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Chapter 15: Problem 35

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.

Short Answer

Expert verified
The equilibrium partial pressures of the components in the reaction are determined by solving a quadratic equation expressing the equilibrium constant of the reaction in terms of changes in the initial pressures of the components. After solving for \(x\), the equilibrium pressures of \(\mathrm{COCl}_{2}\), \(\mathrm{CO}\), and \(\mathrm{Cl}_{2}\) can be determined.

Step by step solution

01

Identify The Initial Conditions

Initially, there is only phosgene, \(\mathrm{COCl}_{2}\), with a partial pressure of 0.760 atm. The initial partial pressures of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) are both 0 since the decomposition has not yet started.
02

Establish The Chemical Equilibrium Expression

The given equilibrium constant \(\(K_{c}\)\) corresponds to the expression \(\(K_{c}\) = \(\frac{[CO][Cl_{2}]}{[COCl_{2}]}\). Given that the initial concentration of \(\mathrm{COCl}_{2}\) decreases by \(x\) at equilibrium, the concentrations of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) increase by \(x\). Hence, the equilibrium concentrations can be expressed as: \(\mathrm{[COCl}_{2}]\) = 0.760 - \(x\), \(\mathrm{[CO]}\) = \(x\), \(\mathrm{[Cl}_{2}]\) = \(x\).
03

Solve For \(x\)

Plug these values into the expression for \(\(K_{c}\)\) to get: \(\frac{x \cdot x}{0.760 - x} = 4.63 \times 10^{-3}\). Solve this quadratic equation for \(x\). The positive root of this quadratic equation gives the equilibrium concentrations of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\), and thus, also the partial pressures since these are proportional to the concentrations.
04

Calculate The Equilibrium Partial Pressures

The equilibrium partial pressures can be calculated by using the relation: \(\(P = n/V \times RT\)\), where \(\(P\)\) is the pressure, \(n\) is the number of moles (concentration after the reaction has reached equilibrium), \(\(V\)\) is the volume, and \(\(R = 0.0821 atm⋅L⋅mol⁻¹⋅K⁻¹\) and \(T\) is the absolute temperature. As the volume and temperature are constant, the pressures at equilibrium will be proportional to the concentrations in moles per liter. Therefore the pressures of \(\mathrm{COCl}_{2}\), \(\mathrm{CO}\), and \(\mathrm{Cl}_{2}\) will be 0.760 atm - \(x\) atm, \(x\) atm, and \(x\) atm respectively.

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Most popular questions from this chapter

Consider the reaction in a closed container: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Initially, 1 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is present. At equilibrium, \(\alpha\) mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) has dissociated to form \(\mathrm{NO}_{2}\). (a) Derive an expression for \(K_{P}\) in terms of \(\alpha\) and \(P,\) the total pressure. (b) How does the expression in (a) help you predict the shift in equilibrium caused by an increase in \(P ?\) Does your prediction agree with Le Châtelier's principle?

A sample of pure \(\mathrm{NO}_{2}\) gas heated to \(1000 \mathrm{~K}\) decomposes: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The equilibrium constant \(K_{P}\) is \(158 .\) Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is 0.25 atm at equilibrium. Calculate the pressure of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in the mixture.

Consider this reaction at equilibrium in a closed container: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What would happen if (a) the volume is increased, (b) some \(\mathrm{CaO}\) is added to the mixture, (c) some \(\mathrm{CaCO}_{3}\) is removed, (d) some \(\mathrm{CO}_{2}\) is added to the mixture, (e) a few drops of an \(\mathrm{NaOH}\) solution are added to the mixture, (f) a few drops of an \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water \(),(\mathrm{g})\) the temperature is increased?

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.)

The dissociation of molecular iodine into iodine atoms is represented as $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ At \(1000 \mathrm{~K}\), the equilibrium constant \(K_{\mathrm{c}}\) for the reaction is \(3.80 \times 10^{-5}\). Suppose you start with 0.0456 mole of \(I_{2}\) in a 2.30 - \(L\) flask at \(1000 \mathrm{~K}\). What are the concentrations of the gases at equilibrium?
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