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Consider the reaction. $$ \mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{FeSCN}^{2+}(a q) $$ Asolution is made containing initial \(\left[\mathrm{Fe}^{3+}\right]=1.0 \times 10^{-3} \mathrm{M}\) and initial \(\left[\mathrm{SCN}^{-}\right]=8.0 \times 10^{-4} \mathrm{M}\). At equilibrium, \(\left[\mathrm{FeSCN}^{2+}\right]=1.7 \times 10^{-4} \mathrm{M}\). Calculate the value of the equilibrium constant. Hint: Use the chemical reaction stoichiometry to calculate the equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\).

Short Answer

Expert verified
The equilibrium constant \(K_c\) for the reaction is 139.

Step by step solution

01

Identify the changes in concentrations

Establish the stoichiometry of the reaction: \(\mathrm{Fe}^{3+}(aq) + \mathrm{SCN}^{-}(aq) \rightleftharpoons \mathrm{FeSCN}^{2+}(aq)\).Write the initial concentrations and the changes that occur as the system reaches equilibrium. We know the equilibrium concentration of \(\mathrm{FeSCN}^{2+}\) is \(1.7 \times 10^{-4} M\), which implies that the same amount of \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\) was used to form \(\mathrm{FeSCN}^{2+}\).
02

Calculate the equilibrium concentrations

Subtract the amount of \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\) that reacted from their initial concentrations to determine the equilibrium concentrations.\(\left[\mathrm{Fe}^{3+}\right]_{\text{eq}} = 1.0 \times 10^{-3} M - 1.7 \times 10^{-4} M\),\(\left[\mathrm{SCN}^{-}\right]_{\text{eq}} = 8.0 \times 10^{-4} M - 1.7 \times 10^{-4} M\).
03

Use the equilibrium concentrations to calculate the equilibrium constant

Use the equilibrium constant expression for the reaction: \(K_c = \frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}\).Insert the equilibrium concentrations into the expression to find the equilibrium constant.
04

Solve for the equilibrium constant

Substitute the equilibrium concentrations into the equilibrium expression:\(K_c = \frac{1.7 \times 10^{-4}}{(1.0 \times 10^{-3} - 1.7 \times 10^{-4})(8.0 \times 10^{-4} - 1.7 \times 10^{-4})}\).Calculate the value of \(K_c\) to obtain the equilibrium constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction—the process that generates the products—is equal to the rate of the reverse reaction, which re-forms the reactants. At this point, even though both reactions are still occurring, there is no net change in the concentrations of reactants and products over time. This balance between two opposing processes is dynamic, meaning it can be adjusted by changing conditions such as temperature, pressure, or concentration of reactants or products.

Understanding equilibrium involves knowing that it does not mean the reactants and products are present in equal amounts. Instead, it indicates that their concentrations have stabilized in a ratio that will remain constant unless disturbed. The equilibrium constant (\(K_c\)) is a number that provides a measure of the reactants and products concentration ratio at equilibrium; it is a unique value for every reaction at a given temperature.
Reaction Stoichiometry
Reaction stoichiometry describes the quantitative relationship between reactants and products in a chemical reaction. It allows us to predict the amounts of products formed from a given amount of reactants and vice versa. Stoichiometry is rooted in the law of conservation of mass, where the total mass of reactants must equal the total mass of products. Stoichiometric coefficients in a balanced chemical equation tell us the proportions in which chemicals react and form.

For example, in a reaction where one mole of iron (III) ions reacts with one mole of thiocyanate ions to form one mole of iron thiocyanate, the coefficients are understood to be 1:1:1. The concept of stoichiometry extends to the idea of the limiting reactant—the reactant that will be completely consumed first, thus limiting the amount of product that can be formed. In the numerical exercise posed, stoichiometry helps to calculate the equilibrium concentrations needed to find the equilibrium constant.
Equilibrium Concentrations
Equilibrium concentrations are the concentrations of the reactants and products in a chemical reaction once it has reached equilibrium. These concentrations do not change over time as long as the system's conditions remain constant. To calculate the equilibrium constant (\(K_c\)), one must know the equilibrium concentrations of the reactants and products.

During the process to reach equilibrium, reactants are converted into products and, in a reversible reaction, products can also revert back into reactants until the system stabilizes. It's important to understand that initial concentrations will differ from the equilibrium concentrations due to these reactions. By applying stoichiometry, it is possible to determine how much of the reactants are left at equilibrium and how much product is formed, as their respective changes in concentration are always proportional to their stoichiometric coefficients. Often, an Ice table (Initial, Change, Equilibrium) is used to organize these data and help in solving for the unknown equilibrium concentrations.

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Most popular questions from this chapter

This reaction is exothermic. $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(g) $$ If you were a chemist trying to maximize the amount of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) produced, which of the following might you try? Assume that the reaction mixture reaches equilibrium. (a) increasing the reaction volume (b) removing \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) from the reaction mixture as it forms (c) lowering the reaction temperature (d) adding \(\mathrm{Cl}_{2}\)

What is the effect of increasing the temperature of an exothermic reaction mixture at equilibrium? Of decreasing the temperature?

Complete the table. Assume that all concentrations are equilibrium concentrations in moles per liter, \(M\). $$ \begin{array}{llll} \text { Compound } & \text { [Cation] } & \text { [Anion] } & \boldsymbol{K}_{\text {eq }} \\ \mathrm{CdS} & 3.7 \times 10^{-15} & 3.7 \times 10^{-15} & \\ \mathrm{BaF}_{2} & & 7.2 \times 10^{-3} & 1.9 \times 10^{-7} \\ \mathrm{Ag}_{2} \mathrm{SO}_{4} & 2.8 \times 10^{-2} & & 1.1 \times 10^{-5} \\\ \hline \end{array} $$

This reaction is exothermic. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ Predict the effect (shift right, shift left, or no effect) of these changes. (a) increasing the reaction temperature (b) decreasing the reaction temperature

What is the effect of increasing the pressure of a reaction mixture at equilibrium if the reactant side has fewer moles of gas particles than the product side?

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