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Chapter 5: Problem 34

(a) Under what condition will the enthalpy change of a process equal the amount of heat transferred into or out of the system? (b) During a constant-pressure process, the system releases heat to the surroundings. Does the enthalpy of the system increase or decrease during the process? (c) In a constant-pressure process, \(\Delta H=0\). What can you conclude about \(\Delta E, q,\) and \(w ?\)

Short Answer

Expert verified
(a) The enthalpy change of a process will equal the amount of heat transferred into or out of the system under the condition of constant pressure, represented as: \(\Delta H = q_P\). (b) In a constant-pressure process where the system releases heat to the surroundings, the enthalpy of the system decreases as \(\Delta H\) will be negative. (c) If \(\Delta H = 0\) in a constant-pressure process, this implies \(q_P = 0\). Based on the first law of thermodynamics, we can conclude that the change in internal energy (\(\Delta E\)) is equal to the negative of the work done by the system (\(w\)), i.e., \(\Delta E = -w\).

Step by step solution

01

Condition for enthalpy change equal to heat transferred

For the enthalpy change to equal the amount of heat transferred into or out of the system, the process needs to happen at a constant pressure. Mathematically, this can be represented as: \[ \Delta H = q_P \] Where \(\Delta H\) is the change in enthalpy, and \(q_P\) is the heat transferred at constant pressure. #b#
02

Enthalpy Change during Constant-Pressure Process

In a constant pressure process, the enthalpy change is given by: \[ \Delta H = q_P \] Since the system is releasing heat to the surroundings, the heat transfer \(q_P\) will be negative, which implies that \(\Delta H\) will also be negative. Therefore, the enthalpy of the system decreases during the process. #c#
03

Conditions for Constant-Pressure Process with no Enthalpy Change

If \(\Delta H = 0\) during a constant-pressure process, the heat transferred at constant pressure must also be zero, i.e., \(q_P = 0\). Now, we can use the first law of thermodynamics which states that the change in internal energy (\(\Delta E\)) is equal to the heat added to the system (\(q\)) minus the work done by the system (\(w\)): \[ \Delta E = q - w \] Under constant pressure, \(q_P = q\), so in this case, \(q = 0\). This simplifies the equation to: \[ \Delta E = -w \] Since \(\Delta H = 0\), we can conclude that the change in the internal energy (\(\Delta E\)) is equal to the negative of the work done by the system (\(w\)).

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Most popular questions from this chapter

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{I}_{2}(s)\) or $1 \mathrm{~mol} \mathrm{I}_{2}(g)$ at the same temperature, (b) \(2 \mathrm{~mol}\) of iodine atoms or \(1 \mathrm{~mol}\) of \(\mathrm{I}_{2},(\mathbf{c}) 1 \mathrm{~mol} \mathrm{I}_{2}(g)\) and $1 \mathrm{~mol} \mathrm{H}_{2}(g)$ at \(25^{\circ} \mathrm{C}\) or \(2 \mathrm{~mol} \mathrm{HI}(g)\) at $25^{\circ} \mathrm{C},(\mathbf{d}) 1 \mathrm{~mol} \mathrm{H}_{2}(g)\( at \)100^{\circ} \mathrm{C}$ or \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(300^{\circ} \mathrm{C}\).

Under constant-volume conditions, the heat of combustion of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is $16.49 \mathrm{~kJ} / \mathrm{g}\(. A \)3.00-\mathrm{g}$ sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.94 to \(24.62^{\circ} \mathrm{C} .(\mathbf{a})\) What is the total heat capacity of the calorimeter? (b) If the size of the sucrose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

The complete combustion of methane, \(\mathrm{CH}_{4}(g)\), to form \(\mathrm{H}_{2} \mathrm{O}(l)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases \(890 \mathrm{~kJ}\) of heat per mole of \(\mathrm{CH}_{4}\). (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

Consider two solutions, the first being \(50.0 \mathrm{~mL}\) of $1.00 \mathrm{M} \mathrm{CuSO}_{4}\( and the second \)50.0 \mathrm{~mL}\( of \)2.00 \mathrm{M} \mathrm{KOH} .$ When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 to \(27.7^{\circ} \mathrm{C} .(\mathbf{a})\) Before mixing, how many grams of Cu are present in the solution of \(\mathrm{CuSO}_{4}\) ? (b) Predict the identity of the precipitate in the reaction. (c) Write complete and net ionic equations for the reaction that occurs when the two solutions are mixed. \((\mathbf{d})\) From the calorimetric data, calculate \(\Delta H\) for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is \(100.0 \mathrm{~mL},\) and that the specific heat and density of the solution after mixing are the same as those of pure water.

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