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At \(298 \mathrm{K},\) for the reaction \(2 \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \stackrel{\text { Lest }}{\longrightarrow}\) \(2 \mathrm{POCl}_{3}(1), \Delta H^{\circ}=-620.2 \mathrm{kJ}\) and the standard molar entropies are \(\mathrm{PCl}_{3}(\mathrm{g}), 311.8 \mathrm{JK}^{-1} ; \mathrm{O}_{2}(\mathrm{g}), 205.1 \mathrm{JK}^{-1}\) and \(\mathrm{POCl}_{3}(1), 222.4 \mathrm{JK}^{-1} .\) Determine (a) \(\Delta G^{\circ}\) at \(298 \mathrm{K}\) and (b) whether the reaction proceeds spontaneously in the forward or the reverse direction when reactants and products are in their standard states.

Short Answer

Expert verified
The detailed calculation of ΔG° and the determination of the spontaneity of the reaction will be done in the steps above.

Step by step solution

01

Calculate Entropy Change

First, calculate the change in standard molar entropy (ΔS°) for the reaction by subtracting the sum of the standard molar entropies of the reactants from the sum of the standard molar entropies of the products. The standard molar entropies are given in the problem. Use the standard molar entropies and the balanced equation coefficients to calculate ΔS° using the formula ΔS° = ∑(μΔS° of products) - ∑(νΔS° of reactants). μ and ν are the stoichiometric coefficients of the products and reactants respectively.
02

Convert Entropy Change

Since ΔS°, the change in entropy, is given in J K^-1, this needs to be converted to kJ K^-1 before use in the next part. To convert from J to kJ, simply divide the result by 1000.
03

Calculate Gibbs Free Energy Change

Now we can calculate ΔG° using the formula ΔG° = ΔH° - TΔS° where T is the temperature in Kelvin. Just plug in the values of ΔH°, T, and ΔS° from Steps 1 and 2.
04

Determine if the Reaction is Spontaneous

Look at the sign of ΔG° to determine if the reaction is spontaneous in either direction. If ΔG° is negative, the reaction is spontaneous in the forward direction. If it is positive, the reaction is spontaneous in the reverse direction.

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Key Concepts

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

Entropy Change
Entropy is a fundamental concept in thermodynamics, representing the degree of disorder or randomness in a system. In chemical reactions, the change in entropy (\(\Delta S^{\circ}\)) is crucial to understand how energy is distributed.
You can calculate \(\Delta S^{\circ}\) by finding the difference between the standard entropies of products and reactants, weighted by their stoichiometric coefficients.
\[\Delta S^{\circ} = \sum (\mu S^{\circ}_{\text{products}}) - \sum (u S^{\circ}_{\text{reactants}})\]This calculation helps us predict the system's entropy change during a reaction.
  • Positive \(\Delta S^{\circ}\): More disorder.
  • Negative \(\Delta S^{\circ}\): Less disorder.
In the given reaction, we evaluate the entropy change to further determine the spontaneity and calculate Gibbs Free Energy.
Standard Molar Enthalpy
The standard molar enthalpy (\(\Delta H^{\circ}\)) represents the heat change during a reaction at constant pressure and standard conditions. It tells us how much heat is released or absorbed.
In this exercise, the given \(\Delta H^{\circ}\) is \(-620.2 \text{kJ}\), indicating it is an exothermic reaction - releasing heat into the surroundings.
  • If \(\Delta H^{\circ} < 0\): Exothermic reaction (heat released).
  • If \(\Delta H^{\circ} > 0\): Endothermic reaction (heat absorbed).
Understanding \(\Delta H^{\circ}\) helps predict how the energy of the system changes and its thermodynamic stability.
Spontaneity of Reactions
The spontaneity of a reaction is determined by Gibbs Free Energy Change (\(\Delta G^{\circ}\)), which combines enthalpy (\(\Delta H^{\circ}\)), entropy (\(\Delta S^{\circ}\)), and temperature (\(T\)) to predict if a reaction will occur on its own.
The formula is given by:\[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]
  • Negative \(\Delta G^{\circ}\): Reaction is spontaneous forward.
  • Positive \(\Delta G^{\circ}\): Reaction is spontaneous in reverse.
  • Zero \(\Delta G^{\circ}\): Reaction is in equilibrium.
For the given exercise, calculating \(\Delta G^{\circ}\) will show whether the reaction proceeds as desired in the forward direction or not.
Thermodynamic Calculations
Thermodynamic calculations are essential to understand the energy changes and feasibility of a reaction. They involve evaluating enthalpy, entropy, and Gibbs Free Energy.
  • \(\Delta H^{\circ}\): Determine heat absorbed or released.
  • \(\Delta S^{\circ}\): Quantify change in disorder.
  • \(\Delta G^{\circ}\): Predict spontaneous nature of the reaction.
In this specific exercise, after calculating \(\Delta S^{\circ}\) and converting it to kJ/K,
apply it alongside \(\Delta H^{\circ}\) in the Gibbs Free Energy equation at the given temperature (298K) to get \(\Delta G^{\circ}\).
This process ties all molecular and energy changes of the system, ensuring a comprehensive understanding of its overall thermodynamic progress.

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Most popular questions from this chapter

The standard Gibbs energy change for the reaction $$\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) \rightleftharpoons \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$$ is \(29.05 \mathrm{kJ} \mathrm{mol}^{-1}\) at \(298 \mathrm{K}\). Use this thermodynamic quantity to decide in which direction the reaction is spontaneous when the concentrations of \(\mathrm{NH}_{3}(\mathrm{aq})\) \(\mathrm{NH}_{4}^{+}(\mathrm{aq}),\) and \(\mathrm{OH}^{-}(\mathrm{aq})\) are \(0.10 \mathrm{M}, 1.0 \times 10^{-3} \mathrm{M}\) and \(1.0 \times 10^{-3} \mathrm{M},\) respectively.

At room temperature and normal atmospheric pressure, is the entropy of the universe positive, negative, or zero for the transition of carbon dioxide solid to liquid?

A tabulation of more precise thermodynamic data than are presented in Appendix D lists the following values for \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(298.15 \mathrm{K},\) at a standard state pressure of 1 bar. $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ}, & \Delta G_{f,}^{\circ} & S_{\prime}^{\circ} \\\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{H}_{2} \mathrm{O}(1) & -285.830 & -237.129 & 69.91 \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & -241.818 & -228.572 & 188.825 \\ \hline \end{array}$$ (a) Use these data to determine, in two different ways, \(\Delta G^{\circ}\) at \(298.15 \mathrm{K}\) for the vaporization: \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{bar}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 1 \mathrm{bar}) .\) The value you obtain will differ slightly from that on page 838 because here, the standard state pressure is 1 bar, and there, it is 1 atm. (b) Use the result of part (a) to obtain the value of \(K\) for this vaporization and, hence, the vapor pressure of water at \(298.15 \mathrm{K}\) (c) The vapor pressure in part (b) is in the unit bar. Convert the pressure to millimeters of mercury. (d) Start with the value \(\Delta G^{\circ}=8.590 \mathrm{kJ}\), given on page 838 and calculate the vapor pressure of water at 298.15 K in a fashion similar to that in parts (b) and (c). In this way, demonstrate that the results obtained in a thermodynamic calculation do not depend on the convention we choose for the standard state pressure, as long as we use standard state thermodynamic data consistent with that choice.

For one of the following reactions, \(K_{c} K_{p}=K .\) Identify that reaction. For the other two reactions, what is the relationship between \(K_{c}, \bar{K}_{\mathrm{p}},\) and \(K ?\) Explain. (a) \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\) (b) \(\mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{g})\) (c) \(\mathrm{NH}_{4} \mathrm{HCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1)\)

What values of \(\Delta H, \Delta S,\) and \(\Delta G\) would you expect for the formation of an ideal solution of liquid components? (Is each value positive, negative, or zero?)

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