Question

Starting with \(0.280 \mathrm{mol} \mathrm{SbCl}_{3}\) and \(0.160 \mathrm{mol} \mathrm{Cl}_{2},\) how many moles of \(\mathrm{SbCl}_{5}, \mathrm{SbCl}_{3},\) and \(\mathrm{Cl}_{2}\) are present when equilibrium is established at \(248^{\circ} \mathrm{C}\) in a 2.50 L flask? $$\begin{aligned} \mathrm{SbCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 2.5 \times 10^{-2} \mathrm{at} \ 248^{\circ} \mathrm{C} \end{aligned}$$

Short answer

To find the final moles of all compounds at equilibrium, create an ICE table, express and solve equilibrium constant expression in terms of X, and substitute the value of X back into the ICE table.

Step-by-step solution

Identify Initial Moles of Compounds

From the problem, we know there are 0.280 mol of SbCl3 and 0.160 mol of Cl2 initially. The problem doesn’t provide the initial amount of SbCl5, so we assume there is a zero mol of SbCl5. This is the 'I' in our ICE table.

Create an ICE Table

We then create an ICE table that represents the Initial moles (I), Change in moles (C) and Equilibrium moles (E) for each compound in the reaction. When the system reaches equilibrium, some amount X (in mol) of SbCl5 forms by using up same amount X of SbCl3 and Cl2. Thus, the 'C' part of ICE table will be +x for SbCl5 and -x for SbCl3 and Cl2.

Calculate Final Moles at Equilibrium

We then fill out the 'E' part of the ICE table. At equilibrium, there will be x mol of SbCl5, (0.280-x) mol of SbCl3, and (0.160-x) mol of Cl2. These values are got by adding the initial moles and change in moles.

Use KC to Find the Value of X

Next, plug the equilibrium concentrations into the expression for the equilibrium constant and solve for x. The equilibrium molar concentration of a compound is its moles at equilibrium divided by the volume of the flask. The equilibrium constant KC is given as 2.5 x 10^-2.

Calculate the Moles of Compounds at Equilibrium

After finding the value of X, substitute it back into the 'E' part of the ICE table to find the final moles of all three compounds at equilibrium.

Key concepts

ICE Table

The ICE table is a powerful tool in chemistry for solving equilibrium problems. ICE stands for Initial, Change, and Equilibrium, and this table helps visualize the changes in concentrations or amounts of reactants and products as a reaction reaches equilibrium.
At the start, we determine the initial moles of each compound. These initial amounts establish the baseline for our calculations. In this reaction, we have 0.280 mol of \( \mathrm{SbCl}_{3} \) and 0.160 mol of \( \mathrm{Cl}_{2} \), and we assume there is 0 mol of \( \mathrm{SbCl}_{5} \) initially.

Next, the 'Change' row represents the shift in moles of reactants and products as the reaction approaches equilibrium. As the reaction progresses, \( x \) moles of \( \mathrm{SbCl}_{5} \) forms, while \( x \) moles are reduced from \( \mathrm{SbCl}_{3} \) and \( \mathrm{Cl}_{2} \). This gives a positive change for \( \mathrm{SbCl}_{5} \) and negative changes for \( \mathrm{SbCl}_{3} \) and \( \mathrm{Cl}_{2} \).
Finally, the 'Equilibrium' row gives all compounds' final amounts. By summing the initials and changes, we have \( x \) moles of \( \mathrm{SbCl}_{5} \), \( (0.280-x) \) moles of \( \mathrm{SbCl}_{3} \), and \( (0.160-x) \) moles of \( \mathrm{Cl}_{2} \) at equilibrium. The ICE table's role is to provide a structured way to keep track of how a system evolves as it moves towards equilibrium.

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a numerical value that characterizes the equilibrium state of a chemical reaction at a certain temperature. It is a crucial concept in understanding how far a reaction proceeds before reaching a state of balance.
For a general reaction, the equilibrium constant is expressed by a fraction, where the concentrations of the products are in the numerator, and those of the reactants are in the denominator, each raised to the power of their respective coefficients from the balanced chemical equation.
In our exercise, the equilibrium constant \( K_c \) for the reaction \( \mathrm{SbCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \) is given as \( 2.5 \times 10^{-2} \) at \( 248^{\circ} \mathrm{C} \).

• This indicates that, at equilibrium, the concentration of products over reactants is relatively low, showing that the reaction doesn't favor the formation of \( \mathrm{SbCl}_{3} \) and \( \mathrm{Cl}_{2} \) extensively under these conditions.
• Using the equilibrium constant formula and the equilibrium concentrations of each compound, we can solve for \( x \), representing the extent of the reaction and fulfilling the balance described by \( K_c \).

Understanding \( K_c \) aids in predicting the direction in which a reaction is likely to proceed when conditions change, such as a shift in concentration or pressure.

Reaction Quotient

The reaction quotient, denoted as \( Q_c \), is a dimensionless quantity similar to the equilibrium constant, \( K_c \). It measures the ratio of the concentrations of products to reactants at any point during the reaction, not necessarily at equilibrium.
Comparing \( Q_c \) to \( K_c \) provides insight into the direction the reaction will proceed to achieve equilibrium.

• If \( Q_c < K_c \), the reaction will proceed in the forward direction, forming more products.
• If \( Q_c > K_c \), more reactants will be formed, as the reaction shifts backward.
• If \( Q_c = K_c \), the system is already at equilibrium.

For the reaction \( \mathrm{SbCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \), calculating \( Q_c \) helps verify if the initial concentrations will automatically adjust to achieve equilibrium, without excess shifts in the reaction direction.
This concept assists in anticipating the effect of concentration changes in a reaction system, ensuring a balanced understanding of chemical dynamics.