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Chapter 5: Problem 31

(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) For a given process at constant pressure, \(\Delta H\) is negative. Is the process endothermic or exothermic?

Short Answer

Expert verified
(a) Measuring change in enthalpy (螖H) is easier than change in internal energy (螖E) because enthalpy change is mainly the heat transferred at constant pressure, which is common in lab experiments, while internal energy change depends on both heat and work done, making it harder to measure accurately. (b) If the change in enthalpy (螖H) at constant pressure is negative, the process is exothermic, as heat is being released by the system to the surroundings. If 螖H were positive, the process would be endothermic.

Step by step solution

01

Answer to (a) - Explanation for measuring ease

The change in enthalpy (螖H) and the change in internal energy (螖E) are both measures of the heat absorbed or released by a system during a process. However, enthalpy change is easier to measure experimentally, as it is mainly the heat transferred at constant pressure, whereas the change in internal energy also depends on the work done by or on the system. In most chemistry laboratory experiments, the processes occur at constant pressure, which makes it easier to measure the heat transferred directly as 螖H. Measuring 螖E, on the other hand, would need the precise measurement of heat and work done, which might not be feasible due to inaccuracies in work done calculations and the instruments available.
02

Answer to (b) - Classifying the process

For a given process at constant pressure, if the change in enthalpy (螖H) is negative, it means that the heat is being released by the system to the surroundings. When heat is released in a process, that process is called exothermic. So, the process is exothermic. Conversely, if the change in enthalpy were positive, indicating heat being absorbed by the system, the process would be endothermic.

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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?

Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\). From the following enthalpy of reaction data and data in Appendix \(C\), calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s):\) \(\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)\) \(\Delta H^{\circ}=-127.2 \mathrm{~kJ}\)

Consider the following hypothetical reactions: $$ \begin{array}{ll} \mathrm{A} \rightarrow \mathrm{B} & \Delta H=+30 \mathrm{~kJ} \\ \mathrm{~B} \longrightarrow \mathrm{C} & \Delta H=+60 \mathrm{~kJ} \end{array} $$ (a) Use Hess's law to calculate the enthalpy change for the reaction A - - - - C. (b) Construct an enthalpy diagram for substances \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\), and show how Hess's law applies.

(a) When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, 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 $$ \begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)-\cdots+2 \mathrm{NO}(g) & \Delta H=+180.7 \mathrm{~kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \cdots-\rightarrow 2 \mathrm{NO}_{2}(g) & \Delta H=-113.1 \mathrm{~kJ} \\ 2 \mathrm{~N}_{2} \mathrm{O}(g)-\cdots 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-163.2 \mathrm{~kJ} \end{aligned} $$ use Hess's law to calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \stackrel{-\cdots} 3 \mathrm{NO}(g) $$
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