/*! 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} Problem 79 Ethanol \(\left(\mathrm{C}_{2} \... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 79

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is currently blended with gasoline as an automobile fuel. (a) Write a balanced equation for the combustion of liquid ethanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming \(\mathrm{H}_{2} \mathrm{O}(g)\) as a product. (c) Calculate the heat produced per liter of ethanol by combustion of ethanol under constant pressure. Ethanol has a density of \(0.789 \mathrm{~g} / \mathrm{mL}\). (d) Calculate the mass of \(\mathrm{CO}_{2}\) produced per \(\mathrm{kJ}\) of heat emitted.

Short Answer

Expert verified
The balanced equation for the combustion of ethanol in air is: \[C_2H_5OH (l) + 3O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (g)\] The standard enthalpy change for this reaction is \(-1367.3 \ \mathrm{kJ/mol}\). The heat produced per liter of ethanol by combustion under constant pressure is \(-23419 \ \mathrm{kJ/L}\). The mass of CO鈧� produced per kJ of heat emitted is \(-0.0327 \ \mathrm{g/kJ}\).

Step by step solution

01

(a) Balanced equation for combustion of ethanol

For the combustion of ethanol, we will need oxygen from the air. The products of ethanol's combustion are carbon dioxide (CO2) and water (H2O). The balanced equation for the combustion of ethanol can be written as: \[C_2H_5OH (l) + 3O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (g)\]
02

(b) Calculate the standard enthalpy change

To find the standard enthalpy change for the reaction, we can use the heat of formation values of the reactants and the products. The standard heat of formation (螖Hf掳) values are: Ethanol: \(-277.7 \ \mathrm{kJ/mol}\) Oxygen: \(0 \ \mathrm{kJ/mol}\) Carbon dioxide: \(-393.5 \ \mathrm{kJ/mol}\) Water (gas): \(-241.8 \ \mathrm{kJ/mol}\) Using the standard enthalpy change formula, we have: 螖H掳 = [Sum of (螖Hf掳 of products)] - [Sum of (螖Hf掳 of reactants)] 螖H掳 = [2(-393.5) + 3(-241.8)] - [(-277.7) + 3(0)] 螖H掳 = \(-1367.3 \ \mathrm{kJ/mol}\)
03

(c) Calculate the heat produced per liter of ethanol

We are given the density of ethanol as \(0.789 \ \mathrm{g/mL}\). To find the heat produced per liter of ethanol, we first need to convert the density to grams per liter, then use the molar mass of ethanol and the standard enthalpy change to find the heat produced. 1: Convert the density of ethanol to grams per liter: Density = \(\frac{0.789 \ \mathrm{g} }{\mathrm{mL}} 脳 1000 \ \mathrm{mL/L} = 789 \ \mathrm{g/L}\) 2: Use the molar mass of ethanol (C: 12.01g/mol, H: 1.01g/mol, and O: 16.00g/mol) to find moles per liter of ethanol: Moles per liter = \(\frac{789 \ \mathrm{g/L}}{2(12.01) + 6(1.01) + 1(16.00) \ \mathrm{g/mol}} = \frac{789}{46.07} = 17.13 \ \mathrm{mol/L}\) 3: Use the standard enthalpy change and moles per liter to find the heat produced per liter of ethanol: Heat produced = \(\frac{-1367.3 \ \mathrm{kJ/mol}}{1 \ \mathrm{mol}} 脳 17.13 \ \mathrm{mol/L} = -23419 \ \mathrm{kJ/L}\)
04

(d) Calculate the mass of CO2 produced per kJ of heat emitted

To find the mass of CO2 produced per kJ of heat emitted, we can use the stoichiometry of the balanced equation (2 moles of CO2 produced per mole of ethanol) and the molar mass of CO2. 1: Calculate moles of CO2 produced per kJ of heat emitted: Moles of CO2 = \(\frac{2 \ \mathrm{mol} \ CO_2}{1 \ \mathrm{mol} \ C_{2}H_{5}OH} 脳 -\frac{1 \ \mathrm{mol} \ C_{2}H_{5}OH}{1367.3 \ \mathrm{kJ}} = -\frac{2}{1367.3} \ \mathrm{mol/kJ}\) 2: Calculate the mass of CO2 produced per kJ of heat emitted using the molar mass of CO2 (C: 12.01g/mol, O: 16.00g/mol): Mass of CO2 per kJ = \(-\frac{2}{1367.3} \ \mathrm{mol/kJ} 脳 (12.01 + 2(16.00)) \ \mathrm{g/mol} = -0.0327 \ \mathrm{g/kJ}\)

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Most popular questions from this chapter

Imagine a book that is falling from a shelf. At a particular moment during its fall, the book has a kinetic energy of \(24 \mathrm{~J}\) and a potential energy with respect to the floor of \(47 \mathrm{~J}\). (a) How do the book's kinetic energy and its potential energy change as it continues to fall? (b) What was the initial potential energy of the book, and what is its total kinetic energy at the instant just before it strikes the floor? (c) If a heavier book fell from the same shelf, would it have the same kinetic energy when it strikes the floor? [Section 5.1]

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{CO}_{2}(s)\) or \(1 \mathrm{~mol} \mathrm{CO} 2(g)\) at the same temperature, (b) \(2 \mathrm{~mol}\) of hydrogen atoms or \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), (c) \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) and \(0.5 \mathrm{~mol} \mathrm{O}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(g)\) at \(25^{\circ} \mathrm{C}\), (d) \(1 \mathrm{~mol} \mathrm{~N}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{~N}_{2}(g)\) at \(300^{\circ} \mathrm{C} .\)

(a) Write an equation that expresses the first law of thermodynamics in terms of heat and work. (b) Under what conditions will the quantities \(q\) and \(w\) be negative numbers?

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) \(1.36\) \(\mathrm{mol}\) of \(\mathrm{O}_{2}\) and (b) \(10.4 \mathrm{~g}\) of \(\mathrm{KCl}\). (c) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\), is likely to be feasible under ordinary conditions? Explain your answer.

The specific heat of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\), is \(2.22 \mathrm{~J} / \mathrm{g}-\mathrm{K}\). (a) How many J of heat are needed to raise the temperature of \(80.0 \mathrm{~g}\) of octane from \(10.0\) to \(25.0^{\circ} \mathrm{C}\) ? (b) Which will require more heat, increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) by a certain amount or increasing the temperature of \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\) by the same amount?
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