Question

In an experiment to study the formation of \(\mathrm{HI}(g)\) $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) were placed in a sealed container at a certain temperature. At equilibrium, \(\left[\mathrm{H}_{2}\right]=6.50 \times 10^{-5} \mathrm{M},\left[\mathrm{I}_{2}\right]=1.06 \times 10^{-3} \mathrm{M},\) and \([\mathrm{HI}]=1.87 \times 10^{-3} \mathrm{M} .\) Calculate \(K_{\mathrm{c}}\) for the reaction at this temperature.

Short answer

The equilibrium constant, \(K_{c}\), is approximately 50.76.

Step-by-step solution

Write the balanced equation

The balanced chemical equation for the reaction is given by: \[ \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2\mathrm{HI}(g) \]

Write the expression for the equilibrium constant

The equilibrium constant expression, \(K_{c}\), for the given reaction is: \[ K_{c} = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]} \]

Substitute the equilibrium concentrations into the expression

Substitute the given equilibrium concentrations into the equilibrium constant expression: \[ [\mathrm{H}_{2}] = 6.50 \times 10^{-5} \mathrm{M}, [\mathrm{I}_{2}] = 1.06 \times 10^{-3} \mathrm{M}, [\mathrm{HI}] = 1.87 \times 10^{-3} \mathrm{M} \] \[ K_{c} = \frac{(1.87 \times 10^{-3})^2}{(6.50 \times 10^{-5})(1.06 \times 10^{-3})} \]

Calculate the value of the equilibrium constant

Calculate the numerator and the denominator separately first: Numerator: \[ (1.87 \times 10^{-3})^2 = 3.4969 \times 10^{-6} \] Denominator: \[ (6.50 \times 10^{-5})(1.06 \times 10^{-3}) = 6.89 \times 10^{-8} \] Now, divide the numerator by the denominator: \[ K_{c} = \frac{3.4969 \times 10^{-6}}{6.89 \times 10^{-8}} \approx 50.76 \]

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction reaches a state where the concentrations of reactants and products no longer change with time. This happens because the rates of the forward and reverse reactions have become equal. Imagine a reaction where substances A and B react to form products C and D. At equilibrium, A and B continue to react to form C and D, but C and D also react to form A and B at the same rate. The system appears static, but it is dynamic at the molecular level where reactions are continuously occurring in both directions.
Understanding chemical equilibrium helps chemists control reactions to produce desired products efficiently.
In the example given, \[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \] the concentrations of hydrogen, iodine, and hydrogen iodide no longer change once equilibrium is reached.

Equilibrium Constant

The equilibrium constant (\text{K}_c\text{) is a value that expresses the ratio of the concentration of products to reactants at equilibrium. Each concentration is raised to the power of its coefficient from the balanced chemical equation.
For a generic reaction: \[ aA + bB \rightleftharpoons cC + dD \] the equilibrium constant expression is given by:
\[ K_c = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \]
It is important because it provides insight into the extent of a reaction at equilibrium. A large \text{K}_c\text{ value indicates that the reaction favors the formation of products, while a small \text{K}_c\text{ value indicates that reactants are favored.
In our example, \[ K_{c} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \] Once you substitute the equilibrium concentrations \[ [\text{H}_2] = 6.50 × 10^{-5} \] \[ [\text{I}_2] = 1.06 × 10^{-3} \] \[ [\text{HI}] = 1.87 × 10^{-3} \] the value calculated is approximately 50.76, showing the product is favored at equilibrium.

Reaction Quotient

The reaction quotient (\text{Q}\text{) serves a similar role to the equilibrium constant but is used to determine the direction in which a reaction will proceed to reach equilibrium.
For the reaction \[ aA + bB \rightleftharpoons cC + dD \] the reaction quotient is calculated using the same expression as the equilibrium constant:
\[ Q = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \]
By comparing \text{Q} with \text{K}_{c}
If \text{Q} < \text{K}_{c}, the reaction will proceed forward to form more products.
If \text{Q} > \text{K}_{c}, the reaction will proceed in reverse to form more reactants.
If \text{Q} = \text{K}_{c}, the reaction is already at equilibrium. This makes \text{Q} a useful tool for predicting changes in a reaction mixture before equilibrium is reached.

Concentration

Concentration is the measure of the amount of a substance in a given volume and is usually expressed in molarity (moles per liter, \text{M}\text{). For a chemical reaction, the concentrations of reactants and products play a crucial role in determining the direction and extent of the reaction.
When measuring concentrations at equilibrium, these values are used to calculate the equilibrium constant. For example, in the exercise, the following concentrations at equilibrium were given:
\[ [\text{H}_2] = 6.50 × 10^{-5} \] \[ [\text{I}_2] = 1.06 × 10^{-3} \] \[ [\text{HI}] = 1.87 × 10^{-3} \] These concentrations are then plugged into the equilibrium constant expression to find \text{K}_{c}.
Understanding how to manipulate and use concentrations helps in predicting how changes in a system (such as adding more reactant or product) can affect the equilibrium position, according to Le Chatelier's principle, providing valuable insight into controlling chemical reactions.