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(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) \(H\) is a state function, but \(q\) is not a state function. Explain. (c) For a given process at constant pressure, \(\Delta H\) is positive. Is the process endothermic or exothermic?

Short Answer

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(a) The change in enthalpy, ΔH, is easier to measure than the change in internal energy, ΔU, because it can be determined experimentally by measuring the heat transferred at constant pressure, which is often more convenient and simpler than directly measuring work done in addition to heat exchange. (b) Enthalpy (H) is a state function because it depends only on the system's current state, while heat (q) is not a state function because it depends on the path taken during the process. (c) For a process with a positive change in enthalpy (ΔH > 0) at constant pressure, the process is endothermic, meaning the system absorbs heat from its surroundings.

Step by step solution

01

(a) Easier measurement of enthalpy change

Enthalpy (H) is a measure of the total energy of a system, including its internal energy and the energy associated with work done by the system against its surroundings (pressure multiplied by volume). The change in enthalpy, ΔH, can be determined experimentally by measuring the heat transferred between the system and its surroundings at constant pressure. On the other hand, the change in internal energy, ΔU, usually involves not only heat exchange but also work done, which can be more difficult to measure. Since most experiments are conducted at constant pressure (e.g., at atmospheric pressure), it is often simpler and more convenient to measure the heat exchange to determine the change in enthalpy, rather than directly measuring the change in internal energy.
02

(b) State functions vs non-state functions

A state function is a property of a system that depends only on its current state (i.e., the values of its variables like temperature, pressure, volume, etc.) and not on the path taken to reach that state. In contrast, non-state functions, or path functions, depend on the specific steps taken to change the system. Enthalpy (H) is a state function because its value is determined solely by the temperature, pressure, and composition of the system. No matter what path the system takes between two states, the change in enthalpy will always be the same. Heat (q) is not a state function because the amount of heat transferred in a process depends on the nature of the path the system takes to go from the initial state to the final state. For example, if a system undergoes a reversible process versus an irreversible process, the amount of heat transferred during these two paths will not be the same. Therefore, heat is a path function and not a state function.
03

(c) Determining endothermic or exothermic processes

For a given process at constant pressure with a positive change in enthalpy (ΔH > 0), the process is endothermic. This is because a positive ΔH value indicates an overall increase in the total energy of the system, which means the system has absorbed heat from its surroundings. If ΔH were negative, the process would be exothermic, meaning the system would be releasing heat to its surroundings. In summary, a positive ΔH corresponds to an endothermic process, while a negative ΔH corresponds to an exothermic process.

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Most popular questions from this chapter

Using values from Appendix C, calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(\mathrm{CaO}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{CaCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (c) \(2 \mathrm{CuO}(s)+\mathrm{NO}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{NO}_{2}(g)\) (d) \(4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

A watt is a measure of power (the rate of energy change) equal to \(1 \mathrm{~J} / \mathrm{s}\). (a) Calculate the number of joules in a kilowatthour. (b) An adult person radiates heat to the surroundings at about the same rate as a 100-watt electric incandescent lightbulb. What is the total amount of energy in kcal radiated to the surroundings by an adult in \(24 \mathrm{~h}\) ?

From the enthalpies of reaction $$ \begin{gathered} 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g) \quad \Delta H=-221.0 \mathrm{~kJ} \\ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g)+4 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{CH}_{3} \mathrm{OH}(g) \quad \Delta H=-402.4 \mathrm{~kJ} \\ \text { calculate } \Delta H \text { for the reaction } \\ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) \end{gathered} $$

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) \(1.36\) \(\mathrm{mol}\) of \(\mathrm{O}_{2}\) and (b) \(10.4 \mathrm{~g}\) of \(\mathrm{KCl}\). (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\), is likely to be feasible under ordinary conditions? Explain your answer.

For the following processes, calculate the change in internal energy of the system and determine whether the process is endothermic or exothermic: (a) A balloon is heated by adding 850 J of heat. It expands, doing $382 \mathrm{~J}\( of work on the atmosphere. (b) A \)50-g$ sample of water is cooled from \(30^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\), thereby losing approximately \(3140 \mathrm{~J}\) of heat. (c) A chemical reaction releases \(6.47 \mathrm{~kJ}\) of heat and does no work on the surroundings.

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