Question

A student mixes \(5.0 \mathrm{mL} 2.00 \times 10^{-3} \mathrm{M} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) with \(5.0 \mathrm{mL} 2.00 \times 10^{-3} \mathrm{M} \mathrm{KSCN}\). She finds that in the equilibrium mixture the concentration of \(\mathrm{FeSCN}^{2+}\) is \(1.2 \times 10^{-4} \mathrm{M}\). Find \(K_{\mathrm{c}}\) for the reaction \(\mathrm{Fe}^{3+}(\mathrm{aq})+\) SCN-(aq) \(=\) FeSCN \(^{2+}\) (aq). Step 1 Find the number of moles \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\) initially present. (Use Eq. 3.) Step 2 How many moles of FeSCN \(^{2+}\) are in the mixture at equilibrium? What is the volume of the equilibrium mixture? (Use Eq. 3.) How many moles of \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\) are used up in making the \(\mathrm{FeSCN}^{2+} ?\) Step 3 How many moles of \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\) remain in the solution at equilibrium? (Usc Eq. 4 and the results of Steps 1 and 2 .) Step 4 What are the concentrations of \(\mathrm{Fe}^{3+}, \mathrm{SCN}^{-}\), and \(\mathrm{FeSCN}^{2+}\) at equilibrium? What is the volume of the equilibrium mixture? (Use Eq. 3 and the results of Step 3.) Step 5 What is the value of \(K_{c}\) for the reaction? (Use Eq. 2 and the results of Step 4.)

Short answer

The value of the equilibrium constant, Kc, for the reaction is approximately \(1.54 \times 10^{-3}\).

Step-by-step solution

Find the initial moles of Fe3+ and SCN-.

To calculate the initial moles of Fe3+ and SCN-, use the given concentration and volume. The concentration of Fe(NO3)3 and KSCN is \(2.00 \times 10^{-3} M\) and the volume is \(5.0 mL\). Use the formula: moles = concentration × volume. For Fe3+ and SCN-: Initial moles = \((2.00 \times 10^{-3} M)(5.00 \times 10^{-3} L)\) = \(1.00 \times 10^{-5}\) moles

Find the moles of FeSCN2+ at equilibrium and the volume of the equilibrium mixture.

To find the moles of FeSCN2+ at equilibrium, use the given equilibrium concentration: \(1.2 \times 10^{-4} M\). The volume of the mixture is the sum of the volumes of Fe(NO3)3 and KSCN solutions, which is \(5.0 mL + 5.0 mL = 10.0 mL\). Moles of FeSCN2+ at equilibrium = equilibrium concentration × volume = \((1.2 \times 10^{-4} M)(10.0 \times 10^{-3} L)\) = \(1.2 \times 10^{-6}\) moles

Find the moles of Fe3+ and SCN- remaining in the solution at equilibrium.

Since one mole of Fe3+ reacts with one mole of SCN- to form one mole of FeSCN2+, the moles of Fe3+ and SCN- used up will be equal to the moles of FeSCN2+ formed. Moles of Fe3+ and SCN- remaining in the solution at equilibrium = initial moles - moles used up = \(1.00 \times 10^{-5}\) moles - \(1.2 \times 10^{-6}\) moles = \(8.8 \times 10^{-6}\) moles

Calculate the equilibrium concentrations of Fe3+, SCN-, and FeSCN2+.

To find the equilibrium concentrations, divide the moles remaining by the volume of the equilibrium mixture. For Fe3+ and SCN-: Equilibrium concentration = \(\frac{8.8 \times 10^{-6} \, moles}{10.0\, \times 10^{-3} L}\) = \(8.8 \times 10^{-4} M\) The equilibrium concentration of FeSCN2+ is given: \(1.2 \times 10^{-4} M\)

Calculate the value of Kc for the reaction.

Using the given equilibrium reaction, we can write the expression for Kc as: \[K_c = \frac{[\mathrm{FeSCN}^{2+}]}{[\mathrm{Fe}^{3+}][\mathrm{SCN}^-]}\] Substitute the equilibrium concentrations into the Kc expression: \[K_c = \frac{(1.2 \times 10^{-4})}{(8.8 \times 10^{-4})(8.8 \times 10^{-4})}\] Calculate Kc: \(K_c \approx 1.54 \times 10^{-3}\) Thus, the value of the equilibrium constant, Kc, for the given reaction is \(1.54 \times 10^{-3}\).

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial when studying reactions in chemistry. In a dynamic state of chemical equilibrium, both the forward and reverse reactions occur at the same rate, resulting in constant concentrations of reactants and products over time. During this process, there is no net change in the amounts of substances involved.

To illustrate, take the equilibrium reaction from our exercise \(\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SCN}^- (\mathrm{aq}) = \mathrm{FeSCN}^{2+} (\mathrm{aq})\). At equilibrium, the rate of formation of \(\mathrm{FeSCN}^{2+}\) equals the rate at which it breaks down into \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^-\). This balance is essential in determining the equilibrium constant, \(K_c\), which quantifies the state of the reaction mixture at a particular temperature.

Significance of \(K_c\) in Predicting Reactions

If \(K_c\) is high, the reaction favors product formation, indicating that, at equilibrium, a significant amount of products is present compared to reactants. Conversely, a low \(K_c\) suggests that reactants predominate. In the exercise, calculating the equilibrium constant helps predict the extent of reaction and the concentrations of ions at equilibrium.

Reaction Stoichiometry

Reaction stoichiometry deals with the quantitative relationship between the amounts of reactants and products in a chemical reaction. It is based on the balanced chemical equation and the conservation of matter principle. By understanding stoichiometry, we can predict the amount of products formed from given amounts of reactants, and vice versa.

For the reaction given in the example, \(1:1:1\) molar ratio is used, meaning one mole of \(\mathrm{Fe}^{3+}\) reacts with one mole of \(\mathrm{SCN}^-\) to produce one mole of \(\mathrm{FeSCN}^{2+}\). When calculating the equilibrium concentration, stoichiometry becomes essential as it helps to determine the moles of reactants consumed and products formed.

Application in Equilibrium Calculations

By employing stoichiometry, one can determine how many moles of \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^-\) remain at equilibrium after the formation of \(\mathrm{FeSCN}^{2+}\). This quantity is then used to calculate the concentrations of each species required to solve for the equilibrium constant.

Concentration Calculation

Concentration calculation is a basic concept in chemistry that denotes how much of a substance (solute) is present in a certain volume of a mixture (solution). Concentrations are most commonly expressed in terms of molarity (M), which is moles of solute per liter of solution.

In the context of our problem, the calculation of concentrations at equilibrium is necessary to find the equilibrium constant \(K_c\). First, you determined the moles of the reactants and products, then calculated their concentrations by dividing the number of moles by the volume of the solution at equilibrium.

Role in Determining Equilibrium Constant

Once the equilibrium concentrations are known for \(\mathrm{Fe}^{3+}\), \(\mathrm{SCN}^-\), and \(\mathrm{FeSCN}^{2+}\), they are plugged into the \(K_c\) expression to find the equilibrium constant. It's important to measure volumes accurately and perform calculations correctly to ensure the reliability of the \(K_c\) value, reflecting on the reaction's behavior in the solution.