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Chapter 17: Problem 80

The equilibrium constant \(K\) for the reaction $$ 2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g) $$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{~K}\) and a \(y\) -intercept of \(-14.51\). Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 79

Short Answer

Expert verified
In conclusion, the standard enthalpy change (螖H掳) for the given reaction is approximately -112428 J/mol, and the standard entropy change (螖S掳) is approximately -120.61 J/mol路K.

Step by step solution

01

Determine the 螖H掳

To find the value of 螖H掳, we can use the given slope of the graph (1.352 脳 10^4 K) and the relationship between the slope of the graph and 螖H掳: Slope = -螖H掳/R We can now solve for 螖H掳: 螖H掳 = -Slope 脳 R 螖H掳 = -(1.352 脳 10^4 K) 脳 (8.314 J/mol路K) 螖H掳 鈮� -112428 J/mol So, the standard enthalpy change (螖H掳) for the reaction is approximately -112428 J/mol.
02

Determine the 螖S掳

To find the value of 螖S掳, we can use the given y-intercept of the graph (-14.51) and the relationship between the y-intercept of the graph and 螖S掳: y-intercept = 螖S掳/R We can now solve for 螖S掳: 螖S掳 = y-intercept 脳 R 螖S掳 = (-14.51) 脳 (8.314 J/mol路K) 螖S掳 鈮� -120.61 J/mol路K So, the standard entropy change (螖S掳) for the reaction is approximately -120.61 J/mol路K. In conclusion, the standard enthalpy change (螖H掳) for the given reaction is approximately -112428 J/mol, and the standard entropy change (螖S掳) is approximately -120.61 J/mol路K.

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Key Concepts

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

Enthalpy Change
The enthalpy change, denoted as \( \Delta H^{\circ} \), represents the heat absorbed or released during a chemical reaction at constant pressure. It is a vital concept in thermodynamics, providing insight into the energy dynamics of reactions. For the reaction involving chlorine gas, calculating \( \Delta H^{\circ} \) involves the slope of the graph of \( \ln(K) \) versus \( 1/T \), as given in the solution formula:
  • The slope of the line corresponds to \( -\Delta H^{\circ}/R \), where \( R \) is the gas constant (8.314 J/mol路K).
  • By rearranging the formula, \( \Delta H^{\circ} = - \text{{Slope}} \times R \).
For our specific problem, plugging in the values gives us \( \Delta H^{\circ} = -(1.352 \times 10^4 \ \text{K}) \times 8.314 \text{ J/mol路K} \), which approximates to -112428 J/mol.
This negative value indicates the reaction releases energy as heat, making it exothermic.
Entropy Change
Entropy change, noted as \( \Delta S^{\circ} \), measures the disorder or randomness within a system during a reaction. A key aspect to understanding these dynamics is considering how energy is distributed among the possible states of the system.
For this problem, the y-intercept from the graph of \( \ln(K) \) versus \( 1/T \) helps determine \( \Delta S^{\circ} \):
  • The y-intercept corresponds to \( \Delta S^{\circ}/R \).
  • The formula \( \Delta S^{\circ} = \text{y-intercept} \times R \) helps calculate the change.
Here, using the given y-intercept \(-14.51\), we find \( \Delta S^{\circ} = (-14.51) \times 8.314 \text{ J/mol路K} \), resulting in approximately -120.61 J/mol路K.
Such a negative entropy change suggests a shift towards decreased randomness in the system during the formation of \( \text{Cl}_2(g) \).
Van't Hoff Equation
The Van鈥檛 Hoff equation links the change in temperature to the equilibrium constant \( K \) of a reaction. It is crucial for understanding how temperature variations affect equilibrium and can predict reaction behavior under different conditions.
This equation is given by:
\[\ln(K) = -\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T}\right) + \frac{\Delta S^{\circ}}{R}\]
The graph of \( \ln(K) \) versus \( 1/T \) showing a straight line illustrates this relationship.
  • The slope \( -\Delta H^{\circ}/R \) allows us to determine \( \Delta H^{\circ} \).
  • The y-intercept \( \Delta S^{\circ}/R \) is used to find \( \Delta S^{\circ} \).
By analyzing these parameters, students can deduce how energy changes influence equilibrium, helping them grasp the fundamental interplay between thermodynamics and chemical reactions.

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Most popular questions from this chapter

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.]

Consider the reaction: $$ \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{~S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ for which \(\Delta H\) is \(-233 \mathrm{~kJ}\) and \(\Delta S\) is \(-424 \mathrm{~J} / \mathrm{K}\). a. Calculate the free energy change for the reaction \((\Delta G)\) at \(393 \mathrm{~K}\). b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

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

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. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

Predict the sign of \(\Delta S\) for each of the following and explain. a. the evaporation of alcohol b. the freezing of water c. compressing an ideal gas at constant temperature d. dissolving \(\mathrm{NaCl}\) in water
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