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A 4.00-L vessel contained 0.0148 mol of phosphorus trichloride, \(0.0126 \mathrm{~mol}\) of phosphorus pentachloride, and 0.0870 mol of chlorine at \(230^{\circ} \mathrm{C}\) in an equilibrium mixture. Calculate the value of \(K_{c}\) for the reaction $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$

Short Answer

Expert verified
The equilibrium constant \(K_c\) is approximately 39.14.

Step by step solution

01

Identify the Reaction and Components

The balanced chemical equation for the reaction is \(\mathrm{PCl}_3(g) + \mathrm{Cl}_2(g) \rightleftharpoons \mathrm{PCl}_5(g)\). The components involved are phosphorus trichloride \(\mathrm{PCl}_3\), chlorine \(\mathrm{Cl}_2\), and phosphorus pentachloride \(\mathrm{PCl}_5\).
02

Write the Expression for Equilibrium Constant \(K_c\)

The expression for the equilibrium constant \(K_c\) is given by:\[K_c = \frac{[\mathrm{PCl}_5]}{[\mathrm{PCl}_3][\mathrm{Cl}_2]}\]Here, \([\mathrm{PCl}_5]\), \([\mathrm{PCl}_3]\), and \([\mathrm{Cl}_2]\) represent the molar concentrations of the respective gases at equilibrium.
03

Calculate Molar Concentrations

The volume of the vessel is 4.00 L. Use the formula for concentration, \([\text{compound}] = \frac{\text{moles of compound}}{\text{volume in liters}}\), to find the molar concentrations:- \([\mathrm{PCl}_3] = \frac{0.0148 \text{ mol}}{4.00 \text{ L}} = 0.0037 \text{ M}\)- \([\mathrm{Cl}_2] = \frac{0.0870 \text{ mol}}{4.00 \text{ L}} = 0.02175 \text{ M}\)- \([\mathrm{PCl}_5] = \frac{0.0126 \text{ mol}}{4.00 \text{ L}} = 0.00315 \text{ M}\)
04

Substitute Concentrations into \(K_c\) Expression

Substitute the calculated molar concentrations into the \(K_c\) expression:\[K_c = \frac{0.00315}{0.0037 \times 0.02175}\]
05

Calculate the Value of \(K_c\)

Perform the calculation:\[K_c = \frac{0.00315}{0.000080475} \approx 39.14\]Thus, the equilibrium constant \(K_c\) for the reaction is approximately 39.14.

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Key Concepts

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

Equilibrium Constant (Kc)
The equilibrium constant, represented as \(K_c\), is a key idea in chemical equilibrium. It provides a snapshot of the relative concentrations of products and reactants at equilibrium, specifically for reactions involving gases. Each chemical reaction has its own unique \(K_c\) value, showing how much the reaction favors products or reactants when it reaches equilibrium.
To express \(K_c\), we use the formula:\[K_c = \frac{[\text{products}]}{[\text{reactants}]}\] This expression is derived from the balanced chemical equation of the reaction. In our example, \(K_c\) is represented by\[K_c = \frac{[\mathrm{PCl}_5]}{[\mathrm{PCl}_3][\mathrm{Cl}_2]}\]Here, \([\mathrm{PCl}_5], [\mathrm{PCl}_3], [\mathrm{Cl}_2]\) are the molar concentrations of phosphorus pentachloride, phosphorus trichloride, and chlorine, respectively, at equilibrium. A larger \(K_c\) means products are favored; a smaller \(K_c\) implies reactants are favored. Calculating \(K_c\) effectively involves determining these concentrations at equilibrium and plugging them into the expression.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products. In the given reaction, phosphorus trichloride \(\mathrm{PCl}_3\) and chlorine \(\mathrm{Cl}_2\) react to form phosphorus pentachloride \(\mathrm{PCl}_5\). This process is reversible and reaches a state of equilibrium.
At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This balance means the concentrations of the reactants and products remain constant over time, though the reaction hasn't stopped. An equilibrium state is dynamic, with ongoing reactions happening in both directions. Still, their effects balance each other out, so there's no net change in the concentration of reactants or products.
Understanding these basics helps in grasping why the equilibrium constant \(K_c\) is meaningful, as it quantifies the balance between reactants and products. Each chemical reaction can have different equilibrium characteristics, altering its \(K_c\). Some reactions may heavily favor products, others reactants, based on their conditions and \(K_c\) values.
Molar Concentration
Molar concentration, or molarity, is a measure of the concentration of a chemical species in a solution, typically expressed in moles per liter (M). It serves as a crucial parameter in determining the equilibrium constant \(K_c\).
To calculate molar concentration, you divide the number of moles of a solute by the volume of the solution in liters. For example, in the problem, you can calculate the concentrations for each component as follows:
  • Phosphorus trichloride \([\mathrm{PCl}_3]\) is \(0.0148\;\text{mol} / 4.00\;\text{L} = 0.0037\;\text{M}\)
  • Chlorine \([\mathrm{Cl}_2]\) is \(0.0870\;\text{mol} / 4.00\;\text{L} = 0.02175\;\text{M}\)
  • Phosphorus pentachloride \([\mathrm{PCl}_5]\) is \(0.0126\;\text{mol} / 4.00\;\text{L} = 0.00315\;\text{M}\)
Once you have these values, they can be substituted into the \(K_c\) equation. Molar concentration effectively links the amount of solute with the space it occupies, offering a clear picture of its role in the reaction and its equilibrium state. Understanding molarity is vital in comprehending how different factors affect chemical equilibrium.

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Most popular questions from this chapter

The amount of nitrogen dioxide formed by dissociation of dinitrogen tetroxide, $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ increases as the temperature rises. Is the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) endothermic or exothermic?

Sulfuryl chloride is used in organic chemistry as a chlorinating agent. At moderately high temperatures it decomposes as follows: $$ \begin{array}{r} \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) \\ \text { with } K_{c}=0.045 \text { at } 650 \mathrm{~K} . \end{array} $$ a. A sample of \(8.25 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(1.00-\mathrm{L}\) reaction vessel and heated to \(650 \mathrm{~K}\). b. What are the equilibrium concentrations of all of the species? What fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) has decomposed? c. If \(5 \mathrm{~g}\) of chlorine is inserted into the reaction vessel, what qualitative effect would this have on the fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that has decomposed?

The following equilibrium was studied by analyzing the equilibrium mixture for the amount of \(\mathrm{H}_{2} \mathrm{~S}\) produced. $$ \mathrm{Sb}_{2} \mathrm{~S}_{3}(s)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{Sb}(s)+3 \mathrm{H}_{2} \mathrm{~S}(g) $$ A vessel whose volume was \(2.50 \mathrm{~L}\) was filled with \(0.0100 \mathrm{~mol}\) of antimony(III) sulfide, \(\mathrm{Sb}_{2} \mathrm{~S}_{3},\) and \(0.0100 \mathrm{~mol}\) \(\mathrm{H}_{2}\). After the mixture came to equilibrium in the closed vessel at \(440^{\circ} \mathrm{C}\), the gaseous mixture was removed, and the hydrogen sulfide was dissolved in water. Sufficient lead(II) ion was added to react completely with the \(\mathrm{H}_{2} \mathrm{~S}\) to precipitate lead(II) sulfide, \(\mathrm{PbS}\). If \(1.029 \mathrm{~g} \mathrm{PbS}\) was obtained, what is the value of \(K_{c}\) at \(440^{\circ} \mathrm{C} ?\)

The equilibrium constant \(K_{c}\) equals 0.0952 for the following reaction at \(227^{\circ} \mathrm{C}\). $$ \mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) $$ What is the value of \(K_{p}\) at this temperature?

The equilibrium-constant expression for a reaction is $$ K_{c}=\frac{\left[\mathrm{NO}_{2}\right]^{4}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]^{2}} $$ What is the equilibrium-constant expression when the equation for this reaction is halved and then reversed?

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