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Chapter 13: Problem 88

For the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ at \(600 . \mathrm{K}\) , the equilibrium constant, \(K_{\mathrm{p}},\) is \(11.5 .\) Suppose that 2.450 \(\mathrm{g} \mathrm{PCl}_{5}\) is placed in an evacuated 500 -mL bulb, which is then heated to \(600 . \mathrm{K}\) . a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the percent dissociation of PCl_ at equilibrium?

Short Answer

Expert verified
a. The initial pressure of PCl鈧� without dissociation is approximately 11.583 atm. b. The partial pressure of PCl鈧� at equilibrium is approximately 8.063 atm. c. The total pressure in the bulb at equilibrium is approximately 15.103 atm. d. The percent dissociation of PCl鈧� at equilibrium is approximately 30.37%.

Step by step solution

01

Calculate the initial pressure of PCl鈧� without dissociation

In order to find the pressure of \(\mathrm{PCl_5}\) without dissociation, we first need to calculate the number of moles of \(\mathrm{PCl_5}\) and then use the ideal gas law, \(PV=nRT\), to find the pressure. Given the mass of \(\mathrm{PCl_5}\) as \(2.450\, g\), we can determine the moles by: \(n=\dfrac{\text{mass}}{\text{molar mass}}\) The molar mass of \(\mathrm{PCl_5}\) is: \(\text{Molar mass} = 1 \times (\text{Atomic mass of P}) + 5 \times (\text{Atomic mass of Cl}) = 1 \times (30.97\, g/mol) + 5 \times (35.45\, g/mol) = 208.22\, g/mol\) Now we can calculate the moles of \(\mathrm{PCl_5}\): \(n=\dfrac{2.450\,g}{208.22\,g/mol} = 0.01176\,mol\) Next, we use the ideal gas law to find the pressure: \(PV = nRT\) \(P = \dfrac{nRT}{V}\) Given the temperature, \(T = 600\, K\), the volume, \(V = 500\, mL = 0.5\, L\), and the gas constant, \(R = 0.0821\, L \cdot atm/(mol \cdot K)\), we can calculate the pressure: \(P = \dfrac{(0.01176\,mol)(0.0821\, L \cdot atm/(mol \cdot K))(600\, K)}{0.5\, L} = 11.583\, atm\) So, the initial pressure of \(\mathrm{PCl_5}\) without dissociation is approximately \(11.583\, atm\).
02

Construct an ICE table for the reaction

Now we need to determine the changes in pressure when the reaction reaches equilibrium. We will use an ICE (Initial, Change, Equilibrium) table for this purpose: \(\begin{array}{c|ccc} & \mathrm{PCl_5} & & \mathrm{PCl_3} & \mathrm{Cl_2} \\ \hline \text{Initial} & 11.583 & & 0 & 0 \\ \text{Change} & -x & & +x & +x \\ \text{Equilibrium} & 11.583-x & & x & x \end{array}\) Here, \(x\) represents the change in pressure of each gas due to the reaction.
03

Solve for the equilibrium pressure of PCl鈧�

Since we have the ICE table and the equilibrium constant, \(K_p = 11.5\), we can write the expression for \(K_p\) and solve for the equilibrium pressure of \(\mathrm{PCl_5}\) (i.e.,\(11.583-x\)): \(K_p = \dfrac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}\) \(11.5 = \dfrac{x^2}{11.583-x}\) Now we can solve for \(x\), which is a quadratic equation. However, if we assume that the dissociation is not too significant (since \(K_p\) is not very large), we can approximate \(11.583-x \approx 11.583\). This simplifies the equation to: \(11.5 = \dfrac{x^2}{11.583}\) \(x^2 = 11.5 \times 11.583\) \(x = \sqrt{11.5 \times 11.583} = 3.52\) So, at equilibrium, the partial pressure of \(\mathrm{PCl_5}\) is approximately \(11.583 - 3.52 = 8.063\, atm\).
04

Calculate the total pressure at equilibrium

From the ICE table, we see that the partial pressures of \(\mathrm{PCl_3}\) and \(\mathrm{Cl_2}\) at equilibrium are both equal to \(x\approx 3.52\, atm\). Thus, the total pressure at equilibrium can be calculated by adding the partial pressures of all gases: \(P_{\text{total}} = [\mathrm{PCl_5}] + [\mathrm{PCl_3}] + [\mathrm{Cl_2}] = 8.063 + 3.52 + 3.52 = 15.103\, atm\) So the total pressure in the bulb at equilibrium is approximately \(15.103\, atm\).
05

Calculate the percent dissociation of PCl鈧� at equilibrium

Finally, we can calculate the percent dissociation of \(\mathrm{PCl_5}\) at equilibrium: \(\text{Percent dissociation} = \dfrac{\text{amount dissociated}}{\text{initial amount}} \times 100\%\) \(\text{Percent dissociation} = \dfrac{x}{11.583} \times 100\%\) \(\text{Percent dissociation} = \dfrac{3.52}{11.583} \times 100\% = 30.37\%\) Thus, the percent dissociation of \(\mathrm{PCl_5}\) at equilibrium is approximately \(30.37\%\).

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation that helps in understanding how gases behave under different conditions. It is expressed as \( PV = nRT \), where \( P \) is the pressure of the gas, \( V \) is the volume it occupies, \( n \) is the number of moles, \( R \) is the universal gas constant \((0.0821\,L\cdot atm/(mol\cdot K))\), and \( T \) is the temperature in Kelvin.

This law assumes that gases are composed of many small particles which are in constant random motion and that they do not interact with each other except for elastic collisions. Although real gases only approximate ideal behavior, especially at high pressures and low temperatures, the Ideal Gas Law is a useful tool for many applications.
  • Calculate pressure, volume, or temperature of the gas using the remaining known variables.
  • Estimate the number of moles (\( n \)) if mass and molar mass are known.
  • Determine relationships in chemical reactions involving gases, such as this exercise with PCl鈧�.
By understanding and applying the Ideal Gas Law, you can predict the behavior of gas phases and solve problems involving gases with accuracy and ease.
Equilibrium Constant
Chemical equilibria involve reactions that do not go to completion. Instead, they proceed to a state where the concentrations of reactants and products remain constant over time. The Equilibrium Constant \( K \) is used to express the ratio of product concentrations to reactant concentrations at equilibrium, raised to the power of their respective coefficients in the balanced chemical equation.

For reactions involving gases, the equilibrium constant \( K_p \) is used in terms of partial pressures. It is calculated by the formula: \[ K_p = \frac{(P_{\text{products}})^{coefficients}}{(P_{\text{reactants}})^{coefficients}} \] where \( P_{\text{products}} \) and \( P_{\text{reactants}} \) represent the partial pressures.
  • The value of \( K_p \) provides insight into the favorability of the reaction; a larger \( K_p \) (>1) suggests products are favored, while a smaller \( K_p \) (<1) suggests reactants are favored.
  • Changes in conditions, such as pressure or temperature, can shift the equilibrium and thus change the position of \( K_p \).
Understanding \( K_p \) is essential for predicting the behavior of chemical reactions involving gases at equilibrium.
Partial Pressure
Partial Pressure is the pressure exerted by an individual gas in a mixture of gases. In any gaseous reaction, each gas contributes to the total pressure with its own partial pressure, helping in understanding the distribution of gases within a system.

Each gas in a mixture behaves independently, showing that its partial pressure is a reflection of its mole fraction multiplied by the total pressure. Mathematically, this is given by:\[ P_{i} = X_{i} \times P_{\text{total}} \] where \( P_{i} \) is the partial pressure of the gas, \( X_{i} \) is its mole fraction, and \( P_{\text{total}} \) is the total pressure of the gas mixture.
  • Helps in identifying the contribution of individual gases towards the total pressure in reactions.
  • Useful in calculating equilibrium constants expressed in terms of pressures \( (K_p) \).
  • Determines how gases will behave and react in different conditions.
Emphasizing partial pressure allows chemists to streamline calculations in reactions involving multiple gaseous components.
ICE Table
The ICE Table is a structured way to keep track of changes in concentrations or pressures of reactants and products as they reach equilibrium. "ICE" stands for Initial, Change, and Equilibrium states. This table simplifies the process of setting up equilibrium expressions and solving for unknowns.

Here's how an ICE Table is structured:
  • **Initial:** This row lists the starting concentration or pressure of each reactant and product before any changes due to reaction progress.
  • **Change:** Represents the change in concentration or pressure as the system moves towards equilibrium. Denoted using \( x \), where \( x \) signifies the shift in terms of moles or pressure.
  • **Equilibrium:** Describes the final concentration or pressure of each reactant and product when the system has reached equilibrium.
Using an ICE table allows for accurate calculation of unknown variables, such as equilibrium pressures or concentrations, by organizing data from the chemical equation and the equilibrium constant. This approach is invaluable when working with gaseous reactions as it keeps the information clear and manageable.

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Most popular questions from this chapter

Consider the reaction $$\mathrm{P}_{4}(g) \rightleftharpoons 2 \mathrm{P}_{2}(g)$$ where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at 1325 \(\mathrm{K}\) . In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at 1325 \(\mathrm{K}\) , the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of 1.00 atm. Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is $$\mathrm{HCO}_{3}^{-}(a q) \leftrightharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) K=5.6 \times 10^{-11}$$ If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into 1.00 \(\mathrm{L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}^{2-2}\) ?

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at \(25^{\circ} \mathrm{C}\) according to the following equation: $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The initial density of the system was recorded as 4.495 \(\mathrm{g} / \mathrm{L}\) . After equilibrium was reached, the density was noted to be 4.086 \(\mathrm{g} / \mathrm{L}\) . a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\operatorname{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K ?\) Explain each answer.

Consider the decomposition of the compound \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) as follows: $$\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g)$$ When a 5.63 -g sample of pure \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g)\) was sealed into an otherwise empty \(2.50-\mathrm{L}\) flask and heated to \(200 .^{\circ} \mathrm{C},\) the pres- sure in the flask gradually rose to 1.63 \(\mathrm{atm}\) and remained at that value. Calculate \(K\) for this reaction.

Le Chatelier's principle is stated (Section 13.7\()\) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.
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