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Chapter 4: Problem 171

(A): The enthalpy of formation of \(\mathrm{H}_{2} \mathrm{O}(l)\) is greater than that of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \((\mathbf{R})\) : Enthalpy change is \(-\) ve for the condensation reaction, \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}\) (l)

Short Answer

Expert verified
A is false, R is true but does not explain A.

Step by step solution

01

Understand Enthalpy of Formation

The enthalpy of formation ( abla_f H) is the heat change that occurs when one mole of a compound forms from its elements in their standard states. abla_f H for liquid water ( abla_f H_{H_2O(l)}) is the energy required to form liquid water, while abla_f H_{H_2O(g)} is for vapor (gaseous water). Generally, bond formation releases energy, hence liquid water would have lower energy content due to additional hydrogen bonds.
02

Analyze Energy Levels

Gaseous water ( abla_f H_{H_2O(g)}) formation requires energy consumption to maintain water molecules in the gas phase, where intermolecular forces are minimal. Consequently, abla_f H_{H_2O(l)} < abla_f H_{H_2O(g)} because turning gas into liquid releases energy (exothermic process), thus liquid water formation is favored energetically.
03

Evaluate Condensation Reaction Enthalpy

The condensation reaction H_2O(g) -> H_2O(l) represents the transition from a high energy state to a lower energy one. The enthalpy change for this reaction ( abla H) is negative ( abla H < 0), as energy is released when water condenses from gas to liquid form, confirming this is an exothermic process.
04

Conclude the Relationship between A and R

Assertion (A) states that abla_f H_{H_2O(l)} > abla_f H_{H_2O(g)} which is incorrect as established in Step 2, whereas Reasoning (R) explains that the condensation is exothermic, correctly noting that energy is released. Thus, R correctly describes an accurate process, but it does not justify the incorrect assertion in (A).

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Key Concepts

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

Enthalpy Change
Enthalpy change is a measure of the amount of energy absorbed or released by a system during a chemical reaction. When we discuss reactions, we often talk about whether they consume or release energy. This energy change is captured in the idea of enthalpy change, denoted as \( \Delta H \).
Typically:
  • If \( \Delta H \) is negative, the reaction releases energy, making it exothermic.
  • If \( \Delta H \) is positive, the reaction absorbs energy, making it endothermic.
Understanding the sign of \( \Delta H \) is crucial because it tells us about the energy profile of a reaction. For example, in the condensation of water vapor to liquid, the enthalpy change is negative. This indicates that energy is being released during this process.
Condensation Reaction
A condensation reaction is a transformation from a gaseous state to a liquid state, here illustrated by the reaction of water vapor changing into liquid water: \( \mathrm{H}_2\mathrm{O(g)} \rightarrow \mathrm{H}_2\mathrm{O(l)} \).
In this reaction type, the molecules in the gas phase lose energy to become more orderly arranged in the liquid state. During condensation, energy is released because less energy is needed to hold molecules together in a liquid than to keep them apart in a gaseous form.
Key Points:
  • Energy is released (exothermic).
  • There is a decrease in the system's energy.
  • The reaction moves toward a more stable (lower energy) state.
Understanding condensation reactions helps in textile manufacturing, air conditioning, and climate science, where water vapor transitions play a crucial role.
Exothermic Process
An exothermic process is any physical or chemical change that releases energy to the surroundings, often in the form of heat. The essence of an exothermic reaction can be captured by observing its enthalpy change. As mentioned, if \( \Delta H \) is negative, the process is exothermic.
For example, the condensation of water vapor is exothermic. When water vapor transitions to a liquid state, it releases stored energy, heating up the surroundings. This is the same principle that powers steam heating systems.
Characteristics of Exothermic Processes:
  • Release energy (heat, light, etc.).
  • Products have lower energy than reactants.
  • Often increase the temperature of the surroundings.
Recognizing exothermic processes is vital in fields like combustion, metallurgy, and even in the biological processes of warm-blooded animals.

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Most popular questions from this chapter

(A): In case of some glassy solids having mixture of isotopes, crystals of \(\mathrm{CO}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO}\) etc. entropy is not zero even at absolute zero temperature (R): These kind of solids do not have perfect order even at absolute zero temperature.

The enthalpy of solution of \(\mathrm{BaCl}_{2}\) (s) and \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) are \(-20.6\) and \(8.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respec- tively. The enthalpy change for the hydration of \(\mathrm{BaCl}_{2}(\mathrm{~s})\) is a. \(29.8 \mathrm{~kJ}\) b. \(-11.8 \mathrm{~kJ}\) c. \(-20.6 \mathrm{~kJ}\) d. \(-29.4 \mathrm{~kJ}\).

Internal energy and entropy are state functions a. internal energy (U) is a state function and \(\Delta \mathrm{U}\) is independent of path. b. In a cyclic process: \(\Delta \mathrm{S}=0\) but \(\Delta \mathrm{U} \neq 0\) c. \(\mathrm{Cv}\) values of \(\mathrm{H}_{2}\) and He are equal at all temperatures d. \(\mathrm{q}\) and \(\mathrm{w}\) are path dependent

The direct conversion of \(\mathrm{A}\) to \(\mathrm{B}\) is difficult, hence it is carried out by the following shown path: Given \(\Delta S_{(\Lambda \rightarrow C)}=50\) e.u \(\Delta \mathrm{S}_{(\mathrm{C} \rightarrow \mathrm{D})}=30 \mathrm{e} . \mathrm{u}\) \(\Delta \mathrm{S}_{(\mathrm{B} \rightarrow \mathrm{D})}=20 \mathrm{e} . \mathrm{u}\) where e.u. is entropy unit then \(\Delta \mathrm{S}(\mathrm{A} \rightarrow \mathrm{B})\) is a. \(+100 \mathrm{e} . \mathrm{u}\). b. \(+60 \mathrm{e} . \mathrm{u}\). c. \(-100\) e.u. d. \(-60\) e.u.

(A): In the case of an ideal gas, the changes in Gibbs and Helmholtz free energies are equal to each other \((\Delta \mathrm{G}=\Delta \mathrm{A})\) for isothermal reversible processes. (R): There is no change in internal energies and enthalpies for ideal gases at constant temperature.
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