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Chapter 15: Problem 19

The two common chlorides of phosphorus, \(\mathrm{PCl}_{3}\) and \(\mathrm{PCl}_{5},\) both important in the production of other phosphorus compounds, coexist in equilibrium through the reaction $$ \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{g}) $$ At \(250^{\circ} \mathrm{C},\) an equilibrium mixture in a \(2.50 \mathrm{L}\) flask contains \(0.105 \mathrm{g} \mathrm{PCl}_{5}, 0.220 \mathrm{g} \mathrm{PCl}_{3},\) and \(2.12 \mathrm{g} \mathrm{Cl}_{2}\) What are the values of (a) \(K_{c}\) and (b) \(K_{\mathrm{p}}\) for this reaction at \(250^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The equilibrium constants are \(K_{c} = 26.2\) and \(K_{p} = 0.108\) at \(250^{\circ} \mathrm{C}\).

Step by step solution

01

Calculate the molar masses of each compound

Begin by finding the molar masses of compounds using the atomic masses on the Periodic Table. The molar mass of PCl3 is \(30.97 + 3(35.45) = 137.32 \: \text{g/mol}\), of PCl5 is \(30.97 + 5(35.45) = 208.22 \: \text{g/mol}\), and of Cl2 is \(2(35.45) = 70.90 \: \text{g/mol}\).
02

Determine the equilibrium concentrations in mol/L

Next, convert grams to moles using the molar masses from Step 1. Then, compute the concentrations in molarity (mol/L) by dividing moles by the volume of the flask, 2.50 L. The concentration of PCl3 is \(0.220 \: \text{g} \times \frac{1 \: \text{mol}}{137.32 \: \text{g}} \div 2.50 \: \text{L} = 0.000644 \: \text{M}\), the concentration of PCl5 is \(0.105 \: \text{g} \times \frac{1 \: \text{mol}}{208.22 \: \text{g}} \div 2.50 \: \text{L} = 0.000201 \: \text{M}\), and concentration of Cl2 is \(2.12 \: \text{g} \times \frac{1 \: \text{mol}}{70.90 \: \text{g}} \div 2.50 \: \text{L} = 0.0120 \: \text{M}\).
03

Compute the equilibrium constant (Kc)

Use the balanced chemical equation to compute Kc. In this case, Kc = \([PCl5]\) / (\([PCl3][Cl2]\), and substituting the concentrations found gives Kc = \(0.000201/ (0.000644 \times 0.0120) = 26.2\).
04

Determine the partial pressure of each gas

Next, calculate the partial pressures from the ideal gas law. The total molar volume of an ideal gas at 250 degrees Celsius (523.15 K) and 1 atmosphere is about 24.27 L. Thus, the partial pressure of PCl3 is \(0.000644 \: \text{mol/L} \times 24.27 \: \text{L} = 0.0156\: \text{atm}\), of PCl5 is \(0.000201 \: \text{mol/L} \times 24.27 \: \text{L} = 0.00488 \: \text{atm}\), and of Cl2 is \(0.0120 \: \text{mol/L} \times 24.27 \: \text{L} = 0.291 \: \text{atm}\).
05

Calculate the equilibrium constant (Kp)

Finally, use the balanced chemical equation to compute Kp. In this case, Kp = \(\frac{(P_{PCl5})}{(P_{PCl3} \times P_{Cl2})}\). Substituting the partial pressures found in the previous step, we get Kp = \(\frac{0.00488}{(0.0156 \times 0.291)} = 0.108\).

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Most popular questions from this chapter

Explain how each of the following affects the amount of \(\mathrm{H}_{2}\) present in an equilibrium mixture in the reaction \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) $$ \Delta H^{\circ}=-150 \mathrm{kJ} $$ (a) Raising the temperature of the mixture; (b) introducing more \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ;\) (c) doubling the volume of the container holding the mixture; (d) adding an appropriate catalyst.

One important reaction in the citric acid cycle is citrate(aq) \(\rightleftharpoons\) aconitate \((\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad K=0.031\) Write the equilibrium constant expression for the above reaction. Given that the concentrations of \([\text { citrate }(\mathrm{aq})]=0.00128 \mathrm{M},[\text { aconitate }(\mathrm{aq})]=4.0 \times\) \(10^{-5} \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]=55.5 \mathrm{M},\) calculate the reaction quotient. Is this reaction at equilibrium? If not, in which direction will it proceed?

The following is an approach to establishing a relationship between the equilibrium constant and rate constants mentioned in the section on page 660 \(\bullet\)Work with the detailed mechanism for the reaction. \(\bullet\) Use the principle of microscopic reversibility, the idea that every step in a reaction mechanism is reversible. (In the presentation of elementary reactions in Chapter \(14,\) we treated some reaction steps as reversible and others as going to completion. However, as noted in Table \(15.3,\) every reaction has an equilibrium constant even though a reaction is generally considered to go to completion if its equilibrium constant is very large.) \(\bullet\) Use the idea that when equilibrium is attained in an overall reaction, it is also attained in each step of its mechanism. Moreover, we can write an equilibrium constant expression for each step in the mechanism, similar to what we did with the steady-state assumption in describing reaction mechanisms. \(\bullet\)Combine the \(K_{\mathrm{c}}\) expressions for the elementary steps into a \(K_{\mathrm{c}}\) expression for the overall reaction. The numerical value of the overall \(K_{c}\) can thereby be expressed as a ratio of rate constants, \(k\) Use this approach to establish the equilibrium constant expression for the overall reaction, $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ The mechanism of the reaction appears to be the following: Fast: \(\quad \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) Slow: \(\quad 2 \mathrm{I}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

For the reaction \(\mathrm{SO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{aq}), K=1.25 \mathrm{at}\) \(25^{\circ} \mathrm{C} .\) Will the amount of \(\mathrm{SO}_{2}(\mathrm{g})\) be greater than or less than the amount of \(\mathrm{SO}_{2}(\mathrm{aq}) ?\)

For the dissociation reaction \(2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{g})+\) \(\mathrm{S}_{2}(\mathrm{g}), K_{\mathrm{p}}=1.2 \times 10^{-2}\) at \(1065^{\circ} \mathrm{C} .\) For this same reaction at \(1000 \mathrm{K},\) (a) \(K_{\mathrm{c}}\) is less than \(K_{\mathrm{p}} ;\) (b) \(K_{\mathrm{c}}\) is greater than \(K_{\mathrm{p}} ;(\mathrm{c}) K_{\mathrm{c}}=K_{\mathrm{p}} ;\) (d) whether \(K_{\mathrm{c}}\) is less than, equal to, or greater than \(K_{\mathrm{p}}\) depends on the total gas pressure.
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