Answer:
The concentration of N2 at the equilibrium will be 0.019 M
Explanation:
Step 1: Data given
Number of moles of NO = 0.10 mol
Number of moles of H2 = 0.050 mol
Number of moles of H2O = 0.10 mol
Volume = 1.0 L
Temperature = 300K
At equilibrium [NO]=0.062M
Step 2: The balanced equation
2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
Step 3: Calculate the initial concentration
Concentration = Moles / volume
[NO] = 0.10 mol / 1L = 0.10 M
[H2] = 0.050 mol / 1L = 0.050 M
[H2O] = 0.10 mol / 1L = 0.10 M
[N2] = 0 M
Step 4: Calculate the concentration at the equilibrium
[NO] at the equilibrium is 0.062 M
This means there reacted 0.038 mol (0.038M) of NO
For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O
This means there will also react 0.038 mol of H2
The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M
There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M
There will be produced 0.038/2 = 0.019 moles of N2
The concentration of N2 at the equilibrium will be 0.019 M
