Problem 9 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

2.0 mol of carbon disulfide and \(4.0 \mathrm{mol}\) of chlorine react at constant temperature according to this equation \\[ \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}) \\] At equilibrium, \(0.30 \mathrm{mol}\) of tetrachloromethane are formed. How much of each of the other components is present in this equilibrium mixture? (Section 15.4 )

Step-by-Step Solution

Answer

At equilibrium: \(\mathrm{CS}_{2} = 1.70\) moles, \(\mathrm{Cl}_{2} = 3.10\) moles, \(\mathrm{S}_{2}\mathrm{Cl}_{2} = 0.30\) moles, \(\mathrm{CCl}_{4} = 0.30\) moles.

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for this reaction is already provided: \[ \mathrm{CS}_{2} + 3 \mathrm{Cl}_{2} \rightleftharpoons \mathrm{S}_{2}\mathrm{Cl}_{2} + \mathrm{CCl}_{4} \] This indicates that 1 mole of \(\mathrm{CS}_{2}\) reacts with 3 moles of \(\mathrm{Cl}_{2}\) to produce 1 mole of \(\mathrm{S}_{2}\mathrm{Cl}_{2}\) and 1 mole of \(\mathrm{CCl}_{4}\).

Step 2: Initial Moles

Initially, there are \(2.0\) moles of \(\mathrm{CS}_{2}\) and \(4.0\) moles of \(\mathrm{Cl}_{2}\) present. Let's denote the initial moles of the products \(\mathrm{S}_{2}\mathrm{Cl}_{4}\) and \(\mathrm{CCl}_{4}\) as \(0\) moles, since the reaction hasn't started.

Step 3: Change in Moles at Equilibrium

Given that \(0.30\) moles of \(\mathrm{CCl}_{4}\) (tetrachloromethane) are formed at equilibrium, we can set up the change in moles from the reaction table: \[ \mathrm{CS}_{2} \text{ : } 2.0 - 0.30 = 1.70 \text{ moles} \] \[ \mathrm{CCl}_{4} \text{ : } 0.30 \text{ moles (formed)} \] The reaction conditions indicate that for each mole of \(\mathrm{CCl}_{4}\), 3 moles of \(\mathrm{Cl}_{2}\) are consumed, leading to a decrease in \(\mathrm{Cl}_{2}\) moles by \(3 \times 0.30 = 0.90\): \[ \mathrm{Cl}_{2} \text{ : } 4.0 - 0.90 = 3.10 \text{ moles} \] Since \(\mathrm{S}_{2}\mathrm{Cl}_{2}\) is produced in the same amount as \(\mathrm{CCl}_{4}\), it also has \(0.30\) moles.

Step 4: Equilibrium Moles

The equilibrium moles for each component are:- \(\mathrm{CS}_{2} = 1.70\) moles - \(\mathrm{Cl}_{2} = 3.10\) moles - \(\mathrm{S}_{2}\mathrm{Cl}_{2} = 0.30\) moles - \(\mathrm{CCl}_{4} = 0.30\) moles. These values represent the moles of each substance present in the equilibrium mixture.

Key Concepts

Equilibrium Calculation

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal. At equilibrium, the concentrations of all reactants and products remain constant over time. To calculate equilibrium, you need to know the initial concentrations and changes that happen during the reaction. In our example, we started with 2.0 moles of carbon disulfide (\(\mathrm{CS}_{2}\)) and 4.0 moles of chlorine (\(\mathrm{Cl}_{2}\)). At equilibrium, 0.30 moles of tetrachloromethane (\(\mathrm{CCl}_{4}\)) are measured. Understanding how these initial amounts change is crucial for calculating the equilibrium conditions. The equilibrium calculation uses these changes in mole numbers to find the final amounts of each reactant and product. Keeping track of these changes helps us set up the reaction table properly, a key part of finding equilibrium points.

Balanced Chemical Equation

The foundation for understanding a chemical reaction is a balanced chemical equation. For the reaction involving carbon disulfide and chlorine, the balanced equation is \( \mathrm{CS}_{2} + 3 \mathrm{Cl}_{2} \rightleftharpoons \mathrm{S}_{2}\mathrm{Cl}_{2} + \mathrm{CCl}_{4} \). A balanced equation satisfies the law of conservation of mass, indicating that atoms are neither created nor destroyed in a chemical reaction. Each side of the equation must have the same number of each type of atom. In this case, 1 mole of \( \mathrm{CS}_{2} \) reacts with 3 moles of \( \mathrm{Cl}_{2} \), producing 1 mole each of \( \mathrm{S}_{2}\mathrm{Cl}_{2} \) and \( \mathrm{CCl}_{4} \). This stoichiometric relationship is essential for determining the amount of each substance involved in the reaction.

Mole Concept

The mole is a standard scientific unit for measuring large quantities of tiny entities like atoms or molecules. It's defined by Avogadro's number, which is \( 6.022 \times 10^{23} \). In chemistry, the mole concept is crucial for converting between atomic-scale measurements to macroscopic amounts of substance. When looking at a balanced chemical equation, the coefficients tell you how many moles of each substance are involved. For example, the reaction \( \mathrm{CS}_{2} + 3 \mathrm{Cl}_{2} \rightleftharpoons \mathrm{S}_{2}\mathrm{Cl}_{2} + \mathrm{CCl}_{4} \) involves moles in a 1:3:1:1 ratio. This means for 2.0 moles of \( \mathrm{CS}_{2} \), you'll need 3 times that amount in moles of \( \mathrm{Cl}_{2} \), which dictates how much product can be formed.

Reaction Stoichiometry

Stoichiometry deals with the quantitative aspects of chemical reactions. It involves calculations based on a balanced chemical equation to predict the amounts of reactants and products. Known as the 'mathematics of chemistry,' stoichiometry allows chemists to plan and optimize reactions. Using our example reaction \( \mathrm{CS}_{2} + 3 \mathrm{Cl}_{2} \rightleftharpoons \mathrm{S}_{2}\mathrm{Cl}_{2} + \mathrm{CCl}_{4} \), for every mole of \( \mathrm{CCl}_{4} \) formed, 0.30 moles of \( \mathrm{CS}_{2} \) are consumed and 0.90 moles of \( \mathrm{Cl}_{2} \) are used. This reflects the equation's 1:3:1:1 stoichiometry. Stoichiometry provides the bridge for calculating reactant consumption and product formation in meaningful quantities.