Problem 6
Question
Write the equilibrium constant expression for the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftarrows \mathrm{Fe}(\mathrm{SCN})^{2+}(a q)\)
Step-by-Step Solution
Verified Answer
The equilibrium constant expression for the reaction \(\mathrm{Fe}^{3+}(aq) + \mathrm{SCN}^{-}(aq) \rightleftarrows \mathrm{Fe}(\mathrm{SCN})^{2+}(aq)\) is given by:
\[K_c = \frac{[\mathrm{Fe}(\mathrm{SCN})^{2+}]}{[\mathrm{Fe}^{3+}] [\mathrm{SCN}^{-}]}\]
1Step 1: Write the balanced chemical equation
For this reaction, the balanced chemical equation is:
\[\mathrm{Fe}^{3+}(aq) + \mathrm{SCN}^{-}(aq) \rightleftarrows \mathrm{Fe}(\mathrm{SCN})^{2+}(aq)\]
2Step 2: Identify the species in the reaction
In this reaction, we have the following species:
- Reactants: \(\mathrm{Fe}^{3+}(aq)\) and \(\mathrm{SCN}^{-}(aq)\)
- Product: \(\mathrm{Fe}(\mathrm{SCN})^{2+}(aq)\)
3Step 3: Write the Equilibrium Constant Expression
The equilibrium constant, denoted as \(K_c\), can be defined as the ratio of concentrations of products to reactants raised to their stoichiometric coefficients. In this case, we have:
\[K_c = \frac{[\mathrm{Fe}(\mathrm{SCN})^{2+}]}{[\mathrm{Fe}^{3+}] [\mathrm{SCN}^{-}]}\]
Here, \([\mathrm{Fe}^{3+}]\), \([\mathrm{SCN}^{-}]\), and \([\mathrm{Fe}(\mathrm{SCN})^{2+}]\) represent the equilibrium concentrations of the respective species in the reaction.
Key Concepts
Chemical EquilibriumStoichiometryAqueous Solutions
Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate. At this point, the concentrations of the reactants and products remain constant over time. This does not mean the reactions have stopped; both the forward and reverse reactions continue to occur, but they do so at the same speed.
One way to think about equilibrium is to imagine a seesaw balanced with equal weights on both sides. It doesn't move in either direction because the forces are balanced, much like the concentrations at equilibrium.
Characteristics of chemical equilibrium include:
One way to think about equilibrium is to imagine a seesaw balanced with equal weights on both sides. It doesn't move in either direction because the forces are balanced, much like the concentrations at equilibrium.
Characteristics of chemical equilibrium include:
- Dynamic Nature: Molecules continuously react, but with no net change in concentration.
- Forward and Reverse Rates are Equal: Once equilibrium is reached, the rate of the forward reaction equals the rate of the reverse reaction.
- Constant Concentrations: At equilibrium, concentration of products and reactants remain constant.
Stoichiometry
Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. It involves calculating the proportions of various substances involved in reactions based on the balanced chemical equation.
For the reaction \[\mathrm{Fe}^{3+}(aq) + \mathrm{SCN}^{-}(aq) \rightleftarrows \mathrm{Fe}(\mathrm{SCN})^{2+}(aq)\]Stoichiometry tells us that 1 mole of \(\mathrm{Fe}^{3+}\) reacts with 1 mole of \(\mathrm{SCN}^{-}\) to produce 1 mole of \(\mathrm{Fe}(\mathrm{SCN})^{2+}\).
This 1:1:1 ratio helps us understand how the reactants combine to form products and how the quantities relate at equilibrium.
The stoichiometric coefficients of a balanced chemical equation are essential in writing the equilibrium constant expression. In our case, all coefficients are one, so this influences the simple form of our equilibrium expression:\[K_c = \frac{[\mathrm{Fe}(\mathrm{SCN})^{2+}]}{[\mathrm{Fe}^{3+}] [\mathrm{SCN}^{-}]}\]Stoichiometry also helps in calculating the equilibrium concentrations if initial amounts and the equilibrium constant are known.
For the reaction \[\mathrm{Fe}^{3+}(aq) + \mathrm{SCN}^{-}(aq) \rightleftarrows \mathrm{Fe}(\mathrm{SCN})^{2+}(aq)\]Stoichiometry tells us that 1 mole of \(\mathrm{Fe}^{3+}\) reacts with 1 mole of \(\mathrm{SCN}^{-}\) to produce 1 mole of \(\mathrm{Fe}(\mathrm{SCN})^{2+}\).
This 1:1:1 ratio helps us understand how the reactants combine to form products and how the quantities relate at equilibrium.
The stoichiometric coefficients of a balanced chemical equation are essential in writing the equilibrium constant expression. In our case, all coefficients are one, so this influences the simple form of our equilibrium expression:\[K_c = \frac{[\mathrm{Fe}(\mathrm{SCN})^{2+}]}{[\mathrm{Fe}^{3+}] [\mathrm{SCN}^{-}]}\]Stoichiometry also helps in calculating the equilibrium concentrations if initial amounts and the equilibrium constant are known.
Aqueous Solutions
Aqueous solutions are solutions where water is the solvent. The term "aqueous" indicates that a substance is dissolved in water and typically given the symbol \((aq)\).
Water is the most common solvent due to its excellent ability to dissolve various substances, which facilitates many chemical reactions. This is because:
In writing equilibrium expressions for reactions in aqueous solutions, it is crucial only to include concentrations within the solution, ignoring pure liquids or solids that might also be involved in the overall reaction.
Water is the most common solvent due to its excellent ability to dissolve various substances, which facilitates many chemical reactions. This is because:
- Water is a polar solvent: It dissolves ionic and polar substances effectively.
- Universal Solvent: Many reactions, such as biological processes, occur in aqueous solutions.
- High Specific Heat: It allows for better temperature control in reactions.
In writing equilibrium expressions for reactions in aqueous solutions, it is crucial only to include concentrations within the solution, ignoring pure liquids or solids that might also be involved in the overall reaction.
Other exercises in this chapter
Problem 4
Write the equilibrium constant expression for the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows \mathrm{CS}_{2}(g)+4 \mathrm{H}
View solution Problem 5
Write the equilibrium constant expression for the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g)\)
View solution Problem 7
At the start of the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g)\) the concentrations are \(\left[\mathrm{H}_{2}\right]=0.10
View solution Problem 9
The equilibrium concentrations for the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftarrows \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) a
View solution