Problem 98 - Chemistry The Molecular Science

Question

Mustard gas was used in chemical warfare in World War I. Mustard gas can be produced according to this reaction: $$ \mathrm{SCl}_{2}(\mathrm{~g})+2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g}) \rightleftharpoons \mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2}(\mathrm{~g}) $$ An evacuated 5.00 -L flask at \(20.0^{\circ} \mathrm{C}\) is filled with \(0.258 \mathrm{~mol} \mathrm{SCl}_{2}\) and \(0.592 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{4}\) and sealed. After equilibrium is established, 0.0349 mol mustard gas is present. (a) Calculate the partial pressure of each gas at equilibrium. (b) Calculate \(K_{\mathrm{c}}\) at \(20.0^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

At equilibrium, partial pressures are 1.08 atm (SCl₂), 2.50 atm (C₂H₄), and 0.168 atm (mustard gas). The equilibrium constant \(K_c\) is 0.145.

Step 1: Identify Initial Moles

We begin by noting the initial moles of the reactants given in the problem. Initially, we have 0.258 mol of \( ext{SCl}_2\) and 0.592 mol of \( ext{C}_2 ext{H}_4\) in a 5.00 L flask.

Step 2: Determine Change in Moles at Equilibrium

Since at equilibrium, 0.0349 mol of mustard gas \( ext{S}( ext{CH}_2 ext{CH}_2 ext{Cl})_2\) is present, we need to find the change in moles of \( ext{SCl}_2\) and \( ext{C}_2 ext{H}_4\). By stoichiometry, this indicates a change of 0.0349 mol of \( ext{SCl}_2\) and 2 x 0.0349 mol of \( ext{C}_2 ext{H}_4\).

Step 3: Calculate Equilibrium Moles of Reactants

Subtract the change from the initial moles to find equilibrium moles:- \( ext{SCl}_2\) remaining = 0.258 - 0.0349 = 0.2231 mol- \( ext{C}_2 ext{H}_4\) remaining = 0.592 - 0.0698 = 0.5222 mol

Step 4: Calculate Equilibrium Partial Pressures

Using the ideal gas law to find partial pressures:- \(P_{\text{SCl}_2} = \frac{nRT}{V} = \frac{0.2231 \times 0.08206 \times 293}{5.00} = 1.08\text{ atm}\)- \(P_{\text{C}_2 ext{H}_4} = \frac{nRT}{V} = \frac{0.5222 \times 0.08206 \times 293}{5.00} = 2.50\text{ atm}\)- \(P_{\text{mustard}} = \frac{0.0349 \times 0.08206 \times 293}{5.00} = 0.168\text{ atm}\)

Step 5: Establish Equilibrium Expression for Kc

The equilibrium expression for the reaction in terms of concentration is:\[ K_c = \frac{[\text{Mustard Gas}]}{[\text{SCl}_2][\text{C}_2 ext{H}_4]^2}\]We will convert pressures to concentrations using the ideal gas law.

Step 6: Calculate Equilibrium Concentrations

Convert the pressures calculated in Step 4 to concentrations:- \([\text{SCl}_2] = \frac{1.088}{0.08206 \times 293} = 0.0450 \text{ M}\)- \([\text{C}_2 ext{H}_4] = \frac{2.4987}{0.08206 \times 293} = 0.104 \text{ M}\)- \([\text{mustard gas}] = \frac{0.1682}{0.08206 \times 293} = 0.00699 \text{ M}\)

Step 7: Calculate Equilibrium Constant Kc

Substitute the equilibrium concentrations into the Kc expression:\[ K_c = \frac{0.00699}{0.0450 \times (0.104)^2} = 0.145\]

Key Concepts

Partial Pressure

Partial pressure is an important concept when dealing with gas reactions and equilibria.
It is the pressure a gas in a mixture would exert if it alone occupied the entire volume of the container.
In the context of chemical equilibrium, calculating partial pressures can help us determine how gases react over time.Let's consider the reaction of mustard gas production. Each gas involved in the reaction contributes to the overall pressure in the flask:

• **SCl₂**
• **C₂H₄**
• **Mustard Gas**

To compute partial pressures at equilibrium, you can use the formula derived from the Ideal Gas Law:\[P = \frac{nRT}{V}\]Here, **P** is the partial pressure, **n** is the number of moles of gas, **R** is the universal gas constant (0.08206 L atm/mol K), **T** is the temperature in Kelvin, and **V** is the volume of the container. This formula allows you to calculate the pressure exerted by each gas once equilibrium is reached.

Equilibrium Constant

The equilibrium constant, noted as **K_c**, gives insight into the ratio of product concentration to reactant concentrations at equilibrium.
It is essential to understanding how far a reaction proceeds.To determine **K_c** for a reaction, we first write the equilibrium expression:\[K_c = \frac{[ ext{Product(s)}]}{[ ext{Reactants}]}\]In our example, the reaction involves the conversion of SCl₂ and C₂H₄ into mustard gas. Thus, the equilibrium expression is:\[K_c = \frac{[ ext{Mustard Gas}]}{[ ext{SCl}_2][ ext{C}_2H_4]^2}\]The equilibrium concentrations of each substance are plugged into this equation to calculate **K_c**. Remember, the concentrations are calculated from the equilibrium partial pressures using the ideal gas law, to ensure we express them in molarity (M). This value is crucial to assess the relative amounts of products and reactants once the reaction reaches equilibrium.

Ideal Gas Law

The Ideal Gas Law is a key equation in chemistry, linking together pressure, volume, temperature, and moles of gas.
The law is expressed as:\[PV = nRT\]In this equation, **P** stands for pressure, **V** is volume, **n** is moles of gas, **R** is the gas constant, and **T** represents temperature in Kelvin.
Within the context of our mustard gas reaction, the Ideal Gas Law helps translate between moles of gas and pressure.**Why use the Ideal Gas Law?**

• **Translate moles to pressure:** Knowing how much pressure each reactant and product contributes at equilibrium helps compute equilibrium concentrations.
• **Link conditions:** Since gases can expand or contract based on conditions, this law helps account for such behaviors.
• **Consistency:** Allows us to maintain consistent units when calculating and comparing physical properties of gases.

Using the Ideal Gas Law simplistically, we can track changes in a chemical reaction involving gases, providing an essential tool for handling equilibrium and partial pressures.