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Chapter 14: Problem 31

The following equilibrium constants were determined at \(1123 \mathrm{~K}\) $$\begin{array}{ll}\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) & K_{P}^{\prime}=1.3 \times 10^{14} \\\\\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) & K_{P}^{\prime \prime}=6.0 \times 10^{-3}\end{array}$$ Write the equilibrium constant expression \(K_{P}\), and calculate the equilibrium constant at \(1123 \mathrm{~K}\) for $$\mathrm{C}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{COCl}_{2}(g)$$

Short Answer

Expert verified
The equilibrium constant \(K_P\) at \(1123 \mathrm{~K}\) for the reaction \(C(s) + CO_2(g) + 2 Cl_2(g) \rightleftharpoons 2 COCl_2(g)\) is \(7.8 \times 10^{11}\).

Step by step solution

01

Analyze the reactions and the respective equilibrium constants

The first reaction is the formation of CO from C and CO2 with an equilibrium constant of \(1.3 \times 10^{14}\). The second reaction shows the formation of COCl2 from CO and Cl2 with an equilibrium constant of \(6.0 \times 10^{-3}\).
02

Write the given reaction juxtaposed with the old reactions

The target reaction is: \(C(s) + CO_2(g) + 2 Cl_2(g) \rightleftharpoons 2 COCl_2(g)\). We observe that the two given reactions can add up to give the target reaction. Thus, the equilibrium constant of target reaction would be the product of equilibrium constants of the reactions that make up the target reaction.
03

Determine the equilibrium constant of the final reaction

The product of the two given equilibrium constants will be the equilibrium constant of the target reaction. Therefore, \(K_P = K_P' \times K_P'' = 1.3 \times 10^{14} \times 6.0 \times 10^{-3} = 7.8 \times 10^{11}\).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction reaches a state where the concentrations of reactants and products do not change with time. This happens when the forward and reverse reaction rates are equal. It is crucial to understand this because it helps predict how the system behaves under different conditions.

In our example exercise, we deal with two reactions that reach equilibrium:
  • The formation of carbon monoxide (CO) from carbon (C) and carbon dioxide (CO鈧�) with an equilibrium constant (\(K_P'\)) of \(1.3 \times 10^{14}\).
  • The formation of phosgene (COCl鈧�) from carbon monoxide and chlorine gas (Cl鈧�) with an equilibrium constant (\(K_P''\)) of \(6.0 \times 10^{-3}\).
Chemical equilibrium is sensitive to changes in temperature, pressure, and concentration. Knowing the equilibrium constant (K) gives us insight into the relative amounts of each species when the system is in equilibrium.
Thermodynamics
Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. It plays a vital role in understanding chemical reactions at equilibrium because it helps determine whether reactions are spontaneous or require energy input.

In the context of chemical equilibrium, the Gibbs free energy change (\(\Delta G\)) is a key concept. For a reaction at equilibrium:
  • \(\Delta G = 0\), meaning there is no free energy change, and the system is stable.
  • If \(\Delta G < 0\), the reaction is spontaneous and tends to proceed forward.
  • If \(\Delta G > 0\), the reaction requires energy to proceed.
Thermodynamics helps us understand the tendencies of a chemical equilibrium system, providing insights into how conditions like temperature and pressure can affect the position of equilibrium.
Reaction Quotient
The reaction quotient (\(Q\)) is a useful tool in determining the direction in which a reaction will proceed to reach equilibrium. It uses the current concentrations of reactants and products at any point during a reaction that is not necessarily at equilibrium.

The main difference between the reaction quotient (\(Q\)) and the equilibrium constant (\(K\)) is that \(Q\) can be calculated at any point in time, not just at equilibrium. By comparing \(Q\) to \(K\), we understand how the system behaves:
  • If \(Q = K\), the system is at equilibrium.
  • If \(Q < K\), the forward reaction is favored and will continue until equilibrium is reached.
  • If \(Q > K\), the reverse reaction is favored to bring the system to equilibrium.
In the exercise given, though we focus on \(K\), having \(Q\) as a concept helps anticipate how changes might occur in reaching equilibrium.

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Most popular questions from this chapter

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: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=260 \mathrm{~kJ} /\mathrm{mol}\end{array}$$ The secondary stage is carried out at about \(1000^{\circ} \mathrm{C}\), in the presence of air, to convert the remaining methane to hydrogen: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stage? (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?

Consider the following equilibrium systems: (a) \(A \Longrightarrow 2 B\) \(\Delta H^{\circ}=20.0 \mathrm{~kJ} / \mathrm{mol}\) (b) \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}\) \(\Delta H^{\circ}=-5.4 \mathrm{~kJ} / \mathrm{mol}\) (c) \(A \Longrightarrow B\) \(\Delta H^{\circ}=0.0 \mathrm{~kJ} / \mathrm{mol}\) Predict the change in the equilibrium constant \(K_{\mathrm{c}}\) that would occur in each case if the temperature of the reacting system were raised.

The equilibrium constant \(K_{\mathrm{c}}\) for the decomposition of phosgene, \(\mathrm{COCl}_{2}\), is \(4.63 \times 10^{-3}\) at \(527^{\circ} \mathrm{C}\) : $$\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$$ Calculate the equilibrium partial pressure of all the components, starting with pure phosgene at 0.760 atm.

Consider the equilibrium $$2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g)$$ What would be the effect on the position of equilibrium of (a) increasing the total pressure on the system by decreasing its volume; (b) adding gaseous I \(_{2}\) to the reaction mixture; and (c) decreasing the temperature at constant volume?

At \(25^{\circ} \mathrm{C},\) the equilibrium partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are 0.15 atm and 0.20 atm, respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.
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