ogen and oxygen react to form nitrous oxide as: 2 N2 (g) + O2 (g) ⇌ 2 N2O (g) Of a mixture of 0.482 mol N2 and 0.933 mol O2 is placed in a reaction vessel of 10.0 L and allowed to from N2O at a temperature for which Kc = 2.0 X 10-37, what will be the composition of the equilibrium mixture?
The given equation is:
2 N2 (g) + O2 (g) ⇌ 2 N2O (g)
Therefore the equilibrium constant Kc, will be given as:
Kc = [N2O] 2 / [N2] 2 [O2]
Initial molar concentrations are:
[N2] = 0.482 mol / 10.0 L = 0.0482 mol L-1
[O2] = 0.933 mol / 10.0 L = 0.0933 mol L-1
Let us assume that x mol of oxygen has reacted with 2x mol of N2 to form 2x mol of N2O, so that equilibrium concentrations are:
[N2] = 0.0482 – 2x
[O2] = 0.0933 – x
[N2O] = 2x
Therefore,
Kc = (2x) 2 / (0.0482 – 2x) 2 (0.0933 – x)
Since, Kc is very small (2.0 X 10-37), we can assume x to be infinitesimally small so that,
0.0482 – 2x ≈ 0.0482 and 0.0933 – x ≈ 0.0933
Kc = 4 x2 / (0.0482) 2 (0.0933)
i.e.
4x2 = (0.0482) 2 (0.0933) X (2.0 X 10-37)
4 x2 = 4.34 X 10-412 = 4.34 X 10-41-41-212] = 0.0482 – 2x ≈ 0.0482
[O2] = 0.0933 – x ≈ 0.0933
[N2O] = 2x = 2 X 3.3 X 10-21-21
To determine the composition of the equilibrium mixture, we first need to calculate the initial moles of each reactant and then use the given equilibrium constant (Kc) to determine the concentration of each species at equilibrium.
Given:
Initial moles of N2 = 0.482 mol
Initial moles of O2 = 0.933 mol
Volume of the reaction vessel = 10.0 L
Equilibrium constant (Kc) = 2.0 x 10^(-37)
Step 1: Calculate the initial concentrations of each species.
Since the volume of the reaction vessel is given, we can convert the initial moles to concentrations using the formula: concentration = moles/volume.
Initial concentration of N2 = 0.482 mol/10.0 L = 0.0482 M
Initial concentration of O2 = 0.933 mol/10.0 L = 0.0933 M
Step 2: Use the equilibrium constant (Kc) expression to calculate the concentrations of each species at equilibrium.
The equilibrium expression for the reaction is:
Kc = [N2O]^2/([N2]^2[O2])
Given that Kc = 2.0 x 10^(-37), we can write the equation as:
2.0 x 10^(-37) = [N2O]^2/([0.0482]^2[0.0933])
Step 3: Solve for [N2O] using the equilibrium constant.
To find the concentration of N2O at equilibrium, rearrange the equation to solve for [N2O]:
[N2O]^2 = 2.0 x 10^(-37) * [0.0482]^2 * [0.0933]
[N2O] = √(2.0 x 10^(-37) * [0.0482]^2 * [0.0933])
Calculating this expression will give you the concentration of N2O at equilibrium.
Step 4: Calculate the concentration of N2 and O2 at equilibrium.
Using the equilibrium concentration of N2O, you can substitute it back into the equilibrium constant expression to solve for the concentrations of N2 and O2 at equilibrium.
[N2] = [N2O]^2/[O2]
[O2] = [N2O]^2/([N2]^2)
Substituting the values obtained in Step 3 into these equations will give you the concentrations of N2 and O2 at equilibrium.
By following these steps, you can determine the composition of the equilibrium mixture in terms of the concentrations of N2, O2, and N2O.