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Chapter 13: Problem 89

At \(25^{\circ} \mathrm{C}\), gaseous \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\) to the extent that \(12.5 \%\) of the original \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is \(0.900\) atm. Calculate the value of \(K_{\mathrm{p}}\) for this system.

Short Answer

Expert verified
The value of \(K_p\) for the decomposition reaction \(\mathrm{SO_2Cl_2} \rightarrow \mathrm{SO_2} + \mathrm{Cl_2}\) at \(25^{\circ}\mathrm{C}\) and 0.900 atm total pressure is 0.0160.

Step by step solution

01

Write the equilibrium expression for the reaction

The equilibrium constant (\(K_p\)) for this reaction can be written as follows: \[K_p = \frac{P_{SO_2} \cdot P_{Cl_2}}{P_{SO_2Cl_2}}\] where \(P_{SO_2}\), \(P_{Cl_2}\), and \(P_{SO_2Cl_2}\) are the partial pressures of the respective gases at equilibrium.
02

Find the remaining moles of \(\mathrm{SO_2Cl_2}\) at equilibrium

Let's assume we initially have n moles of \(\mathrm{SO_2Cl_2}\). 12.5% of the initial moles decompose at equilibrium, so the remaining moles of \(\mathrm{SO_2Cl_2}\) at equilibrium will be: Remaining moles of \(\mathrm{SO_2Cl_2} = 0.875n\)
03

Find the moles of the products at equilibrium

Since the decomposition of \(\mathrm{SO_2Cl_2}\) produces an equal number of moles of \(\mathrm{SO_2}\) and \(\mathrm{Cl_2}\), the moles of the products at equilibrium will be: Moles of \(\mathrm{SO_2} = Moles of \(\mathrm{Cl_2} = 0.125n\)
04

Determine partial pressures by using mole fractions

To find the partial pressures, we'll use the mole fractions and the total pressure at equilibrium (0.900 atm). The mole fraction of a species is the ratio of moles of the species to the total moles in the mixture. Mole fraction of \(\mathrm{SO_2Cl_2} = \frac{0.875n}{n} = 0.875\) Mole fraction of \(\mathrm{SO_2} = \) Mole fraction of \(\mathrm{Cl_2} = \frac{0.125n}{n} = 0.125\) Now, we find the partial pressures: \(P_{SO_2Cl_2} = P_{total} \cdot \) Mole fraction of \(\mathrm{SO_2Cl_2} = 0.900 \cdot 0.875 = 0.7875 \textrm{ atm}\) \(P_{SO_2} = P_{Cl_2} = P_{total} \cdot \) Mole fraction of \(\mathrm{SO_2} = 0.900 \cdot 0.125 = 0.1125 \textrm{ atm}\)
05

Calculate \(K_p\) using the partial pressures

Now we have the values for the partial pressures, we can plug them into the expression for \(K_p\): \[K_p = \frac{P_{SO_2} \cdot P_{Cl_2}}{P_{SO_2Cl_2}} = \frac{0.1125 \cdot 0.1125}{0.7875} = 0.0160\] Thus, the value of \(K_p\) for this system is 0.0160.

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Most popular questions from this chapter

At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ What is the partial pressure of \(\mathrm{NO}\) in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of \(0.80\) and \(0.20\) atm, respectively?

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C}\), equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is $$ \mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \quad K=5.6 \times 10^{-11} $$ If \(0.16 \mathrm{~mol}\) of \(\mathrm{HCO}_{3}^{-}\) is placed into \(1.00 \mathrm{~L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}{ }^{2-}\) ?

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}} \approx 1 \times 10^{-31}\) for the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ a. Calculate the concentration of \(\mathrm{NO}\), in molecules \(/ \mathrm{cm}^{3}\), that can exist in equilibrium in air at \(25^{\circ} \mathrm{C}\). In air, \(P_{\mathrm{N}_{2}}=0.8 \mathrm{~atm}\) and \(P_{\mathrm{O}_{2}}=0.2 \mathrm{~atm}\) b. Typical concentrations of \(\mathrm{NO}\) in relatively pristine environments range from \(10^{8}\) to \(10^{10}\) molecules \(/ \mathrm{cm}^{3}\). Why is there a discrepancy between these values and your answer to part a?

For the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ \(K=2.4 \times 10^{-3}\) at a given temperature. At equilibrium it is found that \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} M\) and \(\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} M\) What is the concentration of \(\mathrm{O}_{2}(g)\) under these conditions?
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