/*! 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 121 If wet silver carbonate is dried... [FREE SOLUTION] | 91影视

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If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$\mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)$$ \(\Delta H^{\circ}\) for this reaction is 79.14 \(\mathrm{kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\) . Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C},\) calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .^{\circ} \mathrm{C}\) (Hint: Manipulate the equation in Exercise 85.)

Short Answer

Expert verified
To prevent the decomposition of Ag鈧侰O鈧 at 110掳C, the partial pressure of CO鈧 must be greater than or equal to \(5.36 \times 10^{-2}\, \mathrm{torr}\).

Step by step solution

01

Find the equilibrium constant (K) at 25掳C

Given that the equilibrium partial pressure of CO鈧 is \(6.23 \times 10^{-3}\) torr at 25掳C. As there are no other gases except carbon dioxide involved in the reaction, we can directly use this partial pressure to represent the equilibrium constant. So, at 25掳C, K鈧 = \(6.23 \times 10^{-3}\).
02

Use Van't Hoff Equation

The Van't Hoff equation relates the change in the equilibrium constant (K) with temperature (T) and enthalpy change (鈭咹掳) as: \[\frac{\mathrm{d} \ln K}{\mathrm{d} T}=\frac{\Delta H^{\circ}}{R T^{2}}\] where R is the gas constant. By integrating the equation, we find the relationship between the equilibrium constants at two different temperatures: \[\ln \frac{K_{2}}{K_{1}}=-\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)\] We are given 鈭咹掳 = 79.14 kJ/mol. So, we convert it to J/mol: 鈭咹掳 = 79140 J/mol. Now, we have all the values to calculate the ratio of the equilibrium constants.
03

Calculate the equilibrium constant (K鈧) at 110掳C

First, we need to convert the given temperatures from Celsius to Kelvin: \(T_1 = 25^{\circ}C + 273.15 = 298.15K\) \(T_2 = 110^{\circ}C + 273.15 = 383.15K\) Now, we can use the Van't Hoff equation: \[\ln \frac{K_{2}}{K_{1}}=-\frac{\Delta H^{\circ}}{R}\left(\frac{1}{383.15}-\frac{1}{298.15}\right)\] Plugging in the given values and solving for K鈧, we get: \[K_{2}=K_{1} \times \mathrm{e}^{-\frac{\Delta H^{\circ}}{R}\left(\frac{1}{383.15}-\frac{1}{298.15}\right)}\] \[K_{2} = 6.23 \times 10^{-3} \times \mathrm{e}^{-\frac{79140}{8.314}\left(\frac{1}{383.15}-\frac{1}{298.15}\right)}\] After calculating the expression, we find that \[K_2 = 5.36 \times 10^{-2}\]
04

Calculate the partial pressure of CO鈧 at 110掳C

We have calculated the value of the equilibrium constant at 110掳C, K鈧 = 5.36 x 10鈦宦. In this case, as only CO鈧 is the gas in the reaction, K鈧 represents the equilibrium partial pressure of CO鈧 at 110掳C. To prevent decomposition, the partial pressure of CO鈧 should be greater than or equal to the equilibrium partial pressure. Therefore, the partial pressure of CO鈧 necessary to prevent decomposition of Ag鈧侰O鈧 at 110掳C is: \[P_{CO鈧倉 \ge 5.36 \times 10^{-2}\, \mathrm{torr}\]

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Key Concepts

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

Van't Hoff Equation: Understanding the Relation Between Temperature and Equilibrium
In the world of chemical reactions, the Van't Hoff Equation is a powerful tool. It helps us see how the equilibrium position of a reaction changes with temperature. This formula is written as:\[\ln \frac{K_{2}}{K_{1}}=-\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)\]Here, \(K\) is the equilibrium constant, \(T\) is the temperature, \(\Delta H^\circ\) is the change in enthalpy, and \(R\) is the gas constant. The equation shows us that if the enthalpy change is large, temperature will have a significant impact on \(K\), shifting the balance of the reaction.
Essentially, this equation connects thermodynamics with chemical equilibrium, revealing how energy changes in reactions.
Partial Pressure: Its Role in Chemical Equilibrium
Partial pressure is a term used to describe the pressure a single gas in a mixture exerts as if it were alone. In our exercise, carbon dioxide (CO鈧) is the gas of interest, so its partial pressure is crucial. Knowing the partial pressure allows us to understand the contributions of CO鈧 to the overall pressure in the system, which is key when considering chemical equilibrium.For a chemical reaction like the decomposition of silver carbonate, the partial pressure of CO鈧 determines whether the reaction will proceed forward or remain stable. The partial pressure directly influences the equilibrium constant \(K\), as higher partial pressures of CO鈧 can shift the equilibrium to favor the formation of reactants over products.
Thus, understanding partial pressure is vital to controlling and predicting the behaviors of gas-involving reactions.
Decomposition Reaction: Breaking Down Complex Compounds
A decomposition reaction involves a single complex compound breaking down into two or more simpler substances. The breakdown of silver carbonate into silver oxide and carbon dioxide is a classic example of such a reaction. This type of reaction is significant in both chemistry and everyday applications. Decomposition reactions are driven by the need to reach a state of lower energy or increased disorder, often influenced by temperature. In the given problem, we prevent the decomposition by ensuring that the partial pressure of carbon dioxide is sufficiently high, thus counteracting the formation of decomposition products.
Understanding decomposition is key in predicting how substances will react under various conditions, and how those conditions, like temperature or pressure, can be manipulated to maintain product stability.
Thermodynamics: The Energy Changes Behind Reactions
Thermodynamics is the study of energy changes within physical systems. In chemical reactions, it looks at how heat and work influence the direction and extent of reactions. When dealing with reactions, like the decomposition of silver carbonate, we use thermodynamic principles to understand and predict
  • Reaction spontaneity
  • Energy requirements
The enthalpy change \(\Delta H^\circ\) is a central thermodynamic quantity here. It tells us whether a reaction absorbs or releases heat. Positive \(\Delta H^\circ\) suggests that the reaction is endothermic, requiring heat from the surroundings to proceed.In our exercise, knowing the enthalpy change helps apply the Van't Hoff Equation to predict how equilibrium shifts with temperature changes, highlighting the profound link between thermodynamics and chemical behavior.

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Most popular questions from this chapter

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

Carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) form ideal solutions. Consider an equimolar solution of \(\mathrm{CC}_{4}\) and \(\mathrm{C}_{6} \mathrm{H}_{6}\) at \(25^{\circ} \mathrm{C}\) . The vapor above the solution is collected and condensed. Using the following data, determine the composition in mole fraction of the condensed vapor.

The following reaction occurs in pure water: $$\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ which is often abbreviated as $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\) . Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 M\) and \(\left[\mathrm{H}^{+}\right]=0.71 M .\)

The deciding factor on why HF is a weak acid and not a strong acid like the other hydrogen halides is entropy. What occurs when HF dissociates in water as compared to the other hydrogen halides?

Consider the following reaction at \(25.0^{\circ} \mathrm{C} :\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03\) kJ/mol and \(-176.6 \mathrm{J} / \mathrm{K}\) . mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\) . Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\) .

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