At 25°C, K = 0.090 for the following reaction:H2O(g) + Cl2O(g) 2 HOCl(g)
Calculate the concentrations of all species at equilibrium for each of the following cases.
(a) 1.2 g H2O and 1.6 g Cl2O are mixed in a 1.4 L flask.
(b) 1.2 mol pure HOCl is placed in a 1.9 L flask
To solve this problem, we will use the given equilibrium constant (K) and the information about the initial amounts of reactants.
(a) 1.2 g H2O and 1.6 g Cl2O are mixed in a 1.4 L flask.
First, let's calculate the number of moles of H2O and Cl2O:
Molar mass of H2O = 2 g/mol
Molar mass of Cl2O = 86.9 g/mol
Number of moles of H2O = 1.2 g / 2 g/mol = 0.6 mol
Number of moles of Cl2O = 1.6 g / 86.9 g/mol = 0.0184 mol
Now we have the initial number of moles for each species. Let's assume the change in moles for H2O and Cl2O is x, and the change in moles for HOCl is 2x (according to the balanced chemical equation).
Using the stoichiometry of the balanced chemical equation, the equilibrium concentrations can be written as follows:
[H2O] = (0.6 - x) mol/1.4 L
[Cl2O] = (0.0184 - x) mol/1.4 L
[HOCl] = (2x) mol/1.4 L
Since K = [HOCl]^2 / ([H2O] [Cl2O]), we can set up the equation:
K = (2x)^2 / ((0.6 - x)(0.0184 - x))
Now rearranging the equation and solving for x:
0.090 = (4x^2) / ((0.6 - x)(0.0184 - x))
Simplify the equation:
0.090(0.6 - x)(0.0184 - x) = 4x^2
Solve for x using the quadratic formula:
4x^2 - 0.1104x + 0.018504 = 0
The solution to this quadratic equation is x = 0.056 mol.
Substituting the value of x back into the expressions for the equilibrium concentrations:
[H2O] = (0.6 - 0.056) mol/1.4 L = 0.388 M
[Cl2O] = (0.0184 - 0.056) mol/1.4 L = 0.0119 M
[HOCl] = (2 * 0.056) mol/1.4 L = 0.080 M
Therefore, the equilibrium concentrations of H2O, Cl2O, and HOCl are 0.388 M, 0.0119 M, and 0.080 M, respectively.
(b) 1.2 mol pure HOCl is placed in a 1.9 L flask.
Since we have a pure sample of HOCl, the initial concentration of HOCl can be calculated directly.
[HOCl] = number of moles / volume
= 1.2 mol / 1.9 L
≈ 0.632 M
The equilibrium concentrations of H2O and Cl2O can be assumed to be zero since they were not initially present.
Therefore, the equilibrium concentrations of H2O, Cl2O, and HOCl are approximately 0 M, 0 M, and 0.632 M, respectively.
To calculate the concentrations of all species at equilibrium for each case, we need to use the equilibrium constant (K) and the given information about the initial amounts of the reactants or products.
Let's start solving each case step by step:
(a) 1.2 g H2O and 1.6 g Cl2O are mixed in a 1.4 L flask.
1. First, we need to calculate the moles of each substance using their molar masses:
Molar mass of H2O = 18.015 g/mol
Molar mass of Cl2O = 86.898 g/mol
moles of H2O = 1.2 g / 18.015 g/mol = 0.0666 mol
moles of Cl2O = 1.6 g / 86.898 g/mol = 0.0184 mol
2. Next, we need to calculate the initial concentrations of each substance by dividing their moles by the volume in liters:
Initial concentration of H2O = 0.0666 mol / 1.4 L = 0.0476 M (rounded to 4 decimal places)
Initial concentration of Cl2O = 0.0184 mol / 1.4 L = 0.0131 M (rounded to 4 decimal places)
3. Using the given equilibrium constant (K = 0.090), we can set up the equilibrium expression:
K = [HOCl]^2 / [H2O][Cl2O]
We want to find the concentrations at equilibrium, so let's assume that x M of HOCl has formed. This means the change in concentration for HOCl will be +2x, and the changes for H2O and Cl2O will be -x.
4. Substitute these values into the equilibrium expression and solve for x:
0.090 = (2x)^2 / (0.0476 - x)(0.0131 - x)
Using this quadratic equation, solve for x. Once you have the value of x, you can determine the equilibrium concentrations by adding x to the initial concentrations of H2O and Cl2O, and subtracting x from the initial concentrations of H2O and Cl2O.
(b) 1.2 mol of pure HOCl is placed in a 1.9 L flask.
In this case, we already have the initial concentration of HOCl:
Initial concentration of HOCl = 1.2 mol / 1.9 L = 0.6316 M (rounded to 4 decimal places)
Since only HOCl is given initially, the initial concentrations of H2O and Cl2O will be zero.
From here, you can follow the same steps as in case (a) to calculate the equilibrium concentrations using the given equilibrium constant.
Remember to check your assumptions and ensure that you are using the appropriate units and conversions throughout the calculations.
You REALLY should put in arrows. How are we to know the products from the reactants? BUT, as I chemist I can GUESS but that's the best I can do.
H2O + Cl2O ==> 2HOCl
Write the Keq expression.
Change 1.2 g H2O and 1.6 g Cl2O to moles and change that to concn using mols/L and the 1.4L flask. Then do an ICE chart, plug into Keq and solve for each.
Same for (b) BUT instead of going to the right to establish equilibrium the reaction must go to the left to establish equilibrium.