K=1.6x10^-5 mol/L for the following reaction
2NOCl(g).... 2 NO(g) + Cl2(g)
Calculate the concentrations of all species at equilibrium for each of the following original mixtures
E) 2.4 mol of NOCl, 2.4 mol of NO, and 1.2 mol Cl2 in a 1.0 L flask
F)1.9 mol/L concentration of all three gases
In order to try and get this spacing right, I will write the equation from top to bottom instead of horizontally.
1st are the equatios. Next column is starting molarityies (mols/L). Third column is change. Fourth column is final concentration. This is just an ICE chart.
2NOCl(g) 2.4 +2x 2.4+2x
|
|
v
2NO(g) 2.4 -2x 2.4-2x
+
Cl2(g) 1.2 -x 1.2-x
You know the reaction will shift to the left BECAUSE Q = (NOCl)^2/(NO)^2*(Cl2) = (2.4)^2/(2.4)^2*(1.2) = 1.2 and that is larger than Keq listed in the problem which means Q is too large and it will sift to the left to reach equilibrium.
The columns indicate changes taking place.
Now substitute all of that (the last column) into Keq expression and solve for x.
Keq = 1.6 x 10^-5 = (NOCl)^2/(NO)^2*(Cl2)^2.
post your work if you get stuck.
To calculate the concentrations of all species at equilibrium for each of the given mixtures, we'll use the equation of the reaction and the given value of the equilibrium constant, K = 1.6 x 10^-5 mol/L.
E) For the mixture with 2.4 mol of NOCl, 2.4 mol of NO, and 1.2 mol of Cl2 in a 1.0 L flask, we can set up an initial moles table:
Species | Initial Moles
---------------------------
NOCl | 2.4 mol
NO | 2.4 mol
Cl2 | 1.2 mol
Total | 2.4 + 2.4 + 1.2 = 6.0 mol
Because there are no products initially, the initial concentrations of NO and Cl2 are zero, as they haven't been formed yet. The concentration of NOCl can be found by dividing the moles by the volume,
[NOCl] = 2.4 mol/1.0 L = 2.4 mol/L
To calculate the concentrations at equilibrium, we need to determine the change in moles for each species based on their stoichiometric coefficients. Let's assume that x moles of NOCl react to form x moles of NO and x moles of Cl2.
Species | Initial Moles | Change in Moles | Equilibrium Moles
-----------------------------------------------------------------------
NOCl | 2.4 mol | -x | 2.4 - x
NO | 2.4 mol | +x | 2.4 + x
Cl2 | 1.2 mol | +x | 1.2 + x
The total moles at equilibrium should remain constant:
(2.4 - x) + (2.4 + x) + (1.2 + x) = 6.0 mol
6.0 + x = 6.0
x = 0
Since x = 0, it means that no NOCl reacts, and the equilibrium concentrations are the same as the initial concentrations:
[NOCl] = 2.4 mol/L
[NO] = 2.4 mol/L
[Cl2] = 1.2 mol/L
F) For the mixture with a concentration of 1.9 mol/L for each gas, we can again use an initial moles table:
Species | Initial Moles
---------------------------
NOCl | 1.9 mol/L x 1.0 L = 1.9 mol
NO | 1.9 mol/L x 1.0 L = 1.9 mol
Cl2 | 1.9 mol/L x 1.0 L = 1.9 mol
Total | 1.9 + 1.9 + 1.9 = 5.7 mol
Using the same approach, we can calculate the equilibrium concentrations by assuming x moles of NOCl react to form x moles of NO and x moles of Cl2:
Species | Initial Moles | Change in Moles | Equilibrium Moles
-----------------------------------------------------------------------
NOCl | 1.9 mol | -x | 1.9 - x
NO | 1.9 mol | +x | 1.9 + x
Cl2 | 1.9 mol | +x | 1.9 + x
Setting up the equilibrium equation:
(1.9 - x) + (1.9 + x) + (1.9 + x) = 5.7
5.7 + 2x = 5.7
2x = 0
x = 0
Since x = 0, it means that no NOCl reacts, and the equilibrium concentrations are the same as the initial concentrations:
[NOCl] = 1.9 mol/L
[NO] = 1.9 mol/L
[Cl2] = 1.9 mol/L
Therefore, for mixture E, the equilibrium concentrations are:
[NOCl] = 2.4 mol/L
[NO] = 2.4 mol/L
[Cl2] = 1.2 mol/L
And for mixture F, the equilibrium concentrations are:
[NOCl] = 1.9 mol/L
[NO] = 1.9 mol/L
[Cl2] = 1.9 mol/L
To calculate the concentrations of all species at equilibrium for each of the given mixtures, we'll use the concept of the reaction's equilibrium constant (K) and the stoichiometry of the balanced equation.
First, let's write the balanced equation for the reaction:
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
Now, let's analyze the given mixtures.
E) 2.4 mol of NOCl, 2.4 mol of NO, and 1.2 mol Cl2 in a 1.0 L flask:
From the balanced equation, we can see that 2 moles of NOCl react to form 2 moles of NO and 1 mole of Cl2. Thus, the initial moles of NOCl, NO, and Cl2 in the mixture are as follows:
NOCl: 2.4 mol
NO: 2.4 mol
Cl2: 1.2 mol
Since the volume of the flask is given as 1.0 L, the initial concentrations of NOCl, NO, and Cl2 can be calculated as:
[NOCl] = 2.4 mol / 1.0 L = 2.4 M (mol/L)
[NO] = 2.4 mol / 1.0 L = 2.4 M (mol/L)
[Cl2] = 1.2 mol / 1.0 L = 1.2 M (mol/L)
Now, to calculate the equilibrium concentrations, we'll use the equilibrium constant expression:
K = [NO]^2 * [Cl2] / [NOCl]^2
Since the stoichiometry of the balanced equation is 2:2:1 for NOCl:NO:Cl2, if the reactants and products have different concentrations at equilibrium, it means that the reaction did not go to completion and reached a dynamic equilibrium.
To solve for the equilibrium concentrations, we'll use an ICE table:
Let x be the change in concentration for NO and Cl2, and 2x be the change in concentration for NOCl.
ICE Table:
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
Initial: 2.4 M 2.4 M 1.2 M
Change: -2x +2x +x
Equilibrium: 2.4 M - 2x 2.4 M + 2x 1.2 M + x
Using the equilibrium concentrations in the equilibrium constant expression, we can write:
K = (2.4 M + 2x)^2 * (1.2 M + x) / (2.4 M - 2x)^2
Now, substitute the value of K (1.6 x 10^-5 mol/L) and solve the equation for x. The resulting x will be the change in concentration for each species at equilibrium.
F) 1.9 mol/L concentration of all three gases:
For this mixture, the initial concentrations of NOCl, NO, and Cl2 are given as:
[NOCl] = 1.9 M (mol/L)
[NO] = 1.9 M (mol/L)
[Cl2] = 1.9 M (mol/L)
Since the initial concentrations are the same for all species, we can directly use these values as the equilibrium concentrations.
Therefore, at equilibrium, the concentrations of NOCl, NO, and Cl2 will all be 1.9 M (mol/L).