Problem 60 - Chemistry

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 -L flask. b. 1.0 mole of pure HOCl is placed in a 2.0 \(\mathrm{L}\) flask.

Step-by-Step Solution

Answer

For case a, the equilibrium concentrations are: \[ [\text{H}_{2}\text{O}] = 0.0381 \, \text{M} \], \[ [\text{Cl}_{2}\text{O}] = 0.0231 \, \text{M} \], and \[ [\text{HOCl}] = 0.0338 \, \text{M} \]. For case b, the equilibrium concentrations are: \[ [\text{H}_{2}\text{O}] = 0.212 \, \text{M} \], \[ [\text{Cl}_{2}\text{O}] = 0.212\, \text{M} \], and \[ [\text{HOCl}] = 0.076 \, \text{M} \].

Step 1: Calculate moles of H2O and Cl2O at the beginning

Use the given masses and molar masses of H2O (18.02 g/mol) and Cl2O (50.45 g/mol) to calculate the moles of each present at the start: Moles of H2O = \( \frac{1.0 \, \text{g}}{18.02 \, \text{g/mol}} = 0.055 \, \text{mol} \) Moles of Cl2O = \( \frac{2.0 \, \text{g}}{50.45 \, \text{g/mol}} = 0.040 \, \text{mol} \)

Step 2: Set up the initial and equilibrium concentrations

Create an ICE (initial, change, equilibrium) table to find the equilibrium concentrations. Initially, the concentration of HOCl is zero because it hasn't been formed yet. | | H2O | Cl2O | HOCl | |-----|-----|------|-------| | I | 0.055 | 0.040 | 0 | | C | -x | -x | +2x | | E | 0.055-x | 0.040-x | 2x |

Step 3: Set up the equilibrium expression and solve for x

Write the equilibrium expression for K and plug in the equilibrium concentrations from the ICE table: \( K = \frac{[\text{HOCl}]^2}{[\text{H}_{2}\text{O}][\text{Cl}_{2}\text{O}]} \) Substitute with the values from the ICE table: \( 0.090 = \frac{(2x)^2}{(0.055-x)(0.040-x)} \) Now, solve the equation for x. Since K is small, we can estimate that x is small compared to 0.055 and 0.040, so we can simplify the expression to: \( 0.090 = \frac{4x^2}{(0.055)(0.040)} \) Solve for x: \( x = 0.0169 \)

Step 4: Calculate the equilibrium concentrations

Substitute the value of x back into the equilibrium values from the ICE table to find the concentrations at equilibrium: \[ [\text{H}_{2}\text{O}] = 0.055 - 0.0169 = 0.0381 \, \text{M} \] \[ [\text{Cl}_{2}\text{O}] = 0.040 - 0.0169 = 0.0231 \, \text{M} \] \[ [\text{HOCl}] = 2(0.0169) = 0.0338 \, \text{M} \] #Case b: Initial mixture of pure HOCl#

Step 1: Calculate initial concentration of HOCl

Calculate the initial concentration of HOCl by dividing the moles by the volume: \[ [\text{HOCl}]_{i} = \frac{1.0 \, \text{mol}}{2.0 \, \text{L}} = 0.5 \, \text{M} \]

Step 2: Set up the initial and equilibrium concentrations

Create an ICE table to find the equilibrium concentrations. Initially, the concentrations of H2O and Cl2O are zero because they haven't been formed yet. | | H2O | Cl2O | HOCl | |-----|-----|------|-------| | I | 0 | 0 | 0.5 | | C | +x | +x | -2x | | E | x | x | 0.5-2x |

Step 3: Set up the equilibrium expression and solve for x

Write the equilibrium expression for K and plug in the equilibrium values from the ICE table: \( K = 0.090 = \frac{x^2}{(0.5-2x)} \) Now, solve the equation for x. Since K is small, we can estimate that 2x is small compared to 0.5, so the equation simplifies to: \( 0.090 = \frac{x^2}{0.5} \) Solve for x: \( x = 0.212 \)

Step 4: Calculate the equilibrium concentrations

Substitute the value of x back into the equilibrium values from the ICE table to find the concentrations at equilibrium: \[ [\text{H}_{2}\text{O}] = x = 0.212 \, \text{M} \] \[ [\text{Cl}_{2}\text{O}] = x = 0.212\, \text{M} \] \[ [\text{HOCl}] = 0.5 - 2x = 0.5 - 2(0.212) = 0.076 \, \text{M} \]

Key Concepts

ICE Table

The ICE table, which stands for Initial, Change, and Equilibrium, is a handy tool that helps us determine the concentrations of different species in a chemical reaction at equilibrium. This method is particularly useful when dealing with reactions at equilibrium, as it allows one to systematically lay out and solve for unknown variables.

To use an ICE table, we start by writing the initial concentrations of the reactants and products. We then define 'x' as the change in concentration of the reactants or products as the system approaches equilibrium. We consider this change for each of the reactants and products, usually based on the stoichiometry of the reaction, and write these changes in the `Change` row. Finally, we calculate the equilibrium concentrations by adding the initial concentration -- adjusted by the change, which gives us the values in the `Equilibrium` row.

This calculation can then be used in the equilibrium expression to solve for x, allowing us to find the exact concentrations at equilibrium. Though the initial setup seems a bit complex, the organized structure of the ICE table makes solving equilibrium problems much more manageable.

Equilibrium Constant

The Equilibrium Constant, denoted by the symbol 'K', is a crucial component in determining the position of equilibrium in a chemical reaction. The value of K is derived from the concentrations of the products and reactants in an equilibrium state.

Specifically, for a general reaction: \[ aA + bB \rightleftharpoons cC + dD \]The equilibrium constant expression is set up as:\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]Where the brackets represent the concentration of each species, and the letters are their stoichiometric coefficients.

If K is large, this indicates a greater concentration of products compared to reactants, suggesting that the reaction proceeds predominantly to the right. Conversely, if K is small, this points to a higher concentration of reactants, indicating that the reaction does not proceed far to the right. The given equilibrium constant is used in conjunction with the ICE table to solve for unknown concentrations at equilibrium.

Reaction Quotient

The Reaction Quotient, symbolized as 'Q', is similar to the equilibrium constant 'K' in form, yet it provides slightly different information. While K is calculated using concentrations at equilibrium, Q can be calculated at any point in time during the reaction. This allows chemists to predict which direction the reaction will shift to reach equilibrium.

The reaction quotient is calculated using the same law of mass action expression as the equilibrium constant:\[ Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]By comparing Q to K:

• If Q = K, the system is at equilibrium, and no net change occurs.
• If Q < K, the reaction will proceed in the forward direction, creating more products to reach equilibrium.
• If Q > K, the reaction will go in the reverse direction, forming more reactants to re-establish equilibrium.

Understanding Q vs. K helps predict how disturbances in concentrations affect the system.

Dynamic Equilibrium

Dynamic Equilibrium is a state reached in a chemical reaction where the forward and reverse reactions occur at the same rate. At this point, the concentrations of reactants and products remain constant over time, even though reactions are still occurring at the molecular level.

This concept is crucial because it shows how reactions can continue to happen without an apparent change. In a dynamic equilibrium, no net change is observed, but the continuous, opposing processes make the system stable.

In the context of chemical reactions, dynamic equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction. While this doesn't mean that the amounts of reactants and products need to be equal, the concentrations stay constant until a disturbance occurs, such as changes in concentration, temperature, or pressure. Using the principles of the Le Chatelier's principle, we can predict how such disturbances might affect the equilibrium state. This forms the basis of many chemical equilibria applications in industries and nature, providing balance in systems.