Problem 118 - Chemistry The Molecular Science

Question

The equilibrium constant \(K_{\mathrm{c}}\) has a value of 3.30 at \(760 . \mathrm{K}\) for the decomposition of phosphorus pentachloride, \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (a) Calculate the equilibrium concentrations of all three species arising from the decomposition of \(0.75 \mathrm{~mol} \mathrm{PCl}_{5}\) in a sealed \(5.00-\mathrm{L}\) vessel. (b) Calculate the equilibrium concentrations of all three species resulting from an initial mixture of \(0.75 \mathrm{~mol} \mathrm{PCl}_{5}\) and \(0.75 \mathrm{~mol} \mathrm{PCl}_{3}\) in a sealed 5.00-L vessel.

Step-by-Step Solution

Answer

(a) [PCl5] = 0.03 M, [PCl3] = 0.12 M, [Cl2] = 0.12 M; (b) [PCl5] = 0.04 M, [PCl3] = 0.26 M, [Cl2] = 0.11 M.

Step 1: Write the equilibrium expression

For the reaction \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g}) + \mathrm{Cl}_2(\mathrm{~g})\), the equilibrium constant expression \( K_c \) is given by: \[ K_c = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]} \] where \([\mathrm{PCl}_3], [\mathrm{Cl}_2], \text{and } [\mathrm{PCl}_5]\) are the equilibrium concentrations in mol/L of the respective species.

Step 2: Calculate initial concentrations

For Part (a), calculate the initial concentration of \( \mathrm{PCl}_5 \): \[ \text{Initial moles of PCl}_5 = 0.75\, \text{mol} \]\[ \text{Initial concentration of PCl}_5 = \frac{0.75}{5.00} = 0.15\, \text{M} \]For Part (b), initial concentrations are:\[ [\mathrm{PCl}_5] = \frac{0.75}{5.00} = 0.15\, \text{M} \]\[ [\mathrm{PCl}_3] = \frac{0.75}{5.00} = 0.15\, \text{M} \].

Step 3: Set up the ICE table for Part (a)

\[\begin{array}{c|c|c|c} & \mathrm{PCl}_5 & \mathrm{PCl}_3 & \mathrm{Cl}_2 \\hline\text{I (Initial)} & 0.15 & 0 & 0 \\text{C (Change)} & -x & +x & +x \\text{E (Equilibrium)} & 0.15-x & x & x \\end{array}\]

Step 4: Solve for x using the equilibrium expression Part (a)

Substitute the equilibrium concentrations into the equilibrium expression: \[ K_c = \frac{x^2}{0.15-x} \]\[ 3.30 = \frac{x^2}{0.15-x} \]Solve the quadratic equation:\[ 3.30 (0.15-x) = x^2 \]\[ 0.495 - 3.30x = x^2 \]Rearrange to form: \[ x^2 + 3.30x - 0.495 = 0 \]Solving using the quadratic formula: \[ x \approx 0.12 \text{ M} \].

Step 5: Calculate equilibrium concentrations Part (a)

Substitute \(x = 0.12\) into the equilibrium concentrations:\[ [\mathrm{PCl}_5] = 0.15 - 0.12 = 0.03\, \text{M} \]\[ [\mathrm{PCl}_3] = 0.12\, \text{M} \]\[ [\mathrm{Cl}_2] = 0.12\, \text{M} \]

Step 6: Set up the ICE table for Part (b)

\[ \begin{array}{c|c|c|c} & \mathrm{PCl}_5 & \mathrm{PCl}_3 & \mathrm{Cl}_2 \\hline\text{I (Initial)} & 0.15 & 0.15 & 0 \\text{C (Change)} & -x & +x & +x \\text{E (Equilibrium)} & 0.15-x & 0.15+x & x \\end{array} \]

Step 7: Solve for x using the equilibrium expression Part (b)

Substitute the equilibrium concentrations into the equilibrium expression:\[ K_c = \frac{(0.15+x)x}{0.15-x} \]\[ 3.30 = \frac{(0.15+x)x}{0.15-x} \]Solving the resulting equation:\[ 0.495 - 3.30x = 0.15x + x^2 \]Rearrange to form:\[ x^2 + 3.45x - 0.495 = 0 \]Solve with the quadratic formula: \[ x \approx 0.11 \text{ M} \].

Step 8: Calculate equilibrium concentrations Part (b)

Substitute \(x = 0.11\) into the equilibrium concentrations:\[ [\mathrm{PCl}_5] = 0.15 - 0.11 = 0.04\, \text{M} \]\[ [\mathrm{PCl}_3] = 0.15 + 0.11 = 0.26\, \text{M} \]\[ [\mathrm{Cl}_2] = 0.11\, \text{M} \]

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the backward reaction.
This means that the concentrations of reactants and products remain constant over time. Although they are not equal to each other, they maintain a stable ratio.

Understanding chemical equilibrium is crucial for predicting how much product will form under given conditions.
Equilibrium is dynamic, meaning the molecules continue to react, but their concentrations remain unchanged.
This does not mean the reaction has stopped, but rather that it has reached a balanced state.

• The equilibrium constant, denoted as \(K_c\), is a value that expresses this balance. It is calculated using concentrations of the products and reactants at equilibrium.
• A large \(K_c\) value (>1) implies that, at equilibrium, products are favored, whereas a small \(K_c\) value (<1) suggests that reactants are favored.

Mastering these concepts helps to solve problems related to chemical equilibria effectively.

ICE Table

An ICE table is a helpful tool used in chemistry to calculate equilibrium concentrations. "ICE" stands for Initial, Change, and Equilibrium.
These tables allow us to visually track changes in concentration throughout a reaction, not only simplifying calculations but also giving us a clear picture of how species change and reach equilibrium.

Setting up an ICE table involves the following steps:

• **Initial:** List the initial concentrations (or pressures) of the reactants and products before the reaction has reached equilibrium. This is generally straightforward as these are usually given in exercises.
• **Change:** Define the change in concentrations using variables like \(-x\) or \(+x\), depending on whether species are being consumed or formed.
• **Equilibrium:** Express the equilibrium concentrations in terms of these changes. At this stage, the equilibrium constant expression can be applied, substituting these expressions to solve for the unknowns.

Using an ICE table makes manipulating and solving for unknowns easier, especially when combined with equations like the quadratic formula.

Quadratic Equation

In chemistry, especially when dealing with equilibrium problems, solving quadratic equations is often necessary.
These equations typically arise when substituting expressions from an ICE table into the equilibrium constant expression.

A quadratic equation is generally in the form \(ax^2 + bx + c = 0\). Solving it can seem challenging, but it boils down to using a straightforward formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This quadratic formula provides solutions for \(x\), which corresponds to changes in concentrations at equilibrium.

Important points to remember when working with quadratic equations:

• Ensure all terms are correctly arranged into a quadratic form before attempting to solve it.
• Use of the quadratic formula requires calculating the discriminant \((b^2 - 4ac)\). If it's positive, two real and distinct solutions exist. If zero, one real solution exists (a repeated root), and if negative, the solutions are complex.

Understanding and applying the quadratic formula is crucial in determining exact concentrations for equilibrium problems.

Reaction Stoichiometry

Reaction stoichiometry involves using ratios from a balanced chemical equation to relate quantities of reactants and products.
In equilibrium problems, stoichiometry is indispensable for determining how changes in one part of the reaction affect others.

When writing balanced equations and using stoichiometry:

• Ensure that each side of the reaction has the same number of each type of atom. This reflects the conservation of mass.
• Stoichiometric coefficients in the balanced equations are used as multipliers for mole calculations.
• When we discuss the "change" part in ICE tables, stoichiometry allows us to relate these changes via the coefficients of the balanced chemical equation.

Example: For the decomposition of \( \mathrm{PCl}_5 (\mathrm{g}) \to \mathrm{PCl}_3 (\mathrm{g}) + \mathrm{Cl}_2 (\mathrm{g}) \), each mole of \( \mathrm{PCl}_5\) decomposed produces one mole of \( \mathrm{PCl}_3 \) and one mole of \( \mathrm{Cl}_2 \).
Having a solid grasp of stoichiometric principles enhances your ability to predict and calculate accurate outcomes in chemical reactions.