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Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ}\) ? a. \(\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)\) b. \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
Reactions b, c, and d are expected to have a negative value for \(\Delta S^{\circ}\).

Step by step solution

01

Compare the number and physical states of reactants to products in a given reaction, which could give rise to order or disorder. More molecules generally indicate a higher entropy state, while gas-phase molecules have higher entropy than liquids and solids due to their increased freedom of movement. #Step 2: Analyzing Reaction a#

In reaction a: \(\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)\), we have one gas and one aqueous compound on the left side, and two gaseous molecules on the right side. In general, gas molecules have higher entropy than other states, so having more gas molecules on the product side would indicate an increase in entropy. Thus, \(\Delta S^{\circ}\) is expected to be positive for reaction a. #Step 3: Analyzing Reaction b#
02

In reaction b: \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\), we have one gas and one solid on the reactant side, and only a solid product. The number of solid molecules is decreasing, and the gas molecules are completely disappearing. As mentioned earlier, gas molecules have higher entropy than other states, so their disappearance signifies a decrease in entropy. Thus, \(\Delta S^{\circ}\) is expected to be negative for reaction b. #Step 4: Analyzing Reaction c#

In reaction c: \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\), we have two gas molecules on the reactant side and one gas molecule on the product side. Since gas molecules have higher entropy than other states, the reduction in the number of gas molecules indicates a decrease in entropy. Thus, \(\Delta S^{\circ}\) is expected to be negative for reaction c. #Step 5: Analyzing Reaction d#
03

In reaction d: \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\), we have one gas and one liquid on the reactant side, and only a liquid product. The gas molecule on the reactant side disappears in the product, signifying a decrease in entropy. Thus, \(\Delta S^{\circ}\) is expected to be negative for reaction d. #Step 6: Analyzing Reaction e#

In reaction e: \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\), we have one solid on the reactant side and one liquid on the product side. Since liquids have higher entropy than solids (though lower than gases), this reaction has an increase in entropy. Thus, \(\Delta S^{\circ}\) is expected to be positive for reaction e. Based on our analysis, reactions b, c, and d are expected to have a negative value for \(\Delta S^{\circ}\).

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Key Concepts

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

Understanding Thermodynamics in Chemical Reactions
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In chemical reactions, thermodynamics provides vital insights into how energy is transferred and transformed. The laws of thermodynamics govern the flow of energy and can predict whether a reaction will occur spontaneously.

One core concept in thermodynamics is the idea of disorder, or entropy. Entropy measures the degree of randomness or chaos in a system. In chemical processes, reactions tend to favor an increase in entropy, meaning they naturally shift towards more disorder. However, this isn't always the case, as some reactions result in a decrease in entropy, signifying an increase in order.

When contemplating a chemical reaction, entropy is a critical component determining spontaneity along with enthalpy, which is the total heat content of a system. These two concepts are combined in the Gibbs free energy equation to establish the feasibility of a reaction under constant temperature and pressure.
Gibbs Free Energy and Reaction Spontaneity
Gibbs free energy, denoted as G, indicates the total amount of energy available to do work during a chemical reaction at a constant temperature and pressure. The change in Gibbs free energy, \( \Delta G^\circ \), is what scientists use to predict whether a reaction will occur spontaneously. The equation \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \) combines enthalpy (\( \Delta H^\circ \)), temperature (\( T \)), and entropy (\( \Delta S^\circ \)) to assess this.

A negative value for \( \Delta G^\circ \) means the reaction is spontaneous, while a positive value means it is not. If \( \Delta G^\circ \) is zero, the system is at equilibrium. Therefore, understanding both the enthalpy change and the entropy change is crucial in predicting the behavior of a chemical reaction.
The Role of Physical States of Reactants and Products
The physical states of reactants and products—solid (s), liquid (l), gas (g), or aqueous (aq)—play an integral role in determining the entropy change during a reaction. Generally, gases have the highest entropy, followed by liquids, with solids having the least because entropy is associated with the freedom of particles to move and spread out.

When a reaction leads to an increase in the number of gas molecules or changes the state from a solid or liquid to a gas, the entropy typically increases. Conversely, if a reaction produces fewer gas molecules or results in the formation of a liquid or solid from a gas, entropy usually decreases.
  • Gas Formation: Increased number of gas molecules usually means higher entropy.
  • Liquid or Solid Formation: A reduction in gas molecules or formation of a more ordered phase typically indicates lower entropy.
An understanding of these physical state changes is vital when determining the overall entropy change for a reaction.
Predicting Entropy Change (\(\Delta S^\circ\))
Entropy change (\(\Delta S^\circ\)) in a reaction can be predicted by analyzing the number and states of reactants and products. This process involves assessing whether the products will be more or less disordered than the reactants.

For instance, if a reaction results in a greater number of product molecules, especially gases, the entropy likely increases. On the other hand, if the total number of gaseous molecules decreases, the entropy likely decreases.
  • More Products or Gases: Indicates an increase in entropy (positive \(\Delta S^\circ\)).
  • Fewer Gases: Suggests a decrease in entropy (negative \(\Delta S^\circ\)).
Using this method, students can systematically approach entropy prediction in chemical reactions, thereby improving their ability to analyze thermodynamic problems effectively. It is important to note, however, that this is a simplified rule of thumb, and actual entropy changes can be influenced by more specific molecular interactions.

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Most popular questions from this chapter

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

From data in Appendix 4, calculate \(\Delta H^{\circ}, \Delta S^{\circ}\), and \(\Delta G^{\circ}\) for each of the following reactions at \(25^{\circ} \mathrm{C}\). a. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)\) c. \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(s)\) d. \(\mathrm{HCl}(\mathrm{g})+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\)

Consider the reaction $$ 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) $$ a. Calculate \(\Delta G^{\circ}\) for this reaction. The \(\Delta G_{\mathrm{f}}^{\circ}\) values for \(\mathrm{POCl}_{3}(g)\) and \(\mathrm{PCl}_{3}(g)\) are \(-502 \mathrm{~kJ} / \mathrm{mol}\) and \(-270 . \mathrm{kJ} / \mathrm{mol}\), respectively. b. Is this reaction spontaneous under standard conditions at \(298 \mathrm{~K}\) ? c. The value of \(\Delta S^{\circ}\) for this reaction is \(179 \mathrm{~J} / \mathrm{K} \cdot\) mol. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

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 following energy levels, each capable of holding two particles: $$ \begin{array}{l} E=2 \mathrm{~kJ}\\\ \boldsymbol{E} \quad E=1 \mathrm{~kJ}\\\ E=0 \quad \quad X X \end{array} $$ Draw all the possible arrangements of the two identical particles (represented by \(X\) ) in the three energy levels. What total energy is most likely, that is, occurs the greatest number of times? Assume that the particles are indistinguishable from each other.

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