consider the following equilibrium.
4NH3(g) +3O2(g)<-> 2N2(g) + 6H2O(g)
suppose 0.30mol of NH3 and 0.40mol of oxygen are added to a 5.0L container. If x mol of water is present at equilibrium, what is the equilibrium concentration of oxygen?
What's K?
K is not given.
To find the equilibrium concentration of oxygen, we need to calculate the concentrations of the other substances (NH3 and water) at equilibrium first.
Let's start by writing the balanced equation and constructing an ICE (Initial, Change, Equilibrium) table:
4NH3(g) + 3O2(g) <-> 2N2(g) + 6H2O(g)
Initial:
NH3: 0.30 mol
O2: 0.40 mol
N2: 0 mol (not given)
H2O: x mol (unknown)
Change:
NH3: -4x (since the stoichiometric coefficient of NH3 is 4)
O2: -3x (since the stoichiometric coefficient of O2 is 3)
N2: +2x (since the stoichiometric coefficient of N2 is 2)
H2O: +6x (since the stoichiometric coefficient of H2O is 6)
Equilibrium:
NH3: 0.30 - 4x
O2: 0.40 - 3x
N2: 2x
H2O: x + 6x = 7x
The total volume of the container is 5.0 L. Therefore, the equilibrium concentrations can be calculated using the formula:
concentration = moles/volume
NH3 concentration: (0.30 - 4x) / 5.0
O2 concentration: (0.40 - 3x) / 5.0
N2 concentration: (2x) / 5.0
H2O concentration: (7x) / 5.0
Now, we need to set up an expression for the equilibrium constant (Kc) using the balanced equation:
Kc = [(N2)^2 * (H2O)^6] / [(NH3)^4 * (O2)^3]
Plugging in the concentrations from above:
Kc = [(2x/5.0)^2 * (7x/5.0)^6] / [(0.30 - 4x/5.0)^4 * (0.40 - 3x/5.0)^3]
To find the equilibrium concentration of oxygen, we need to solve for x and then substitute it back into the expression we derived for O2 concentration:
O2 concentration = (0.40 - 3x) / 5.0
Solving for x involves setting up an expression for Kc and solving it. This requires a quadratic equation, which can be a bit complex to solve step-by-step. If you'd like, I can provide the value of x and the corresponding O2 concentration using a numerical method or approximation.
To find the equilibrium concentration of oxygen, we need to solve for the value of x in the equation.
Step 1: Write the balanced equation and identify the stoichiometric coefficients.
4NH3(g) + 3O2(g) ↔ 2N2(g) + 6H2O(g)
Step 2: Determine the initial moles of NH3 and O2.
NH3: 0.30 mol
O2: 0.40 mol
Step 3: Determine the total moles at equilibrium.
The number of moles of water, H2O, formed is equal to 6 times the number of moles of oxygen, O2, consumed.
Moles of H2O = 6x
Moles of O2 at equilibrium = 0.40 - (3/4) * 6x (using the stoichiometric coefficients)
Step 4: Use the ideal gas law to calculate the equilibrium concentration.
PV = nRT
P is the pressure (assumed constant)
V is the volume (5.0 L)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature (assumed constant)
The concentration of oxygen, [O2], can be calculated as:
[O2] = (moles of O2 at equilibrium) / (volume of the container)
Therefore:
[O2] = (0.40 - (3/4) * 6x) / 5.0
This equation gives us the equilibrium concentration of oxygen in terms of x. We need more information to solve for x or to find the numerical value of [O2].