Question

A toxicologist studying mustard gas, \(\mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2},\) a blistering agent, prepares a mixture of \(0.675 M \mathrm{SCl}_{2}\) and \(0.973 \mathrm{M}\) \(\mathrm{C}_{2} \mathrm{H}_{4}\) and allows it to react at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) $$ \mathrm{SCl}_{2}(g)+2 \mathrm{C}_{2} \mathrm{H}_{4}(g) \rightleftharpoons \mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2}(g) $$ At equilibrium, \(\left[\mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2}\right]=0.350 \mathrm{M}\). Calculate \(K_{\mathrm{p}}\)

Short answer

The equilibrium constant \(K_p\) is approximately 0.025.

Step-by-step solution

- Write the Balanced Equation

The balanced chemical equation is: \[ \mathrm{SCl}_{2}(g) + 2 \mathrm{C}_{2} \mathrm{H}_{4}(g) \rightleftharpoons \mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2}(g) \]

- Calculate the Initial Concentrations

Given initial concentrations are \[ \mathrm{SCl}_{2} = 0.675 \mathrm{M} \quad \text{and} \quad \mathrm{C}_{2} \mathrm{H}_{4} = 0.973 \mathrm{M} \]

- Determine Changes in Concentrations

At equilibrium, the concentration of \( \mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2} \) is given as 0.350 M. This means 0.350 M of \( \mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2} \) is formed, so the change in concentration for \( \mathrm{SCl}_{2} \) and \( \mathrm{C}_{2} \mathrm{H}_{4} \) can be calculated accordingly.

- Calculate the Change in Reactant Concentrations

The change in concentration for \( \mathrm{SCl}_{2} \) is: \[ \Delta [\mathrm{SCl}_{2}] = -0.350 \text{ M} \] The change for \( \mathrm{C}_{2} \mathrm{H}_{4} \) is: \[ \Delta [\mathrm{C}_{2} \mathrm{H}_{4}] = -2 \times 0.350 = -0.700 \text{ M} \]

- Calculate Equilibrium Concentrations

The equilibrium concentration for \( \mathrm{SCl}_{2} \) and \( \mathrm{C}_{2} \mathrm{H}_{4} \) can be calculated as: \[ [\mathrm{SCl}_{2}]_{eq} = 0.675 - 0.35 = 0.325 \text{ M} \] \[ [\mathrm{C}_{2} \mathrm{H}_{4}]_{eq} = 0.973 - 0.70 = 0.273 \text{ M} \]

- Write the Expression for the Equilibrium Constant \( K_c \)

The expression for the equilibrium constant in terms of concentration is: \[ K_c = \frac{[\mathrm{S}(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl})_2]}{[\mathrm{SCl}_{2}][\mathrm{C}_{2} \mathrm{H}_{4}]^2} \]

- Substitute Equilibrium Concentrations and Solve for \( K_c \)

Substitute the equilibrium concentrations into the \( K_c \) expression: \[ K_c = \frac{0.350}{0.325 \times (0.273)^2} \] Calculate \( K_c \): \[ K_c = \frac{0.350}{0.325 \times 0.074529} \approx 14.55 \]

- Relate \( K_c \) to \( K_p \)

Use the equation that relates \( K_c \) to \( K_p \): \[ K_p = K_c(RT)^{\Delta n} \] Here, \( \Delta n = (\text{moles of gas product}) - (\text{moles of gas reactants}) \), and \( \Delta n \) for this reaction is: \[ \Delta n = 1 - 3 = -2 \]

- Calculate \( RT \) at Given Conditions

Given temperature is 20.0°C, which is 293.15 K. The gas constant \( R \) is 0.0821 L·atm/(K·mol).\[ RT = 0.0821 \times 293.15 = 24.06 \text{ atm·L/mol} \]

- Substitute and Calculate \( K_p \)

Substitute the values into the \( K_p \) expression: \[ K_p = 14.55 \times (24.06)^{-2} \approx 0.025 \]

Key concepts

chemical equilibrium

Chemical equilibrium occurs in a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of products and reactants remain constant. It's a dynamic state where reactants and products continually convert into each other, but their overall quantity stays the same.
To illustrate, imagine two people passing a ball back and forth without any net change in the number of passes. This reflects how, in equilibrium, the amount of each substance doesn't change even though the molecules keep reacting.
Understanding this balance is key when calculating equilibrium constants, as seen in the provided exercise. Here, the reaction reaches equilibrium and specific concentrations are measured to determine the equilibrium constant.

reaction kinetics

Reaction kinetics involves the study of the rates of chemical processes. It helps us understand how quickly a reaction reaches equilibrium.
The rate at which reactants turn into products and vice versa depends on several factors, including concentration, temperature, and the presence of catalysts.
In our sample exercise, kinetics lets us determine how the concentrations of the reacted and products change over time to reach equilibrium. This involves monitoring the forward and reverse reaction rates until they become equal.
A mastery of reaction kinetics paves the way for understanding more complex concepts like activation energy and reaction mechanisms.

equilibrium constant Kc and Kp

The equilibrium constant, expressed as either \( K_c \) or \( K_p \), quantifies the concentrations of reactants and products at equilibrium. \( K_c \) uses molar concentrations, whereas \( K_p \) involves partial pressures.
The expression for \( K_c \) is given by the formula:
\[ K_c = \frac{[\text{Products}]}{[\text{Reactants}]} \]
For gases, we can relate \( K_c \) to \( K_p \) using the ideal gas law in the equation:
\[ K_p = K_c(RT)^{\triangle n} \]
Here, \( \triangle n \) represents the change in the number of moles of gas from reactants to products.
In the example exercise, calculating \( K_p \) required first finding \( K_c \) and then using the provided temperature to adjust for gas behavior.

stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in chemical reactions. It allows us to make accurate calculations based on balanced chemical equations.
In the presented exercise, stoichiometry helps us determine how much of each reactant is consumed and how much of each product is produced.
This involves using molar ratios derived from the balanced equation. For instance, the reaction equation provided:
\[ \text{SCl}_{2}(g) + 2 \text{C}_2 \text{H}_4(g) \rightleftharpoons \text{S(CH}_2\text{CH}_2\text{Cl})_2(g) \]
shows a 1:2:1 ratio. By knowing one concentration change, we can easily find the others through stoichiometry, simplifying the calculation of equilibrium concentrations and constants.