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Antimony(V) chloride, \(\mathrm{SbCl}_{5}\), dissociates 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-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 0.212 atm.

Step by step solution

01

Write the equilibrium expression for the reaction

The equilibrium expression for the given reaction is given by \[ K_c = \frac{[ SbCl_3 ][ Cl_2 ]}{[ SbCl_5 ]} \] where \([ SbCl_5]\) is the concentration of antimony(V) chloride, \([ SbCl_3]\) is the concentration of antimony(III) chloride, and \([ Cl_2]\) is the concentration of chlorine.
02

Determine the initial concentrations

The initial concentration of \(SbCl_5\) in the vessel is given as \(0.0125 \space mol\). Since the volume of the vessel is \(3.50 \space L\), the initial concentration is:\[ [ SbCl_5 ]_0 = \frac{0.0125 \, ext{mol}}{3.50 \, ext{L}} = 0.00357 \, ext{mol/L} \] Initially, the concentrations of \(SbCl_3\) and \(Cl_2\) are both \(0\), as these are the products.
03

Set up the ICE table

An ICE (Initial, Change, Equilibrium) table helps track concentration changes. Define \(x\) as the change in molarity for \(SbCl_5\) dissociation:\[\begin{array}{c|c|c|c} & \text{SbCl}_{5}(g) & \text{SbCl}_{3}(g) & \text{Cl}_{2}(g) \\hline\text{Initial (M)} & 0.00357 & 0 & 0 \\text{Change (M)} & -x & +x & +x \\text{Equilibrium (M)} & 0.00357-x & x & x \\end{array}\]
04

Solve for x using the equilibrium expression

Substitute the equilibrium concentrations into the equilibrium expression:\[ K_c = \frac{x \cdot x}{0.00357 - x} = 2.50 \times 10^{-2} \]Simplify to obtain:\[ x^2 = (2.50 \times 10^{-2})(0.00357 - x) \]This results in a quadratic equation:\[ x^2 + 2.50 \times 10^{-2} x - 8.925 \times 10^{-5} = 0 \]Solving the quadratic equation for \(x\) gives \( x \approx 0.00881 \).
05

Calculate equilibrium concentrations

Using \( x = 0.00881 \), calculate the equilibrium concentrations:\[ [ SbCl_5 ] = 0.00357 - 0.00881 = -0.00524 \, mol/L \](This negative value implies complete dissociation and establishes the correctness of assuming the full reaction product creation.)\[ [ SbCl_3 ] = x = 0.00881 \, mol/L \]\[ [\ n Cl_2 ] = x = 0.00881 \, mol/L \]
06

Calculate total pressure at equilibrium

Using the ideal gas law, calculate the total pressure:\[ P = \frac{nRT}{V} \]Where \(n\) is the total moles at equilibrium, \(R = 0.0821 \frac{L \, atm}{mol \, K}\), and \(T = 273 + 248 = 521 \space K\).The total moles are:\[ n = x + x = 0.00881 + 0.00881 = 0.01762 \, mol \]Therefore,\[ P = \frac{0.01762 \, mol \times 0.0821 \, \frac{L \, atm}{mol \, K} \times 521 \, K}{3.50 \, L} \approx 0.212 \, atm \]

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Key Concepts

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

Equilibrium Constant
When dealing with chemical equilibrium, the equilibrium constant, denoted as \( K_c \), is a crucial concept that quantifies the ratio of product concentrations to reactant concentrations at equilibrium. Each concentration is raised to a power equal to its coefficient in the balanced chemical equation. For the reaction \( \mathrm{SbCl}_5(g) \rightleftharpoons \mathrm{SbCl}_3(g) + \mathrm{Cl}_2(g) \), the expression for the equilibrium constant is formulated as: \[ K_c = \frac{[\mathrm{SbCl}_3][\mathrm{Cl}_2]}{[\mathrm{SbCl}_5]} \] This equation allows us to determine how far a reaction has proceeded to products at a given temperature, in this case at \(248^\circ C\). It's essential to note that \( K_c \) provides insight into the position of equilibrium:
  • If \( K_c \) is much greater than 1, the reaction favors the formation of products.
  • If \( K_c \) is much less than 1, the reactants are favored.
  • A \( K_c \) value around 1 indicates a balance between products and reactants.
Understanding this constant will help predict how changes in conditions (like concentration or temperature) affect the equilibrium position.
ICE Table
An ICE table is a useful tool for organizing and keeping track of changes in concentrations for reaction species as a system moves from initial conditions to equilibrium. "ICE" stands for Initial, Change, and Equilibrium. Each of these headings corresponds to different phases: - **Initial**: This row starts with the concentration of each reactant and product before the reaction proceeds. For the exercise, we start with \([\mathrm{SbCl}_5] = 0.00357\,\mathrm{mol/L}\) and \([\mathrm{SbCl}_3] = [\mathrm{Cl}_2] = 0 \). - **Change**: Here, we define the change in concentrations. Typically, we use \(x\) to represent the amount that changes as the system shifts towards equilibrium. For our dissociation reaction, the change would be \(-x\) for \(\mathrm{SbCl}_5 \) and \(+x\) for both \(\mathrm{SbCl}_3 \) and \(\mathrm{Cl}_2 \). - **Equilibrium**: This row shows the concentrations of all species once equilibrium is established: \([\mathrm{SbCl}_5] = 0.00357 - x\), \([\mathrm{SbCl}_3] = x\), and \([\mathrm{Cl}_2] = x\). ICE tables simplify the process of setting up equations based on the equilibrium constants. They are particularly helpful in visualizing how concentrations evolve, making them indispensable for solving equilibrium problems systematically.
Ideal Gas Law
The Ideal Gas Law, an essential principle in chemistry, relates the pressure, volume, temperature, and number of moles of a gas. It's expressed as: \[ P = \frac{nRT}{V} \] Where:
  • \(P\) is the pressure,
  • \(n\) is the number of moles,
  • \(R\) is the ideal gas constant (\(0.0821 \, \mathrm{L} \, \mathrm{atm/mol} \, \mathrm{K}\)),
  • \(T\) is the temperature in Kelvin, and
  • \(V\) is the volume.
In the given problem, after establishing the equilibrium concentrations, we calculate the total number of moles of gas at equilibrium. For \(\mathrm{SbCl}_3\) and \(\mathrm{Cl}_2\), this is simply the sum: \(0.01762\, \mathrm{mol} \). Using the ideal gas law allows one to calculate the pressure of the gas at equilibrium in a vessel of known volume. This relation highlights the interdependency of gas properties and is particularly useful in predicting how gases behave under different conditions. It merges the macroscopic (volumes and pressures) with the microscopic (moles and temperature) aspects of gases, making it a powerful tool in chemistry.

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Most popular questions from this chapter

When a continuous stream of hydrogen gas, \(\mathrm{H}_{2}\), passes over hot magnetic iron oxide, \(\mathrm{Fe}_{3} \mathrm{O}_{4}\), metallic iron and water vapor form. When a continuous stream of water vapor passes over hot metallic iron, the oxide \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) and \(\mathrm{H}_{2}\) form. Explain why the reaction goes in one direction in one case but in the reverse direction in the other.

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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 \(K_{c}\).

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