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 0.05-0.038 = ??
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 = ??
To find the equilibrium concentrations of H2, N2, and H2O, we can use the information given about the initial amounts and the equilibrium amount of NO.
Given:
Initial amount of NO = 0.10 mol
Initial amount of H2 = 0.050 mol
Initial amount of H2O = 0.10 mol
Equilibrium concentration of NO = 0.062 M
First, let's assume the change in the amount of NO is x mol. This means that the number of moles of H2, N2, and H2O formed will also be x mol since the stoichiometric coefficient for NO, H2, N2, and H2O in the balanced equation is 2.
The equilibrium amount of NO can be calculated as follows:
Equilibrium amount of NO = Initial amount of NO - x mol
0.062 = 0.10 - x
Solving the above equation, we get:
x = 0.10 - 0.062
x = 0.038 mol
So, at equilibrium, the amount of H2, N2, and H2O will be:
Amount of H2 = Initial amount of H2 + x mol
Amount of H2 = 0.050 + 0.038
Amount of H2 = 0.088 mol
Amount of N2 = x mol
Amount of N2 = 0.038 mol
Amount of H2O = Initial amount of H2O + x mol
Amount of H2O = 0.10 + 0.038
Amount of H2O = 0.138 mol
Now, we can calculate the equilibrium concentrations by dividing the amounts by the volume (1 liter) of the vessel:
Equilibrium concentration of H2 = Amount of H2 / Volume
Equilibrium concentration of H2 = 0.088 mol / 1 L
Equilibrium concentration of H2 = 0.088 M
Equilibrium concentration of N2 = Amount of N2 / Volume
Equilibrium concentration of N2 = 0.038 mol / 1 L
Equilibrium concentration of N2 = 0.038 M
Equilibrium concentration of H2O = Amount of H2O / Volume
Equilibrium concentration of H2O = 0.138 mol / 1 L
Equilibrium concentration of H2O = 0.138 M
So, the equilibrium concentrations are:
[H2] = 0.088 M
[N2] = 0.038 M
[H2O] = 0.138 M
To determine the equilibrium concentrations of H2, N2, and H2O, we can use the balanced equation and the given equilibrium concentration of NO.
The balanced equation for the reaction is:
2 NO + 2 H2 ⇌ N2 + 2 H2O
Let's denote the equilibrium concentrations of H2, N2, and H2O as [H2], [N2], and [H2O], respectively.
According to the stoichiometry of the balanced equation, the change in concentration of NO is -2x (as 2 moles of NO are consumed for every mole of N2 produced).
Since the initial concentration of NO is 0.10 mol and the equilibrium concentration is 0.062 M, the change in concentration of NO is:
-2x = 0.10 M - 0.062 M
-2x = 0.038 M
Thus, the equilibrium concentration of NO is 0.062 M.
Now, let's use the stoichiometry of the balanced equation to find the equilibrium concentrations of H2, N2, and H2O.
For every mole of N2 produced, 2 moles of H2 are also consumed, and 2 moles of H2O are produced.
Therefore, the change in concentration of H2 is -2x, the change in concentration of N2 is +x, and the change in concentration of H2O is +2x.
The equilibrium concentrations can be expressed as follows:
[H2] = initial concentration of H2 - 2x
[N2] = x
[H2O] = initial concentration of H2O + 2x
Given that the initial concentration of H2 is 0.050 mol and the initial concentration of H2O is 0.10 mol, we can substitute these values into the expressions above:
[H2] = 0.050 M - 2x
[N2] = x
[H2O] = 0.10 M + 2x
To find the value of x, we need to use the equilibrium expression for the reaction, which is given by the ratio of the product concentrations to the reactant concentrations, each raised to their respective stoichiometric coefficients:
Kc = ([N2] * [H2O]^2) / ([NO]^2 * [H2]^2)
Substituting the known equilibrium concentrations, we can solve for x:
Kc = ([N2] * [H2O]^2) / ([NO]^2 * [H2]^2)
Kc = (x * (0.10 + 2x)^2) / (0.062^2 * (0.050 - 2x)^2)
The value of Kc for this reaction would need to be given in order to solve for x. Once Kc is known, solving the equation above will yield the value of x, which can then be used to calculate the equilibrium concentrations of H2, N2, and H2O using the expressions we obtained earlier.
Without the value of Kc, we cannot determine the exact equilibrium concentrations of H2, N2, and H2O in this particular case.