, at equilibrium:
[H2] = 0.062 M
[N2] = 0.031 M
[H2O] = 0.124 M