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Chapter 17: Problem 39

For ammonia \(\left(\mathrm{NH}_{3}\right)\), the enthalpy of fusion is \(5.65 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of fusion is \(28.9 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). a. Will \(\mathrm{NH}_{3}(s)\) spontaneously melt at \(200 . \mathrm{K}\) ? b. What is the approximate melting point of ammonia?

Short Answer

Expert verified
a. Yes, NH₃(s) will spontaneously melt at 200 K since the calculated ∆G is -130 J/mol, which is negative. b. The approximate melting point of ammonia is 195.5 K.

Step by step solution

01

Write down the Gibbs free energy equation

The equation for Gibbs free energy is given by: ∆G = ∆H - T∆S where ∆G is the change in Gibbs free energy, ∆H is the change in enthalpy, T is the temperature in Kelvin, and ∆S is the change in entropy.
02

Substitute given values into the equation

We are given the enthalpy of fusion (∆H) for ammonia as 5.65 kJ/mol and the entropy of fusion (∆S) as 28.9 J/K⋅mol. The temperature, T, is given as 200 K. Substitute these values into the Gibbs free energy equation: ∆G = (5.65 kJ/mol) - (200 K)(28.9 J/K⋅mol) Note that the units for ∆H and ∆S are different. Convert ∆H to J/mol: 5.65 kJ/mol × (1000 J/1 kJ) = 5650 J/mol Now, substitute the converted ∆H value into the equation: ∆G = (5650 J/mol) - (200 K)(28.9 J/K⋅mol)
03

Calculate ∆G

Now, calculate the value of ∆G: ∆G = 5650 J/mol - (200 K)(28.9 J/K⋅mol) ≈ 5650 J/mol - 5780 J/mol ≈ -130 J/mol Since the calculated ∆G is negative, the melting of NH₃(s) at 200 K is spontaneous. #b. Finding the approximate melting point#
04

Set ∆G equal to zero

At the melting point, the change in Gibbs free energy, ∆G, equals zero. We can use the same Gibbs free energy equation from part a and set ∆G to zero: 0 = ∆H - T∆S
05

Solve for the temperature, T

Now, rearrange the equation and solve for the temperature, T: T = ∆H / ∆S Substitute the given values for ∆H and ∆S: T = (5650 J/mol) / (28.9 J/K⋅mol) ≈ 195.5 K The approximate melting point of ammonia is 195.5 K.

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Key Concepts

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

Gibbs Free Energy
Gibbs Free Energy is a key concept in determining the spontaneity of a process. It is denoted by \( \Delta G \) and is calculated using the formula:
  • \( \Delta G = \Delta H - T\Delta S \)
Where:
  • \( \Delta H \) is the change in enthalpy
  • \( T \) is the temperature in Kelvin
  • \( \Delta S \) is the change in entropy
The sign of \( \Delta G \) indicates whether a reaction is spontaneous:
  • Negative \( \Delta G \): The process is spontaneous
  • Positive \( \Delta G \): The process is non-spontaneous
  • Zero \( \Delta G \): The system is in equilibrium
In the context of melting, if melting is spontaneous at a given temperature, \( \Delta G \) will be negative. For example, for ammonia at 200 K, substituting the given values for enthalpy and entropy into the equation shows a negative \( \Delta G \), meaning NH₃ will spontaneously melt at this temperature.
Entropy of Fusion
Entropy of fusion, \( \Delta S \), is a measure of the disorder or randomness added to a system as a solid turns into a liquid. During fusion, disorder increases because the molecular structure of the solid breaks down to form a more randomized liquid structure.
  • It is usually expressed in units of J/K⋅mol.
  • A higher \( \Delta S \) implies a greater increase in disorder.
For ammonia, \( \Delta S \) of fusion is given as 28.9 J/K⋅mol. This value signifies how much disorder is introduced per mole of ammonia as it changes from a solid to a liquid. Entropy increases indicate the likelihood of spontaneous melting given sufficient heat (enthalpy). As seen in the calculation for ammonia at 200 K, the entropy contributes to the spontaneity of the melting process when considered with enthalpy in the Gibbs Free Energy equation.
Melting Point
The melting point of a substance is the temperature at which it transitions from a solid state to a liquid state. At this temperature, a system is in equilibrium, meaning there is no net change in the amount of solid and liquid:
  • Gibbs Free Energy \( \Delta G = 0 \)
  • The enthalpy \( \Delta H \) balances out the term \( T\Delta S \).
To find the melting point of a substance using thermodynamic data,we set the Gibbs Free Energy equation to zero and solve for \( T \). For ammonia, given its enthalpy of fusion and entropy, the approximate melting point is calculated to be 195.5 K. This temperature indicates the point at which solid ammonia will coexist with liquid ammonia under normal atmospheric pressure. Understanding this balance helps predict phase changes in various molecular and chemical systems.

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Most popular questions from this chapter

Consider the reactions $$ \begin{aligned} \mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) & \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) \\ \mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) & \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q) \end{aligned} $$ where $$ \text { en }=\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2} $$ The \(\Delta H\) values for the two reactions are quite similar, yet \(\mathrm{K}_{\text {reaction } 2}>K_{\text {reaction } 1} .\) Explain.

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

As \(\mathrm{O}_{2}(l)\) is cooled at \(1 \mathrm{~atm}\), it freezes at \(54.5 \mathrm{~K}\) to form solid \(\mathrm{I}\). At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the \(\mathrm{I} \rightarrow\) II phase transition is \(-743.1 \mathrm{~J} / \mathrm{mol}\), and \(\Delta S\) for the same transition is \(-17.0 \mathrm{~J} / \mathrm{K} \cdot\) mol. At what temperature are solids I and II in equilibrium?

For a liquid, which would you expect to be larger, \(\Delta S_{\text {fusion }}\) or \(\Delta S_{\text {evaporation }}\) ? Why?

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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