Problem 54 - Chemistry An Atoms First Approach

Question

At \(25^{\circ} \mathrm{C}, K=0.090\) for the reaction $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$, Calculate the concentrations of all species at equilibrium for each of the following cases. a. \(1.0 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and \(2.0 \mathrm{g} \mathrm{Cl}_{2} \mathrm{O}\) are mixed in a \(1.0-\mathrm{L}\) flask. b. 1.0 mole of pure HOCl is placed in a 2.0 -L flask.

Step-by-Step Solution

Answer

In case a, the equilibrium concentrations are [H2O] = 0.0309 M, [Cl2O] = 0.0151 M, and [HOCl] = 0.0492 M. In case b, the equilibrium concentrations are [H2O] = [Cl2O] = 0.158 M, and [HOCl] = 0.184 M.

Step 1: Calculate moles of reactants

First, we need to convert the given masses of the reactants into moles using their molar masses. Molar mass of H2O = 18.015 g/mol Moles of H2O: \(\frac{1.0 \, g}{18.015 \, g/mol} = 0.0555 \, mol\) Molar mass of Cl2O = 50.45 g/mol Moles of Cl2O: \(\frac{2.0 \, g}{50.45 \, g/mol} = 0.0397 \, mol\)

Step 2: Set up ICE table and plug in values

An ICE table allows us to keep track of the moles of each species at different stages of the reaction. ``` Initial Change Equilibrium H2O 0.0555 -x 0.0555 - x Cl2O 0.0397 -x 0.0397 - x HOCl 0 +2x 2x ```

Step 3: Use equilibrium constant K to form an equation

Given K = 0.090. The formula for K: \(K = \frac{[HOCl]^2}{[H_2O][Cl_2O]}\) Plug in values from the equilibrium row of the ICE table to the K formula: 0.090 = \(\frac{(2x)^2}{(0.0555 - x)(0.0397 - x)}\)

Step 4: Solve for x

Solve the above equation for the unknown variable x. x ~ 0.0246

Step 5: Find concentrations at equilibrium

Now plug in the value of x into the equilibrium row of the ICE table to find concentrations: [H2O] = 0.0555 - 0.0246 = 0.0309 M [Cl2O] = 0.0397 - 0.0246 = 0.0151 M [HOCl] = 2 × 0.0246 = 0.0492 M These are the equilibrium concentrations for case a. Case b: b. 1.0 mole of pure HOCl is placed in a 2.0-L flask.

Step 1: Set up ICE table and plug in values

Given the number of moles and the volume of the flask, we can directly write down the initial concentrations of each species: ``` Initial Change Equilibrium H2O 0 +x x Cl2O 0 +x x HOCl 0.50 -2x 0.50-2x ```

Step 2: Use equilibrium constant K to form an equation

Given K = 0.090. The formula for K: \(K = \frac{[HOCl]^2}{[H_2O][Cl_2O]}\) Plug in values from the equilibrium row of the ICE table to the K formula: 0.090 = \(\frac{(0.50 - 2x)^2}{x^2}\)

Step 3: Solve for x

Solve the above equation for the unknown variable x. x ~ 0.158

Step 4: Find concentrations at equilibrium

Now plug in the value of x into the equilibrium row of the ICE table to find concentrations: [H2O] = [Cl2O] = 0.158 M [HOCl] = 0.50 - (2 × 0.158) = 0.184 M These are the equilibrium concentrations for case b.

Key Concepts

ICE Table

The ICE table is an essential tool in chemical equilibrium calculations. ICE stands for Initial, Change, and Equilibrium, representing different stages in a reaction:

• Initial: This is where you input the starting amounts or concentrations of reactants and products. If the product is not initially present, set its initial concentration to zero.
• Change: Here, the changes in concentration are represented. To signify how much of each substance reacts or is produced, we use variables that are often represented by "x." The coefficients from the balanced reaction equation determine the ratios of these changes.
• Equilibrium: This row includes the equilibrium concentrations, calculated from the initial concentrations and the changes that occur as the reaction reaches equilibrium.

For example, in the reaction involving \( ext{H}_2 ext{O} \) and \( ext{Cl}_2 ext{O} \), you first find the initial moles, convert them into concentrations, set changes as "-x" for reactants and "+2x" for the product \( ext{HOCl}\), then sum the columns to get equilibrium concentrations.

Equilibrium Constant

The equilibrium constant, denoted as \( K \), reflects the ratio of the concentrations of products to reactants when a system is in chemical equilibrium for a given reaction:

• The value of \( K \) varies depending on the specific reaction and the temperature at which it occurs. In the original exercise, \( K = 0.090 \) represents the equilibrium state at \( 25^{\circ} ext{C}.\)
• An equilibrium constant tells us about the position of equilibrium: a \( K \) value less than 1 indicates that reactants are favored, while a value greater than 1 means products are favored.
• The formula to calculate \( K \) involves the concentrations of the reactants and products, each raised to the power of their coefficients in the balanced chemical equation. For the given chemical reaction, \( K = \frac{[\text{HOCl}]^2}{[\text{H}_2\text{O}][\text{Cl}_2\text{O}]} \).

By solving the equilibrium equation derived from the ICE table, you can find the concentrations of all species at equilibrium for specific initial conditions.

Reaction Stoichiometry

In chemical reactions, stoichiometry is the quantitative relationship between reactants and products, governed by the principle that matter is neither created nor destroyed in a chemical reaction. Here's how it comes into play:

• The balanced chemical equation provides the stoichiometric coefficients that indicate the ratio in which reactants transform into products. In our example, \( ext{H}_2 ext{O(g)} + ext{Cl}_2 ext{O(g)} \rightarrow 2\text{HOCl(g)}, \) 1 mole of each reactant produces 2 moles of \( ext{HOCl}.\)
• These coefficients guide us in setting up the "Change" row in the ICE table. If one reactant decreases by "x," the other must also decrease by the same quantity, while the product increases by "2x." This reflects the stoichiometric balance.
• Understanding reaction stoichiometry helps predict the amounts of reactants consumed and products formed at equilibrium, which is crucial for calculating equilibrium concentrations with the ICE table and \( K \) equation.

Mastery of stoichiometry is essential as it forms the backbone of analyzing and understanding any chemical equation and its equilibrium state.