Problem 36 - Chemistry and Chemical Reactivity

Question

Equal numbers of moles of \(\mathrm{H}_{2}\) gas and \(\mathrm{I}_{2}\) vapor are mixed in a flask and heated to \(700^{\circ} \mathrm{C}\). The initial concentration of each gas is \(0.0088 \mathrm{mol} / \mathrm{L},\) and \(78.6 \%\) of the \(\mathrm{I}_{2}\) is consumed when equilibrium is achieved according to the equation $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ Calculate \(K_{c}\) for this reaction.

Step-by-Step Solution

Answer

The equilibrium constant \(K_{c}\) for the reaction is approximately 54.97.

Step 1: Write the balanced chemical equation

The balanced chemical equation is given as: \(\mathrm{H}_{2}(\mathrm{g}) + \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\).

Step 2: Define the initial concentrations and changes

The initial concentration of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) is \(0.0088 \mathrm{mol} / \mathrm{L}\). Since 78.6% of \(\mathrm{I}_{2}\) is consumed, the change in concentration for \(\mathrm{I}_{2}\) is \(-0.0088 \times 0.786 = -0.00693 \mathrm{mol}/\mathrm{L}\).

Step 3: Calculate the change in concentrations of reactants and products

The change in \(\mathrm{I}_{2}\) is \(-0.00693 \mathrm{mol}/\mathrm{L}\), so the change in \(\mathrm{H}_{2}\) is also \(-0.00693 \mathrm{mol}/\mathrm{L}\) because they react in a 1:1 ratio. The formation of \(\mathrm{HI}\) is \(2 \times 0.00693 = 0.01386 \mathrm{mol}/\mathrm{L}\) because 2 moles of \(\mathrm{HI}\) are produced for every 1 mole each of \(\mathrm{H}\) and \(\mathrm{I}_{2}\).

Step 4: Determine equilibrium concentrations

The equilibrium concentration of \(\mathrm{I}_{2}\) is \(0.0088 - 0.00693 = 0.00187 \mathrm{mol}/\mathrm{L}\), \(\mathrm{H}_{2}\) is also \(0.00187 \mathrm{mol}/\mathrm{L}\) and \(\mathrm{HI}\) is \(0.01386 \mathrm{mol}/\mathrm{L}\).

Step 5: Write the expression for the equilibrium constant

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

Step 6: Substitute equilibrium concentrations into the equilibrium expression

Substitute the equilibrium concentrations into the expression:\[K_{c} = \frac{(0.01386)^2}{(0.00187) \times (0.00187)}\]Calculate to find \(K_{c}\).

Step 7: Calculate the equilibrium constant

Perform the calculation:\[K_{c} = \frac{0.0001922596}{0.0000034969} \approx 54.97\] Thus, \(K_{c} = 54.97\).

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented as \(K_c\), is a critical concept in understanding chemical equilibrium. It provides a quantifiable measure of the ratio between the concentrations of products and reactants at equilibrium for a given chemical reaction. For the reaction \[ \mathrm{H}_{2}(\mathrm{g}) + \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \]the equilibrium constant expression is: \[ K_{c} = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}] \times [\mathrm{I}_{2}]} \]This expression shows how \(K_c\) is determined by squaring the concentration of \(\mathrm{HI}\) and dividing it by the product of the concentrations of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\). The value of \(K_c\) tells us about the position of equilibrium:

• A large \(K_c\) value indicates that, at equilibrium, the reaction favors the formation of products.
• A small \(K_c\) value suggests that reactants are favored at equilibrium.

In this exercise example, we calculated a \(K_c\) of approximately 54.97, suggesting a dominance of product formation under the given conditions.

Chemical Reaction

A chemical reaction involves the rearrangement of atoms in reactants to form products. For the reaction under consideration:\[ \mathrm{H}_{2}(\mathrm{g}) + \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \]it represents a dynamic equilibrium situation. The forward reaction (\(\mathrm{H}_{2} + \mathrm{I}_{2} \rightarrow 2\mathrm{HI}\)) and the reverse reaction (\(2\mathrm{HI} \rightarrow \mathrm{H}_{2} + \mathrm{I}_{2}\)) compete until a state of balance is achieved.The reaction is balanced when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant, though they are not necessarily equal. Understanding this balance is crucial because it shows how chemical systems respond to changes in conditions (like changes in temperature, pressure, or concentrations). This balance leads us to the concept of the equilibrium constant, which provides insight into the position of this equilibrium.

Concentration Change

The concept of concentration change in chemical equilibrium is essential to predict how a system will behave when disturbed. Initially, for this reaction, the concentrations of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) are each \(0.0088\,\text{mol/L}\). As the reaction proceeds to equilibrium, these concentrations change due to the consumption of reactants and formation of products.In the problem we are working on, 78.6% of \(\mathrm{I}_{2}\) is consumed during the reaction, leading to a new lower concentration.

• The change in \(\mathrm{I}_{2}\) concentration is calculated as \(-0.00693\,\text{mol/L}\) based on initial amounts.
• This same change applies to \(\mathrm{H}_{2}\) because of the 1:1 stoichiometry.
• For \(\mathrm{HI}\), the change is twice that of \(\mathrm{H}_{2}\) or \(\mathrm{I}_{2}\) because two moles of \(\mathrm{HI}\) are produced in the reaction, resulting in an increase of \(0.01386\,\text{mol/L}\).

Understanding these changes allows us to determine the concentrations at equilibrium, which are used in calculating the equilibrium constant, \(K_c\). This demonstrates how stoichiometry and initial conditions shape the composition of equilibrium mixtures.