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Chapter 15: Problem 21

The equilibrium constant \(K_{P}\) for the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ is 1.05 at \(250^{\circ} \mathrm{C}\). The reaction starts with a mixture of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) at pressures of \(0.177 \mathrm{~atm}\) 0.223 atm, and 0.111 atm, respectively, at \(250^{\circ} \mathrm{C}\). When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.

Short Answer

Expert verified
To know which pressures increased or decreased at equilibrium, we calculate the change in pressure (x). By plugging equilibrium pressures into the equilibrium constant formula and solving for x (change in pressure), we can determine whether pressures of reactants or products have increased or decreased.

Step by step solution

01

Definition of equilibrium constant

The equilibrium constant for the given reaction at 250 degrees Celsius is denoted as \(K_{P}\) and it equals 1.05. The formula to calculate the equilibrium constant for a reaction is given by \(K_{p} = \frac{[PCl_{3}]_{eq}[Cl_{2}]_{eq}}{[PCl_{5}]_{eq}}\), where [PCl_{5}]_{eq}, [PCl_{3}]_{eq} and [Cl_{2}]_{eq} represent the equilibrium pressures of PCl_{5}, PCl_{3} and Cl_{2}, respectively.
02

Application of Change in Pressure

Since we know the initial pressures and the equilibrium constant, let's denote the change in the pressure of the products (PCl_{3} and Cl_2) as x. Therefore, at equilibrium, the pressure of PCl_{5} decreases by x to become (0.177-x) atm. The pressures of PCl_{3} and Cl_{2} increase by x each to become (0.223+x) atm and (0.111+x) atm, respectively.
03

Calculating the value of x

Now substitute the equilibrium pressures into the \(K_{P}\) formula and solve for x. \[1.05 = \frac{(0.223+x)(0.111+x)}{0.177-x}\]. By solving this equation for x, we find the value that will bring this system to equilibrium.
04

Interpretation

Once we have calculated x, we can determine which pressures have decreased and increased. If x is positive, then the pressures of PCl_{3} and Cl_{2} have increased, while the pressure of PCl_{5} has decreased. If x is negative, the opposite would be true. This would be due to Le Chatelier's Principle, which says the reaction will shift in the direction that reduces any stress (such as changes in pressure or temperature).

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that helps us understand how systems at equilibrium respond to external changes. When a system at equilibrium experiences a change in pressure, temperature, or concentration, it will adjust itself to counteract that change and return to equilibrium. This principle is like a balancing act, ensuring the system stays stable.
  • If a reactant or product is added, the system shifts to consume the added substances.
  • If the temperature of an exothermic reaction is increased, the equilibrium will shift to favor the endothermic direction to absorb the extra heat.
  • For reactions involving gases, a change in pressure will cause the equilibrium to shift according to the number of moles of gas on each side of the reaction.
In our reaction involving phosphorus pentachloride (\(\text{PCl}_5\)), phosphorus trichloride (\(\text{PCl}_3\)), and chlorine (\(\text{Cl}_2\)), Le Chatelier's Principle means that if pressure changes, the reaction will shift to either form more reactants or products to alleviate that change based on how many moles are involved.
Change in Pressure
For reactions involving gases, changes in pressure can significantly impact the equilibrium position. In our example reaction, we start with certain initial pressures for each gas. When pressure changes, the system will shift its equilibrium to balance that change.
  • Increasing overall pressure will generally shift the equilibrium towards the side with fewer moles of gas.
  • Decreasing pressure does the opposite, shifting towards the side with more moles.
In our situation, all reactants and products are gases, and changes in their individual pressures are connected. When the system shifts to reach equilibrium, some gases will experience an increase in pressure while others decrease.
Determining exactly how these pressures change involves calculations, but the essence is that the reaction will adjust, seeking a new equilibrium that aligns with the current conditions, following the guidance of Le Chatelier's Principle.
Equilibrium Pressures
Equilibrium pressures refer to the pressures of gases in a chemical reaction when the system is at equilibrium. This state is where the forward and reverse reactions occur at the same rate, so the concentrations of reactants and products remain stable.

In our specific reaction of \(\text{PCl}_5\), \(\text{PCl}_3\), and \(\text{Cl}_2\), we calculate the equilibrium pressures using the equilibrium constant, \(K_P\). The relationship among the equilibrium pressures is represented by the equation \(K_P = \frac{[PCl_{3}]_{eq}[Cl_{2}]_{eq}}{[PCl_{5}]_{eq}}\).
To figure out the equilibrium pressures:
  • Identify the initial pressures of each gas.
  • Set up the equation to solve for the change in pressure, \(x\).
  • Apply this change to the initial pressures to find the equilibrium pressures.
By solving the equilibrium equation, we determine which gas pressures increased or decreased. This revision helps identify how the gases react to reach equilibrium, illustrating the core concepts of equilibrium chemistry and Le Chatelier's Principle in action.

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Most popular questions from this chapter

Explain the difference between physical equilibrium and chemical equilibrium. Give two examples of each.

Baking soda (sodium bicarbonate) undergoes thermal decomposition as $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Would we obtain more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel?

The decomposition of ammonium hydrogen sulfide $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ is an endothermic process. A \(6.1589-\mathrm{g}\) sample of the solid is placed in an evacuated 4.000 - \(L\) vessel at exactly \(24^{\circ} \mathrm{C}\). After equilibrium has been established, the total pressure inside is \(0.709 \mathrm{~atm}\). Some solid \(\mathrm{NH}_{4} \mathrm{HS}\) remains in the vessel. (a) What is the \(K_{P}\) for the reaction? (b) What percentage of the solid has decomposed? (c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel?

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}\) : $$ \mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftharpoons \mathrm{NaCl}(l)+\mathrm{K}(g) $$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C}\), respectively.)

About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: $$ \begin{aligned} \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g) &+3 \mathrm{H}_{2}(g) \\ \Delta H^{\circ} &=206 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ The secondary stage is carried out at about \(1000^{\circ} \mathrm{C}\) in the presence of air, to convert the remaining methane to hydrogen: \(\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)\) $$ \Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol} $$ (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stages? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?
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