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The equilibrium constant for a certain reaction decreases from \(8.84\) to \(3.25 \times 10^{-2}\) when the temperature increases from \(25^{\circ} \mathrm{C}\) to \(75^{\circ} \mathrm{C}\). Estimate the temperature where \(K=1.00\) for this reaction. Estimate the value of \(\Delta S^{\circ}\) for this reaction. (Hint: Manipulate the equation in Exercise \(79 .\) )

Short Answer

Expert verified
The temperature where K = 1.00 for this reaction is approximately 403.48 K. The value of 螖S掳 for this reaction is approximately 30.40 J/(mol K).

Step by step solution

01

Write down the given information

We are given: - Initial equilibrium constant, K1 = 8.84 - Final equilibrium constant, K2 = 3.25 脳 10鈦宦 - Initial temperature, T1 = 25掳C (convert to Kelvin: 25 + 273.15 = 298.15 K) - Final temperature, T2 = 75掳C (convert to Kelvin: 75 + 273.15 = 348.15 K) We want to find the temperature where K = 1.00 and the value of 螖S掳.
02

Write down the Van 't Hoff equation

The Van 't Hoff equation is: \( \ln \frac{K_2}{K_1} = - \frac{ \Delta H ^{\circ}}{R} (\frac{1}{T_2} - \frac{1}{T_1}) \), where R is the gas constant, approximately 8.314 J/(mol K).
03

Solve for 螖H掳

We will first solve for 螖H掳 with the given information: 1. \( \ln \frac{K_2}{K_1} = \ln \frac{3.25 \times 10^{-2}}{8.84} \) 2. \( \ln \frac{3.25 \times 10^{-2}}{8.84}=- \frac{\Delta H^{\circ}}{8.314}(\frac{1}{348.15}-\frac{1}{298.15}) \) 3. Solve for 螖H掳: \( \Delta H掳 \approx -9.56 \times 10^3 \, \mathrm{J/mol} \) Now we have the value of 螖H掳.
04

Find the temperature where K=1.00

We will now find the temperature where K=1.00. Let's call it T3. 1. Replace all the known values in the Van 't Hoff equation: \( \ln \frac{1}{8.84} = - \frac{-9.56 \times 10^3}{8.314}(\frac{1}{T_3}-\frac{1}{298.15}) \) 2. Solve for T3: \( T_3 \approx 403.48 \, \mathrm{K } \) Thus, the temperature where K = 1.00 for this reaction is approximately 403.48 K.
05

Estimate the value of 螖S掳

Now, we can estimate the value of 螖S掳 using the relationship between 螖H掳, 螖S掳, and the equilibrium constant: 1. 螖G掳 = 螖H掳 - T螖S掳, and 螖G掳 = -RT ln K 2. At T1 (298.15 K): -RT ln K1 = 螖H掳 - T螖S掳 3. Solve for 螖S掳: \( \Delta S ^{\circ} \approx 30.40 \, \mathrm{J/(mol \cdot K)} \) So, the value of 螖S掳 for this reaction is approximately 30.40 J/(mol K).

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, often denoted as \( K \), is a crucial concept in chemistry that helps us understand the balance of reactants and products in a chemical reaction at equilibrium. It provides insight into the extent of a reaction under specific conditions.
  • If \( K \) is greater than 1, the reaction favors the formation of products.
  • If \( K \) is less than 1, the reaction favors the formation of reactants.
In the given exercise, the equilibrium constant changes significantly with temperature. This indicates that the reaction is temperature dependent. At 25掳C, \( K = 8.84 \), meaning more products are formed. However, at 75掳C, \( K \) decreases to \( 3.25 \times 10^{-2} \), suggesting a shift towards reactants. Understanding this shift is vital for predicting reaction behavior in different conditions.
Temperature Dependence
The change in the equilibrium constant with temperature can be explained through the Van 't Hoff equation. This relationship highlights how equilibrium constitutes a response to temperature changes, governed by enthalpy.The Van 't Hoff equation used in the solution is:\[ \ln \frac{K_2}{K_1} = - \frac{ \Delta H ^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]This equation describes how the equilibrium constant \( K \) varies inversely with temperature if the reaction is exothermic, as seen by an increase in \( K \) with decreasing temperature.Using this equation, we can determine how different temperatures affect the reaction, helping us predict the position of equilibrium based on thermal conditions. It's a useful tool for chemists dealing with reactions sensitive to temperature changes.
Enthalpy Change
Enthalpy change (\( \Delta H^{\circ} \)) is a measure of the total heat content of a system. It reflects the energy change of a reaction and indicates whether a reaction is exothermic or endothermic.
  • An exothermic reaction releases heat and has a negative \( \Delta H^{\circ} \).
  • An endothermic reaction absorbs heat and has a positive \( \Delta H^{\circ} \).
In this exercise, we calculate \( \Delta H^{\circ} \) to be approximately \(-9.56 \times 10^3 \, \mathrm{J/mol} \), suggesting the reaction releases heat as it proceeds. This energy release results in lower equilibrium constants with increasing temperature, consistent with the Van 't Hoff equation's prediction.Understanding \( \Delta H^{\circ} \) gives us insight into the energetic dynamics of the reaction and its equilibrium behavior under various temperatures.
Entropy Change
Entropy change (\( \Delta S^{\circ} \)) is a crucial concept as it relates to the disorder or randomness in a system. It, along with enthalpy, determines the spontaneity and direction of chemical reactions.
  • Positive \( \Delta S^{\circ} \) indicates an increase in disorder.
  • Negative \( \Delta S^{\circ} \) indicates a decrease in disorder.
In the exercise, \( \Delta S^{\circ} \) is estimated using the change in equilibrium constant and enthalpy. It is found to be \( 30.40 \, \mathrm{J/(mol \cdot K)} \), hinting at increased randomness as the reaction reaches equilibrium.The interplay of \( \Delta H^{\circ} \) and \( \Delta S^{\circ} \) determines the Gibbs free energy change (\( \Delta G^{\circ} \)), a key factor in assessing reaction feasibility. This detailed understanding assists in predicting how reactions behave in varying thermal environments.

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Most popular questions from this chapter

Ethanethiol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ;\right.\) also called ethyl mercaptan \()\) is commonly added to natural gas to provide the "rotten egg" smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is \(27.5 \mathrm{~kJ} / \mathrm{mol}\). What is the entropy of vaporization for this substance?

At \(100 .{ }^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}, \Delta H^{\circ}=40.6 \mathrm{~kJ} / \mathrm{mol}\) for the vaporiza- tion of water. Estimate \(\Delta G^{\circ}\) for the vaporization of water at \(90 .{ }^{\circ} \mathrm{C}\) and \(110 .{ }^{\circ} \mathrm{C}\). Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) do not depend on temperature.

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a \(100 \%\) yield of pure \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

Hydrogen cyanide is produced industrially by the following exothermic reaction: Is the high temperature needed for thermodynamic or kinetic reasons?

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ a. Without referring to Appendix 4 , predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperaturedependent. Predict the sign of \(\Delta S_{\text {surr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K}\). mol at \(298 \mathrm{~K}\). Using these values and data in Appendix 4 , calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C}\). The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .{ }^{\circ} \mathrm{C}\). f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\). The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\). g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\). Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\). The boiling point for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(29.0 \mathrm{~kJ} / \mathrm{mol}\). [Hint: The phase change reaction and the corresponding equilibrium expression are $$ \mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{Ni}(\mathrm{CO})} $$ \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) will liquefy when the pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is greater than the \(K\) value.]

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