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Antimony(V) chloride, \(\mathrm{SbCl}_{5}\), decomposes on heating to give antimony(III) chloride, \(\mathrm{SbCl}_{3},\) and chlorine. $$ \mathrm{SbCl}_{5}(g) \rightleftharpoons \mathrm{SbCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ A closed \(3.50-\mathrm{L}\) vessel initially contains \(0.0125 \mathrm{~mol} \mathrm{SbCl}_{5}\) What is the total pressure at \(248^{\circ} \mathrm{C}\) when equilibrium is achieved? The value of \(K_{c}\) at \(248^{\circ} \mathrm{C}\) is \(2.50 \times 10^{-2}\).

Short Answer

Expert verified
The total pressure at equilibrium is approximately 0.169 atm.

Step by step solution

01

Identify Initial Conditions

The closed vessel has a volume of \(3.50\, \text{L}\) and initially contains \(0.0125\, \text{mol}\) of \(\text{SbCl}_5\). There is no \(\text{SbCl}_3\) or \(\text{Cl}_2\) initially present.
02

Set Up the Reaction Table

Define \(x\) as the moles of \(\text{SbCl}_5\) that decompose. At equilibrium, moles of \(\text{SbCl}_5 = 0.0125 - x\), \(\text{SbCl}_3 = x\), and \(\text{Cl}_2 = x\).
03

Use Equilibrium Constant Expression

The expression for \(K_c\) is\[K_c = \frac{[\text{SbCl}_3][\text{Cl}_2]}{[\text{SbCl}_5]}\]= \(2.50 \times 10^{-2}\). Convert moles to concentration (moles/L) using the vessel volume.
04

Substitute Concentrations into Kc Expression

Substitute the concentrations into the expression:\[2.50 \times 10^{-2} = \frac{x \cdot x}{(0.0125 - x)} \cdot \frac{1}{3.50}\]Solve the quadratic equation for \(x\).
05

Solve the Quadratic Equation

The quadratic equation is:\[x^2 = 2.50 \times 10^{-2} \times (0.0125 - x) \times 3.50\]Solving yields \(x \approx 0.001247 \text{ mol/L}\).
06

Calculate Total Moles at Equilibrium

Total moles are \(0.0125 - x + x + x = 0.0125 + x\). Given \(x = 0.001247\), total moles = \(0.0125 + 0.001247 = 0.013747\, \text{mol}\).
07

Apply Ideal Gas Law for Total Pressure

Use the ideal gas law formula: \(PV = nRT\). Convert temperature to Kelvin: \(248^\circ \text{C} = 521 \text{K}\).\[P = \frac{nRT}{V}\]Where \(n = 0.013747\, \text{mol}, \ R = 0.0821 \text{ L atm/mol K}\), and \(V = 3.50 \text{L}\).
08

Calculate Total Pressure

Substitute the values into the equation:\[P = \frac{0.013747 \times 0.0821 \times 521}{3.50} \approx 0.169 \text{ atm}\]Thus, the total pressure at equilibrium is approximately 0.169 atm.

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Key Concepts

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

Antimony(V) chloride decomposition
Antimony(V) chloride, denoted as \( \text{SbCl}_5 \), undergoes thermal decomposition. This process transforms \( \text{SbCl}_5 \) into antimony(III) chloride, \( \text{SbCl}_3 \), and chlorine gas, \( \text{Cl}_2 \), when heated. The chemical equation representing this reversible reaction is:
  • \( \text{SbCl}_5 (g) \rightleftharpoons \text{SbCl}_3 (g) + \text{Cl}_2 (g) \)
Understanding this decomposition is crucial as it forms the basis of achieving chemical equilibrium in the reaction system.
This reaction takes place in a gaseous state within a closed vessel, which is essential because it allows gases to reach equilibrium without losing any material to the surroundings.
This setup also ensures that the system’s volume remains constant, which is important for calculating concentrations and pressures during the reaction.
Equilibrium constant (Kc)
The equilibrium constant, \( K_c \), is a crucial part of understanding chemical equilibria. It is used to quantify the concentrations of reactants and products at equilibrium in a chemical reaction. For the decomposition of \( \text{SbCl}_5 \), the expression for \( K_c \) is given by:
  • \( K_c = \frac{[\text{SbCl}_3][\text{Cl}_2]}{[\text{SbCl}_5]} \)
Here,
  • \([\text{SbCl}_3]\), \([\text{Cl}_2]\), and \([\text{SbCl}_5]\) represent the molar concentrations of each substance.
Given that \( K_c \) is \( 2.50 \times 10^{-2} \), this tells us that at 248°C, the system favors the decomposition of \( \text{SbCl}_5 \), since a lower value of \( K_c \) suggests the reactants are still significantly present at equilibrium.
Calculating the equilibrium position involves setting and solving equations based on the initial amounts and changes (using an 'ICE' - Initial, Change, Equilibrium - table), to find the concentrations of each component at equilibrium.
Ideal Gas Law
The Ideal Gas Law is a fundamental principle used in chemistry to relate the state variables of a gas. The formula is given by:
  • \( PV = nRT \)
Where
  • \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
In the context of the antimony chloride decomposition, it enables the calculation of the total pressure in the vessel at equilibrium. After determining the amount of decomposition that occurs, the Ideal Gas Law is used to calculate the final pressure by substituting the equilibrium moles and system conditions into the equation.
This application highlights the interdependence of gas properties, such as volume and temperature, with the pressure and quantity of gas, which are modulated to maintain equilibrium under defined conditions.
Phase equilibrium
Phase equilibrium describes the balance between different phases of matter, such as solid, liquid, and gas, within a system. In chemical reactions like the decomposition of \( \text{SbCl}_5 \), phase equilibrium is crucial to ensure that all reactants and products can exist and interact.
  • Here, the entire reaction occurs in the gas phase, where variables such as pressure and volume play a decisive role.
  • This kind of equilibrium is particularly sensitive to temperature and pressure changes, fundamental in processes occurring at high temperatures, like 248°C in this case.
Achieving phase equilibrium in closed systems helps in maintaining the consistency of the reaction environment, thereby enabling the calculation of equilibrium constants and application of the Ideal Gas Law.
Understanding phase equilibrium is essential for predicting the behavior of reactions under different conditions, which aids in controlling industrial chemical processes.

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Most popular questions from this chapter

The major constituents of the atmosphere are nitrogen, \(\mathrm{N}_{2}\), and oxygen, \(\mathrm{O}_{2}\). Dry atmospheric air at a pressure of 1.000 atm has a partial pressure of \(\mathrm{N}_{2}\) of 0.781 atm and a partial pressure of \(\mathrm{O}_{2}\) of 0.209 atm. At high temperatures, \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) react to produce nitrogen monoxide (nitric oxide), NO. $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose a sample of dry air is raised to \(2127^{\circ} \mathrm{C}\) in a hot flame and then rapidly cooled to fix the amount of \(\mathrm{NO}\) formed. If \(K_{p}\) for this reaction at this temperature is 0.0025 , how many grams of NO would be produced from \(100.0 \mathrm{~g}\) of dry air?

The reaction $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ has \(K_{p}\) equal to 6.55 at \(627^{\circ} \mathrm{C}\). What is the value of \(K_{c}\) at this temperature?

A chemist put 1.18 mol of substance \(A\) and 2.85 mol of substance \(\mathrm{B}\) into a 10.0 - \(\mathrm{L}\) flask, which she then closed. A and B react by the following equation: $$ \mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons 3 \mathrm{C}(g)+\mathrm{D}(g) $$ She found that the equilibrium mixture at \(25^{\circ} \mathrm{C}\) contained \(0.376 \mathrm{~mol}\) of \(\mathrm{D}\). How many moles of \(\mathrm{B}\) are in the flask at equilibrium at \(25^{\circ} \mathrm{C} ?\) a. \(2.47 \mathrm{~mol}\) b. \(3.60 \mathrm{~mol}\) c. \(2.52 \mathrm{~mol}\) d. \(2.10 \mathrm{~mol}\) e. \(2.41 \mathrm{~mol}\)

For the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ show that $$ K_{c}=K_{p}(R T)^{2} $$ Do not use the formula \(K_{p}=K_{c}(R T)^{\Delta n}\) given in the text. Start from the fact that \(P_{i}=[i] R T,\) where \(P_{i}\) is the partial pressure of substance \(i\) and \([i]\) is its molar concentration. Substitute into \(K_{c}\)

Iodine monobromide, IBr, occurs as brownish-black crystals that vaporize with decomposition: $$ 2 \mathrm{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) $$ The equilibrium constant \(K_{c}\) at \(100^{\circ} \mathrm{C}\) is 0.026 . If \(0.010 \mathrm{~mol}\) \(\mathrm{IBr}\) is placed in a 1.0 - \(\mathrm{L}\) vessel at \(100^{\circ} \mathrm{C}\), what are the moles of substances at equilibrium in the vapor?

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