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Chapter 14: Problem 26

Pure phosgene gas \(\left(\mathrm{COCl}_{2}\right), 3.00 \times 10^{-2} \mathrm{~mol},\) was placed in a 1.50-L container. It was heated to \(800 \mathrm{~K}\), and at equilibrium the pressure of \(\mathrm{CO}\) was found to be 0.497 atm. Calculate the equilibrium constant \(K_{P}\) for the reaction $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g)$$

Short Answer

Expert verified
The equilibrium constant \(K_P\) for the reaction is 0.497.

Step by step solution

01

Calculate initial pressure of phosgene gas

Before any reaction happens, the only species present is \(\mathrm{COCl}_{2}\). Using the ideal gas law \(P=\frac{nRT}{V}\), we can calculate the initial pressure where \(n=3.00 \times 10^{-2}\) mol, \(R=0.0821\) L.atm/K.mol, \(T=800\) K and \(V=1.50\) L. This gives us \(P_i =0.0333\) atm.
02

Calculate change in pressure

At equilibrium, the pressure of \(\mathrm{CO}\) which is one of the products, is given, i.e. \(P_{CO}=0.497\) atm. This is equal to the equilibrium pressure of \(\mathrm{Cl}_{2}\) and change in pressure of \(\mathrm{COCl}_{2}\) (\(∆P\)) as per the stoichiometry of the reaction. Therefore, \(P_i - ∆P = P_{COCl_{2}}\) and \( ∆P = P_{i} - P_{COCl_{2}} = 0.0333 - 0.497 = -0.4637\) atm as phosgene gas (\(\mathrm{COCl}_{2}\)) is being consumed in the reaction.
03

Calculate equilibrium pressures

At equilibrium, \(P_{CO} = P_{Cl_{2}} = ∆P\) and \(P_{COCl_{2}} = P_i - ∆P\). Now we can substitute the values getting \(P_{CO} = P_{Cl_{2}} = 0.497\) atm and \(P_{COCl_{2}} = 0.0333 - (-0.4637) = 0.497\) atm.
04

Calculate equilibrium constant

The equilibrium constant \(K_P\) is defined as \[K_P = \frac{P_{products}}{P_{reactants}}\] Substituting the pressures of products and reactants at equilibrium from step 3, we get \[K_P = \frac{P_{CO}.P_{Cl_{2}}}{P_{COCl_{2}}} = \frac{0.497*0.497}{0.497} = 0.497\]

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Most popular questions from this chapter

What is the rule for writing the equilibrium constant for the overall reaction involving two or more reactions?

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?

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

Does the addition of a catalyst have any effects on the position of an equilibrium?

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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