At 3745°C, K = 0.093 for the following reaction.
N2(g) + O2(g) equilibrium reaction arrow 2 NO(g)
Calculate the concentrations of all species at equilibrium for each of the following cases.
(a) 2.0 g N2 and 3.0 g O2 are mixed in a 1.3-L flask.
[N2]
[O2]
[NO]
(b) 1.5 mol pure NO is placed in a 1.8-L flask.
[N2]
[O2]
[NO]
mols N2 = 2.0/28 = about 0.07
(N2) = 0.07/1.3L = about 0.05
mols O2 = 3.0/32 = about 0.09
(O2) = a0.09/1.L = about 0.07
All of the above need to be recalculated more accurately.
.......N2 + O2 ==> 2NO
I....0.05..0.07......0
C.....-x...-x.......2x
E..0.05-x..0.07-x....2x
Substitute the E line into the Keq expression and solve for x, then evaluate the N2 and O2.
The second part is done the same way but you start with NO as 1.5/1.8 = ? and N2 and O2 are zero. The E line will be x for N2, x for O2, and ?-2x for NO.
Post your work if you get stuck.
To calculate the concentrations of all species at equilibrium for each case, we need to use the concept of the equilibrium constant (K) and the stoichiometry of the reaction.
Given information:
- K = 0.093 at 3745°C
- Reaction: N2(g) + O2(g) ⇌ 2NO(g)
(a) 2.0 g N2 and 3.0 g O2 in a 1.3-L flask:
To calculate the concentrations, we need to convert the given masses of N2 and O2 to moles, and then divide by the total volume of the flask to get the concentrations.
1. Calculate the moles of N2:
Molar mass of N2 = 2 * atomic mass of N = 2 * 14.01 g/mol = 28.02 g/mol
moles of N2 = mass / molar mass = 2.0 g / 28.02 g/mol
2. Calculate the moles of O2:
Molar mass of O2 = 2 * atomic mass of O = 2 * 16.00 g/mol = 32.00 g/mol
moles of O2 = mass / molar mass = 3.0 g / 32.00 g/mol
3. Calculate the concentrations:
[ N2 ] = moles of N2 / volume of the flask
[ O2 ] = moles of O2 / volume of the flask
[ NO ] = 0 (since no NO is present initially, as stated in the question)
(b) 1.5 mol pure NO in a 1.8-L flask:
In this case, we already have the moles of NO. Since the stoichiometry of the reaction is 1:1 for N2 and O2, and 2:1 for NO, we can use the mole ratio to calculate the concentrations.
1. Calculate the concentrations:
[ N2 ] = [ O2 ] = 0 (since no N2 or O2 is present initially, as stated in the question)
[ NO ] = moles of NO / volume of the flask
Remember to always divide the moles by the total volume of the flask to obtain the concentrations.
Now that we've explained the method, let's do the calculations:
(a) 2.0 g N2 and 3.0 g O2 in a 1.3-L flask:
1. moles of N2 = 2.0 g / 28.02 g/mol ≈ 0.0714 mol
2. moles of O2 = 3.0 g / 32.00 g/mol ≈ 0.0938 mol
3. [ N2 ] = 0.0714 mol / 1.3 L ≈ 0.0550 M
[ O2 ] = 0.0938 mol / 1.3 L ≈ 0.0721 M
[ NO ] = 0 M (since no NO is present initially)
(b) 1.5 mol pure NO in a 1.8-L flask:
1. [ N2 ] = [ O2 ] = 0 (since no N2 or O2 is present initially)
2. [ NO ] = 1.5 mol / 1.8 L ≈ 0.833 M
Therefore, the concentrations at equilibrium for each case are as follows:
(a) 2.0 g N2 and 3.0 g O2 in a 1.3-L flask:
[ N2 ] ≈ 0.0550 M
[ O2 ] ≈ 0.0721 M
[ NO ] = 0 M
(b) 1.5 mol pure NO in a 1.8-L flask:
[ N2 ] = [ O2 ] = 0 M
[ NO ] ≈ 0.833 M
Please note that the calculation assumes the given temperature is valid for the equilibrium constant (K) you provided.