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A researcher put \(0.400 \mathrm{~mol} \mathrm{PCl}_{3}\) and \(0.600 \mathrm{~mol} \mathrm{Cl}_{2}\) into a 5.00-L vessel at a given temperature to produce phosphorus pentachloride, \(\mathrm{PCl}_{5}\) : $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ What will be the composition of this gaseous mixture at equilibrium? \(K_{c}=25.6\) at the temperature of this experiment.

Short Answer

Expert verified
At equilibrium, \([\mathrm{PCl}_3] = 0.0289 \text{ M}\), \([\mathrm{Cl}_2] = 0.0689 \text{ M}\), and \([\mathrm{PCl}_5] = 0.0511 \text{ M}\).

Step by step solution

01

Write the equilibrium constant expression

For the given reaction \(\mathrm{PCl}_3(g) + \mathrm{Cl}_2(g) \rightleftharpoons \mathrm{PCl}_5(g)\), the equilibrium constant \(K_c\) is expressed as:\[K_c = \frac{[\mathrm{PCl}_5]}{[\mathrm{PCl}_3][\mathrm{Cl}_2]}\]where \([\mathrm{PCl}_5]\), \([\mathrm{PCl}_3]\), and \([\mathrm{Cl}_2]\) are the equilibrium concentrations of the respective species.
02

Determine initial concentrations

The initial concentrations of the reactants in the 5.00 L vessel are calculated as follows:- For \(\mathrm{PCl}_3\): \([\mathrm{PCl}_3]_0 = \frac{0.400 \text{ mol}}{5.00 \text{ L}} = 0.0800 \text{ M}\)- For \(\mathrm{Cl}_2\): \([\mathrm{Cl}_2]_0 = \frac{0.600 \text{ mol}}{5.00 \text{ L}} = 0.1200 \text{ M}\)Initially, the concentration of \(\mathrm{PCl}_5\): \([\mathrm{PCl}_5]_0 = 0\text{ M}\).
03

Set up changes in concentration

Assume the change in concentration of \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) as \(-x\), meaning they decrease by \(x\), while \(\mathrm{PCl}_5\) increases by \(x\). The concentration changes at equilibrium are:- \([\mathrm{PCl}_3] = 0.0800 - x\)- \([\mathrm{Cl}_2] = 0.1200 - x\)- \([\mathrm{PCl}_5] = x\).
04

Substitute into the equilibrium expression

Substituting the equilibrium concentrations into the \(K_c\) expression gives:\[25.6 = \frac{x}{(0.0800 - x)(0.1200 - x)}\]Solving this equation will allow us to determine the value of \(x\).
05

Solve the quadratic equation

Multiply both sides by \((0.0800 - x)(0.1200 - x)\) to clear the fraction:\[25.6(0.0800 - x)(0.1200 - x) = x\]Expanding the left-hand side, we have:\[25.6(0.00960 - 0.200x + x^2) = x\]\[0.24576 - 5.12x + 25.6x^2 = x\]Rearrange to form a quadratic equation:\[25.6x^2 - 6.12x + 0.24576 = 0\].Solve using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 25.6\), \(b = -6.12\), and \(c = 0.24576\).
06

Find roots of the quadratic equation

Calculate the discriminant:\[b^2 - 4ac = (-6.12)^2 - 4 \times 25.6 \times 0.24576 = 37.4544 - 25.19328 = 12.26112\]Thus, the roots are:\[x = \frac{6.12 \pm \sqrt{12.26112}}{2 \times 25.6}\].
07

Calculate valid value of x

Finding the solution to the quadratic equation, we have two possible \(x\) values:\[x_1 = \frac{6.12 + 3.5015}{51.2} \approx 0.1887\]\[x_2 = \frac{6.12 - 3.5015}{51.2} \approx 0.0511\]Since \(x_1 = 0.1887\) leads to negative concentrations of reactants, we use \(x_2 = 0.0511 \text{ M}\).
08

Determine equilibrium concentrations

Substitute \(x = 0.0511\) into the expressions for equilibrium concentrations:- \([\mathrm{PCl}_3] = 0.0800 - 0.0511 = 0.0289 \text{ M}\)- \([\mathrm{Cl}_2] = 0.1200 - 0.0511 = 0.0689 \text{ M}\)- \([\mathrm{PCl}_5] = 0.0511 \text{ M}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Concentrations
In a chemical reaction, equilibrium concentrations are the amounts of reactants and products present when the reaction reaches equilibrium. At this stage, the concentrations do not change over time despite the reaction still occurring in both forward and reverse directions. The dynamic balance ensures that the rate of the forward reaction equals the rate of the reverse reaction.

When a researcher sets up the reaction in a 5-liter vessel with certain moles of reactants, initially, we calculate the concentrations before any product is formed. These are the initial concentrations. As the reaction progresses towards equilibrium, the concentrations begin to change.

To find the equilibrium concentrations, we assume a variable change, represented by \(x\), based on stoichiometry. For instance, in the reaction \(\text{PCl}_3 + \text{Cl}_2 \rightleftharpoons \text{PCl}_5\), both \(\text{PCl}_3\) and \(\text{Cl}_2\) decrease by \(x\), while \(\text{PCl}_5\) forms and increases by \(x\). Thus, at equilibrium, concentrations are adjusted by this shift to solve for \(x\) and substituting back to find each equilibrium value.
Equilibrium Constant
The equilibrium constant, \(K_c\), is fundamental in predicting the direction and extent of a chemical reaction at a given temperature. It reflects the proportion between product and reactant concentrations at equilibrium and is represented by a specific formula derived from the balanced chemical equation.

For the reaction \(\text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g)\), \(K_c\) is given by:
  • \[K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]}\]
This relationship indicates that the larger \(K_c\), the more products favored at equilibrium, signaling a shift towards more product formation. In this instance, \(K_c = 25.6\) shows that the formation of \(\text{PCl}_5\) is quite favorable compared to its decomposition into \(\text{PCl}_3\) and \(\text{Cl}_2\).

Understanding \(K_c\) allows us to predict if a reaction will go to completion, or if significant reactants will remain in the system at equilibrium. A value close to 1 indicates that neither reactants nor products are heavily favored, indicating a balance.
Reaction Quotient
The Reaction Quotient, \(Q_c\), serves as an exploratory tool to predict the direction in which a chemical reaction will proceed to reach equilibrium. It is calculated in the same way as \(K_c\), but it can be evaluated at any point in the reaction, not just equilibrium.

The formula for the reaction quotient for the reaction \(\text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g)\) is:
  • \[Q_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]}\]
By comparing \(Q_c\) with \(K_c\):
  • If \(Q_c < K_c\), the reaction shifts to the right, favoring product formation until equilibrium is reached.
  • If \(Q_c > K_c\), the reaction shifts to the left, favoring reactant formation.
By analyzing where the current system stands by using \(Q_c\), clarity is gained on whether more of the products or reactants are required to achieve balance, ultimately helping us predict changes and states in systems moving towards equilibrium.

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Most popular questions from this chapter

The equilibrium constant \(K_{c}\) for the synthesis of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\) $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ is \(4.3\) at \(250^{\circ} \mathrm{C}\) and \(1.8\) at \(275^{\circ} \mathrm{C}\). Is this reaction endothermic or exothermic?

The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ equals 49 at \(230^{\circ} \mathrm{C}\). If \(0.400 \mathrm{~mol}\) each of phosphorus trichloride and chlorine are added to a 4.0-L reaction vessel, what is the equilibrium composition of the mixture at \(230^{\circ} \mathrm{C}\) ?

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Suppose liquid water and water vapor exist in equilibrium in a closed container. If you add a small amount of liquid water to the container, how does this affect the amount of water vapor in the container? If, instead, you add a small amount of water vapor to the container, how does this affect the amount of liquid water in the container?

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