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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 constantpressure process, \(\Delta H=0 .\) What can you conclude about \(\Delta E, q\) and \(w ?\)

Short Answer

Expert verified
(a) The enthalpy change (\(\Delta H\)) is equal to the heat transfer (\(q\)) under constant pressure conditions, such as \(\Delta H = q_P\). (b) When the system releases heat to the surroundings during a constant-pressure process, the enthalpy of the system decreases (\(\Delta H < 0\)). (c) For a constant-pressure process with \(\Delta H = 0\), we can conclude that \(\Delta E = 0\), \(q = 0\), and \(w = 0\).

Step by step solution

01

a) Condition for enthalpy change equal to heat transfer

The enthalpy change (\(\Delta H\)) is equal to the heat transfer (\(q\)) under constant pressure conditions. Mathematically, this can be expressed as follows: \[\Delta H = q_P\] where \(P\) denotes constant pressure.
02

b) Enthalpy change during heat release in a constant-pressure process

When the system releases heat to the surroundings during a constant-pressure process, the heat transfer (\(q\)) is negative (i.e., \(q < 0\)), as energy is flowing out of the system. According to the relationship in part (a), we have \(\Delta H = q_P\), so if \(q_P < 0\), then \(\Delta H < 0\). This means that the enthalpy of the system decreases during the process.
03

c) Conclusion for \(\Delta H = 0\) in a constant-pressure process

When \(\Delta H = 0\) in a constant-pressure process, the heat transfer (\(q_P\)) equals zero as well, since \(\Delta H = q_P\). We can use this information and the first law of thermodynamics to determine the values or relationships of \(\Delta E\) and \(w\). The first law of thermodynamics can be expressed as: \[\Delta E = q + w\] Since \(\Delta H = 0\) and we know that \(\Delta H = q_P\), we have: \[q + w = 0\] Since \(q = 0\) (no heat transfer), the only way for the equation to be satisfied is when \(w = 0\). Therefore, for a constant-pressure process with \(\Delta H = 0\), we can conclude that \(\Delta E = 0\), \(q = 0\), and \(w = 0\).

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Most popular questions from this chapter

A 1.800 -g sample of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) was burned in a bomb calorimeter whose total heat capacity is \(11.66 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter plus contents increased from \(21.36^{\circ} \mathrm{C}\) to \(26.37{ }^{\circ} \mathrm{C}\). (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of phenol? Per mole of phenol?

Using values from Appendix \(\mathrm{C},\) 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) \(4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Identify the force present and explain whether work is being performed in the following cases: (a) You lift a pencil off the top of a desk. (b) A spring is compressed to half its normal length.

(a) What are the units of molar heat capacity? (b) What are the units of specific heat? (c) If you know the specific heat of copper, what additional information do you need to calculate the heat capacity of a particular piece of copper pipe?

At the end of 2009 , global population was about 6.8 billion people. What mass of glucose in kg would be needed to provide 1500 Cal/person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) according to the following thermochemical equation: \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H^{\mathrm{o}}=-2803 \mathrm{~kJ}\)
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