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Chapter 5: Problem 33

(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

Expert verified
(a) The change in enthalpy (\(\Delta H\)) is easier to measure than the change in internal energy (\(\Delta U\)) because most experiments occur at constant pressure, making \(\Delta H\) equal to the heat absorbed or released. Measuring work, which is required to calculate \(\Delta U\), is difficult in practice. (b) Enthalpy (\(H\)) is a state function as it depends only on the system's current state (being the sum of \(U, P,\) and \(V\), all state functions). Heat (\(q\)) is not a state function as it depends on the path taken. (c) A process with a positive change in enthalpy (\(\Delta H > 0\)) at constant pressure is endothermic, as it involves absorption of heat by the system.

Step by step solution

01

(a) Comparison between measuring change in enthalpy and change in internal energy

Enthalpy (\(H\)) is a thermodynamic property defined as the sum of the system's internal energy (\(U\)) and the product of its pressure (\(P\)) and volume (\(V\)): \(H = U + PV\). The change in enthalpy (\(\Delta H\)) is usually easier to measure than the change in internal energy (\(\Delta U\)) because most practical experiments and reactions occur at constant pressure. Under constant pressure conditions, the change in enthalpy (\(\Delta H\)) is equal to the heat absorbed or released by the system (\(q_p\)): \(\Delta H = q_p\). However, the change in internal energy (\(\Delta U\)) involves measuring the changes in both heat (\(q\)) and work (\(w\)) done by or on the system, since \(\Delta U = q + w\). Measuring work is quite difficult in practice, especially for chemical reactions, while measuring heat absorbed or released is relatively easier. Therefore, it is more convenient to measure the change in enthalpy than the change in internal energy.
02

(b) Enthalpy as a state function and heat not being a state function

State functions are properties that depend only on the current state of the system and are independent of the path taken to reach that state. Enthalpy (\(H\)) is a state function because it is defined as the sum of the system's internal energy (\(U\), also a state function) and the product of pressure (\(P\), a state function) and volume (\(V\), a state function): \(H = U + PV\). On the other hand, heat (\(q\)) is not a state function because it is a form of energy transfer that depends on the path taken. The amount of heat transferred between two states of a system is not solely determined by the initial and final states but depends on the process through which the system changes.
03

(c) Determining whether a process is endothermic or exothermic based on the change in enthalpy

A process at constant pressure has a positive change in enthalpy (\(\Delta H > 0\)). Based on the definition of enthalpy and its relationship with heat at constant pressure conditions (\(\Delta H = q_p\)), a positive value for \(\Delta H\) implies that the heat has been absorbed by the system from its surroundings. Since endothermic processes involve absorption of heat by the system, a process with a positive change in enthalpy at constant pressure is endothermic.

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Most popular questions from this chapter

(a) When a \(0.235-\mathrm{g}\) sample of benzoic acid is combusted in a bomb calorimeter (Figure 5.19 ), the temperature rises \(1.642{ }^{\circ} \mathrm{C}\). When a 0.265 -g sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~N}_{4}\), is burned, the temperature rises \(1.525^{\circ} \mathrm{C}\). Using the value \(26.38 \mathrm{~kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to 0.001 \(\mathrm{g}\), what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

Given the data $$ \begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g) & & \Delta H=+180.7 \mathrm{~kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) & & \Delta H=-113.1 \mathrm{~kJ} \\ 2 \mathrm{~N}_{2} \mathrm{O}(g) & \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H &=-163.2 \mathrm{~kJ} \end{aligned} $$ use Hess's law to calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g) $$

Consider a system consisting of the following apparatus, in which gas is confined in one flask and there is a vacuum in the other flask. The flasks are separated by a valve. Assume that the flasks are perfectly insulated and will not allow the flow of heat into or out of the flasks to the surroundings. When the valve is opened, gas flows from the filled flask to the evacuated one. (a) Is work performed during the expansion of the gas? (b) Why or why not? (c) Can you determine the value of \(\Delta E\) for the process?

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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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