Chapter 6: Problem 9
Describe two exothermic processes and two endothermic processes.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 6: Problem 9
Describe two exothermic processes and two endothermic processes.
All the tools & learning materials you need for study success - in one app.
Get started for free
Calculate the standard enthalpy change for the reaction $$ 2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ given that $$ \begin{aligned} 2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow & \mathrm{Al}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-1669.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow & \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-822.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$
Lime is a term that includes calcium oxide \((\mathrm{CaO}\) also called quicklime) and calcium hydroxide \(\left[\mathrm{Ca}(\mathrm{OH})_{2},\right.\) also called slaked lime \(] .\) It is used in the steel industry to remove acidic impurities, in airpollution control to remove acidic oxides such as \(\mathrm{SO}_{2}\), and in water treatment. Quicklime is made industrially by heating limestone \(\left(\mathrm{CaCO}_{3}\right)\) above \(2000^{\circ} \mathrm{C}\) : $$ \begin{aligned} \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \\ \Delta H^{\circ} &=177.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Slaked lime is produced by treating quicklime with water: $$ \begin{aligned} \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s) \\ \Delta H^{\circ} &=-65.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ The exothermic reaction of quicklime with water and the rather small specific heats of both quicklime \(\left(0.946 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) and slaked lime \(\left(1.20 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) make it hazardous to store and transport lime in vessels made of wood. Wooden sailing ships carrying lime would occasionally catch fire when water leaked into the hold. (a) If a 500 -g sample of water reacts with an equimolar amount of \(\mathrm{CaO}\) (both at an initial temperature of \(\left.25^{\circ} \mathrm{C}\right)\), what is the final temperature of the product, \(\mathrm{Ca}(\mathrm{OH})_{2} ?\) Assume that the product absorbs all of the heat released in the reaction. (b) Given that the standard enthalpies of formation of \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O}\) are \(-635.6 \mathrm{~kJ} / \mathrm{mol}\) and \(-285.8 \mathrm{~kJ} / \mathrm{mol}\), respectively, cal- culate the standard enthalpy of formation of \(\mathrm{Ca}(\mathrm{OH})_{2}\).
From the following data, $$ \begin{array}{c} \mathrm{C} \text { (graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-3119.6 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ calculate the enthalpy change for the reaction $$ 2 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) $$
You are given the following data: $$ \begin{array}{c} \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) \quad \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{Br}(g) \quad \Delta H^{\circ}=192.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g) \\ \Delta H^{\circ}=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}(g)+\operatorname{Br}(g) \longrightarrow \operatorname{HBr}(g) $$
A gas expands in volume from \(26.7 \mathrm{~mL}\) to \(89.3 \mathrm{~mL}\) at constant temperature. Calculate the work done (in joules) if the gas expands (a) against a vacuum, (b) against a constant pressure of \(1.5 \mathrm{~atm},\) and (c) against a constant pressure of 2.8 atm.
What do you think about this solution?
We value your feedback to improve our textbook solutions.