Problem 33 - Chemistry and Chemical Reactivity

Question

Carbon tetrachloride can be produced by the following reaction: $$ \mathrm{CS}_{2}(\mathrm{g})+3 \mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows \mathrm{S}_{2} \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{CCl}_{4}(\mathrm{g}) $$ Suppose 1.2 mol of \(\mathrm{CS}_{2}\) and 3.6 mol of \(\mathrm{Cl}_{2}\) are placed in a \(1.00-\mathrm{L}\). flask. After equilibrium has been achieved, the mixture contains 0.90 mol \(\mathrm{CCl}_{4}\). Calculate \(K_{c}\).

Step-by-Step Solution

Answer

The equilibrium constant \(K_c\) is approximately 3.705.

Step 1: Write the Equilibrium Expression

The equilibrium constant expression for the reaction \(\mathrm{CS}_{2}(\mathrm{g}) + 3 \mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows \mathrm{S}_{2}\mathrm{Cl}_{2}(\mathrm{g}) + \mathrm{CCl}_{4}(\mathrm{g})\) is given by the equation: \[ K_c = \frac{[\mathrm{S}_{2}\mathrm{Cl}_{2}][\mathrm{CCl}_{4}]}{[\mathrm{CS}_{2}][\mathrm{Cl}_{2}]^3} \] where the square brackets indicate the concentration of each species at equilibrium.

Step 2: Calculate Initial Concentrations

Since the volume of the flask is 1.00 L, the initial concentrations are equivalent to the number of moles. Thus, initially, \([\mathrm{CS}_{2}] = 1.2\;\text{mol/L}\) and \([\mathrm{Cl}_{2}] = 3.6\;\text{mol/L}\). The concentration of products, \(\mathrm{S}_{2}\mathrm{Cl}_{2}\) and \(\mathrm{CCl}_{4}\), is 0.

Step 3: Determine Change in Concentrations

At equilibrium, \([\mathrm{CCl}_{4}] = 0.90\;\text{mol/L}\). Because the stoichiometry is in a 1:1 relationship with \(\mathrm{S}_{2}\mathrm{Cl}_{2}\), \([\mathrm{S}_{2}\mathrm{Cl}_{2}] = 0.90\;\text{mol/L}\). The change in concentration for \(\mathrm{CS}_{2}\) is \(-0.90\) mol and for \(\mathrm{Cl}_{2}\) is \(-3\times 0.90 = -2.70\) mol due to the 1:3 stoichiometry.

Step 4: Calculate Equilibrium Concentrations

Subtract the changes from the initial concentrations to find the equilibrium concentrations. Therefore, \([\mathrm{CS}_{2}]_{eq} = 1.2 - 0.90 = 0.30\;\text{mol/L}\) and \([\mathrm{Cl}_{2}]_{eq} = 3.6 - 2.70 = 0.90\;\text{mol/L}\).

Step 5: Substitute into Equilibrium Expression

Substitute the equilibrium concentrations into the expression: \[ K_c = \frac{(0.90)(0.90)}{(0.30)(0.90)^3} \] Simplify to find \(K_c\).

Step 6: Calculate and Simplify

Calculate the value: \[ K_c = \frac{0.81}{0.30 \times 0.729} = \frac{0.81}{0.2187} \approx 3.705 \] Thus, the equilibrium constant \(K_c\) is approximately 3.705.

Key Concepts

Chemical Equilibrium

In a chemical reaction, chemical equilibrium occurs 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 over time. This does not mean the concentrations are equal, but they are balanced in such a way that the overall composition does not change.

Equilibrium is dynamic, meaning reactions are still happening, but there is no net change in concentrations. The equilibrium constant, denoted as \( K_c \), is a measure of this balance and varies depending on factors like temperature. A large \( K_c \) favors product formation, while a small \( K_c \) favors reactants. Understanding chemical equilibrium helps in predicting how a system responds to changes in conditions, which is crucial in industrial processes and laboratory settings.

Stoichiometry

Stoichiometry is the mathematical relationship between reactants and products in a chemical reaction. It involves using balanced equations to determine the quantitative relationships between substances involved in these reactions.

In the case of our exercise: - One molecule of \( \mathrm{CS}_2 \) reacts with three molecules of \( \mathrm{Cl}_2 \).- This results in one molecule of \( \mathrm{S}_2\mathrm{Cl}_2 \) and one molecule of \( \mathrm{CCl}_4 \).
The stoichiometric coefficients (the numbers in front of the molecules in the chemical equation) tell us about the proportions of each substance needed for the reaction to proceed without any left over.

Stoichiometry is vital in calculating how much of a reactant is needed or how much product will be formed. Hence, for equilibrium calculations, understanding these ratios helps in determining changes in concentration.

Reaction Quotient

The reaction quotient, \( Q_c \), helps predict the direction in which a chemical reaction will proceed to reach equilibrium. While \( K_c \) describes a system at equilibrium, \( Q_c \) is calculated in the same manner as \( K_c \), but using initial concentrations at any point before reaching equilibrium.

To find \( Q_c \), use the expression:
\[Q_c = \frac{[\text{Products}]^n}{[\text{Reactants}]^m}\]where \( n \) and \( m \) represent the stoichiometric coefficients.

If \( Q_c \) is less than \( K_c \), the reaction will proceed forward to form more products. If \( Q_c \) is greater, the reaction will reverse, producing more reactants. When they are equal, the reaction is at equilibrium. Understanding \( Q_c \) is crucial for predicting and understanding the changes in concentrations within the reaction.

Le Chatelier's Principle

Le Chatelier’s Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore balance. This principle is a powerful tool in predicting the effects of changes in temperature, pressure, and concentration on a reaction.

For example, in the reaction provided:- Increasing the concentration of \( \mathrm{CS}_2 \) or \( \mathrm{Cl}_2 \) will favor the forward reaction, forming more \( \mathrm{S}_2\mathrm{Cl}_2 \) and \( \mathrm{CCl}_4 \).- Removing some \( \mathrm{CCl}_4 \) will also shift the equilibrium right to produce more products.

On the other hand, increasing the pressure on reactions involving gases may shift the equilibrium towards the side with fewer moles of gas. Le Chatelier’s Principle provides a qualitative tool for understanding how changes will influence the shifts in equilibrium, though the exact quantitative change requires calculations.