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Chapter 15: Problem 53

\(1.00 \mathrm{mol}\) each of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) are introduced into an evacuated 1.75 L flask, and the following equilibrium is established at \(668 \mathrm{K}\). $$ \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{COCl}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=22.5 $$ For this equilibrium, calculate (a) the partial pressure of \(\mathrm{COCl}_{2}(\mathrm{g}) ;\) (b) the total gas pressure.

Short Answer

Expert verified
The partial pressure of COCl2 is 0.427 atm and the total gas pressure is 0.715 atm.

Step by step solution

01

Establish the Initial Conditions

Start by calculating the initial partial pressure of both CO and Cl2. Initially there is no COCl2 present. The partial pressures of CO and Cl2 can be calculated using the ideal gas law, \(P = n/V\), where n is the amount in moles and V is the volume in liters. Here, \(P_{CO} = P_{Cl2} = 1.00\, mol / 1.75\, L = 0.571\, atm\).
02

Setup the ICE table

Now setup an ICE (Initial, Change, Equilibrium) table to understand the changes in moles (and thus pressure) of each entity. Let \(x\) be the moles of COCl2 formed when equilibrium is achieved. Then, for CO and Cl2 it would be a decrease by `x` moles and for COCl2 it would be an increase by `x` moles. The expression in the ICE table would be: \n\nInitial: \(0.571\, atm, 0.571\, atm, 0\, atm\) for CO, Cl2 and COCl2 respectively\nChange: \(-x, -x, +x\)\nEquilibrium: \(0.571 - x, 0.571 - x, x\)
03

Setup the Equation using Equilibrium Constant

Using the equilibrium constant expression \(K_p = [COCl2] / ([CO][Cl2]) \) and the values from the ICE table set up an equation. Equilibrium constant Kp is given as 22.5, thus, \n\(22.5 = x / ((0.571 - x)(0.571 - x))\). This forms a quadratic equation which can be solved to find the value of \(x\).
04

Calculate the partial and total pressure

Solving the equation yields \(x\) = 0.427 atm, (discarding the negative root as pressure cannot be negative). This means that the partial pressure of COCl2 at equilibrium is 0.427 atm. The partial pressures of CO and Cl2 are 0.571 - 0.427 = 0.144 atm each. The total pressure is simply the sum of the partial pressures, i.e., \(2(0.144) + 0.427 = 0.715\, atm\).

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Most popular questions from this chapter

Equilibrium is established at \(1000 \mathrm{K},\) where \(K_{\mathrm{c}}=281\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) .\) The equilibrium amount of \(\mathrm{O}_{2}(\mathrm{g})\) in a \(0.185 \mathrm{L}\) flask is 0.00247 mol. What is the ratio of \(\left[\mathrm{SO}_{2}\right]\) to \(\left[\mathrm{SO}_{3}\right]\) in this equilibrium mixture?

In the Ostwald process for oxidizing ammonia, a variety of products is possible- \(\mathrm{N}_{2}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO},\) and \(\mathrm{NO}_{2}-\) depending on the conditions. One possibility is $$\begin{aligned} \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) &+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ K_{\mathrm{p}} &=2.11 \times 10^{19} \mathrm{at} 700 \mathrm{K} \end{aligned}$$ For the decomposition of \(\mathrm{NO}_{2}\) at \(700 \mathrm{K}\) $$\mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.524$$ (a) Write a chemical equation for the oxidation of \(\mathrm{NH}_{3}(\mathrm{g})\) to \(\mathrm{NO}_{2}(\mathrm{g})\) (b) Determine \(K_{\mathrm{p}}\) for the chemical equation you have written.

Is a mixture of \(0.0205 \mathrm{mol} \mathrm{NO}_{2}(\mathrm{g})\) and \(0.750 \mathrm{mol}\) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) in a \(5.25 \mathrm{L}\) flask at \(25^{\circ} \mathrm{C},\) at equilibrium? If not, in which direction will the reaction proceed toward products or reactants? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3} \mathrm{at} 25^{\circ} \mathrm{C}$$

An equilibrium mixture at 1000 K contains an equilibrium mixter \(0.276\ \mathrm{mol}\ \mathrm{H}_{2}, 0.276 \mathrm{mol}\ \mathrm{CO}_{2}, 0.224\ \mathrm{mol}\ \mathrm{CO},\) and \(0.224\ \mathrm{mol}\ \mathrm{H}_{2} \mathrm{O}\) $$\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ (a) Show that for this reaction, \(K_{\mathrm{c}}\) is independent of the reaction volume, \(V\) (b) Determine the value of \(K_{\mathrm{c}}\) and \(K_{\mathrm{p}}\)

The standard enthalpy of reaction for the decomposition of calcium carbonate is \(\Delta H^{\circ}=813.5 \mathrm{kJmol}^{-1}\) As temperature increases, does the concentration of calcium carbonate increase, decrease, or remain the same? Explain.
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