/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 107 The following equilibrium was st... [FREE SOLUTION] | 91Ó°ÊÓ

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The following equilibrium was studied by analyzing the equilibrium mixture for the amount of \(\mathrm{H}_{2} \mathrm{~S}\) produced. $$ \mathrm{Sb}_{2} \mathrm{~S}_{3}(s)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{Sb}(s)+3 \mathrm{H}_{2} \mathrm{~S}(g) $$ A vessel whose volume was \(2.50 \mathrm{~L}\) was filled with \(0.0100 \mathrm{~mol}\) of antimony(III) sulfide, \(\mathrm{Sb}_{2} \mathrm{~S}_{3}\), and \(0.0100 \mathrm{~mol} \mathrm{H}_{2}\). After the mixture came to equilibrium in the closed vessel at \(440^{\circ} \mathrm{C}\), the gaseous mixture was removed, and the hydrogen sulfide was dissolved in water. Sufficient lead(II) ion was added to react completely with the \(\mathrm{H}_{2} \mathrm{~S}\) to precipitate lead(II) sulfide, \(\mathrm{PbS}\). If \(1.029 \mathrm{~g}\) PbS was obtained, what is the value of \(K_{c}\) at \(440^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
\( K_c \approx 1.73 \) at 440°C.

Step by step solution

01

Write the balanced equation

The balanced chemical equation is given as: \( \mathrm{Sb}_{2} \mathrm{~S}_{3}(s) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{Sb}(s) + 3 \mathrm{H}_{2} \mathrm{~S}(g) \). We know that solids do not appear in the expression for the equilibrium constant, therefore only concentrations of gases are considered for calculating \( K_c \).
02

Find moles of PbS

First, calculate the number of moles of \( \mathrm{PbS} \) using its mass. The molar mass of \( \mathrm{PbS} \) (Lead Sulfide) is \( 239.26 \text{ g/mol} \). The moles of \( \mathrm{PbS} \) are calculated as:\[\text{Moles of } \mathrm{PbS} = \frac{1.029 \text{ g}}{239.26 \text{ g/mol}} \approx 0.0043 \text{ mol}\]
03

Equilibrium moles of H2S

Since the reaction \( \mathrm{Pb^{2+} (aq) + H_2S (g) \rightarrow PbS (s) + 2H^+ (aq) } \) is complete, the moles of \( \mathrm{H_2S} \) equal the moles of \( \mathrm{PbS} \) precipitated. Thus, at equilibrium, there are approximately \( 0.0043 \) mol of \( \mathrm{H_2S} \) in the vessel.
04

Calculate initial and change in moles of H2

Initially, there were 0.0100 mol of \( \mathrm{H_2} \). The reaction consumes 3 mol of \( \mathrm{H_2} \) for every 2 mol of \( \mathrm{H_2S} \) produced. Since \( 0.0043 \) mol of \( \mathrm{H_2S} \) was produced, \( (3/2) \times 0.0043 = 0.00645 \) mol of \( \mathrm{H_2} \) was consumed, leaving \( 0.0100 - 0.00645 = 0.00355 \) mol \( \mathrm{H_2} \) at equilibrium.
05

Calculate equilibrium concentrations

The equilibrium concentration of \( \mathrm{H_2S} \) is \( \frac{0.0043 \text{ mol}}{2.50 \text{ L}} = 0.00172 \text{ M} \), and the equilibrium concentration of \( \mathrm{H_2} \) is \( \frac{0.00355 \text{ mol}}{2.50 \text{ L}} = 0.00142 \text{ M} \).
06

Determine Kc

The expression for \( K_c \) is given by:\[ K_c = \frac{[\mathrm{H_2S}]^3}{[\mathrm{H_2}]^3} = \frac{(0.00172)^3}{(0.00142)^3} \]Calculate \( K_c \):\[ K_c = \frac{(0.00172)^3}{(0.00142)^3} = \frac{4.98 \times 10^{-9}}{2.87 \times 10^{-9}} \approx 1.73 \]
07

Conclusion

The equilibrium constant \( K_c \) for the reaction at \( 440^{\circ} \mathrm{C} \) is approximately \( 1.73 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
In chemical reactions, the equilibrium constant, represented as \( K_c \), plays a crucial role. It tells us how the concentrations of products and reactants compare at equilibrium, a state where the forward and reverse reactions balance each other. This concept simplifies understanding chemical systems by transforming complex interactions into something more predictable.

For the reaction \( \mathrm{Sb}_{2} \mathrm{~S}_{3}(s) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{Sb}(s) + 3 \mathrm{H}_{2} \mathrm{~S}(g) \), only gases are included in the \( K_c \) equation. Solid substances like antimony and antimony sulfide do not affect \( K_c \) because their concentrations remain constant. The equilibrium constant for a gas reaction is calculated with the formula:

\[ K_c = \frac{[\mathrm{H_2S}]^3}{[\mathrm{H_2}]^3} \]

This expression uses the concentrations of gaseous hydrogen sulfide \([\mathrm{H_2S}]\) and hydrogen \([\mathrm{H_2}]\) at equilibrium, raised to the power of their coefficients from the balanced chemical equation. Understanding \( K_c \) helps predict how the system will respond to changes in concentration, pressure, or temperature.
Le Chatelier's Principle
Le Chatelier's Principle is a handy guideline to predict the behavior of a chemical system when it is disturbed from its state of equilibrium. According to this principle, if an external change is applied to a system at equilibrium, the system will adjust itself to counteract this change, shifting the equilibrium position accordingly.

Consider the scenario where the concentration of \( \mathrm{H_{2}S} \) increases. In response, the system will try to flatten this increase by favoring the backward reaction, thus producing more \( H_2 \) and \( \mathrm{Sb}_2 \mathrm{S}_3 \). In other words, the effect of pressure changes, concentration variations, or temperature shifts can all be predicted by Le Chatelier's Principle, helping chemists control and optimize reactions.

One practical implication of this principle is in industrial processes, where reaction conditions are manipulated to maximize yield. By understanding how a reaction shifts when disturbed, better product formation and efficiency are achieved. It's like tweaking the perfect recipe to get the best outcome.
Chemical Reactions
A chemical reaction involves the transformation of substances, called reactants, into new substances known as products, through the breaking and forming of bonds. Chemical equations symbolize these reactions, showing the rearrangement of atoms. In the given equilibrium reaction, \( \mathrm{Sb}_{2} \mathrm{~S}_{3}(s) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{Sb}(s) + 3 \mathrm{H}_{2} \mathrm{~S}(g) \), the reactants include solid antimony sulfide and gaseous hydrogen, while the products are elemental antimony and hydrogen sulfide gas.

Chemical reactions can be dynamic and reversible, meaning they can proceed in both forward and backward directions. Equilibrium is reached when the forward and backward reactions occur at the same rate, resulting in stable concentrations of reactants and products. This particular reaction shows such behavior, where the equilibrium can be shifted by changing reaction conditions such as pressure, concentration, or temperature.

Understanding chemical reactions involves not only writing and balancing equations but also comprehending the conditions under which they occur and the factors influencing their equilibrium. This offers a fundamental insight into processes ranging from biological mechanisms to industrial production.

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Most popular questions from this chapter

On the basis of the value of \(K_{c}\), decide whether or not you expect nearly complete reaction at equilibrium for each of the following: a. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) ; K_{c}=4.6 \times 10^{-31}\) b. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(g) ; K_{c}=1.3 \times 10^{21}\)

At \(850^{\circ} \mathrm{C}\) and \(1.000\) atm pressure, a gaseous mixture of carbon monoxide and carbon dioxide in equilibrium with solid carbon is \(90.55 \%\) CO by mass. $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Calculate \(K_{c}\) for this reaction at \(850^{\circ} \mathrm{C}\).

During an experiment with the Haber process, a researcher put \(1 \mathrm{~mol} \mathrm{~N}_{2}\) and \(1 \mathrm{~mol} \mathrm{H}_{2}\) into a reaction vessel to observe the equilibrium formation of ammonia, \(\mathrm{NH}_{3}\). $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ When these reactants come to equilibrium, assume that \(x\) mol \(\mathrm{H}_{2}\) react. How many moles of ammonia form?

At high temperatures, a dynamic equilibrium exists between carbon monoxide, carbon dioxide, and solid carbon. $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) ; \Delta H^{\circ}=172.5 \mathrm{~kJ} $$ At \(900^{\circ} \mathrm{C}, K_{c}\) is \(0.238\). a. What is the value of \(K_{p}\) ? b. Some carbon dioxide is added to the hot carbon and the system is brought to equilibrium at \(900^{\circ} \mathrm{C}\). If the total pressure in the system is \(6.40 \mathrm{~atm}\), what are the partial pressures of \(\mathrm{CO}\) and \(\mathrm{CO}_{2} ?\) c. How will the equilibrium system respond if additional carbon at \(900{ }^{\circ} \mathrm{C}\) is added to the reaction system? Why?

How is it possible for a catalyst to give products from a reaction mixture that are different from those obtained when no catalyst or a different catalyst is used? Give an example.

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