/*! 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 90 Acetylene \(\left(\mathrm{C}_{2}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 90

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) are gaseous fuels with enthalpies of combustion of \(-49.9 \mathrm{~kJ} / \mathrm{g}\) and \(-49.5 \mathrm{~kJ} / \mathrm{g}\), respectively. Compare the energy available from the combustion of a given volume of acetylene to the combustion energy from the same volume of butane at the same temperature and pressure.

Short Answer

Expert verified
For a given volume of acetylene and butane at the same temperature and pressure, we can calculate the combustion energy by using the energy per mole and the ratio of moles in the given volume. The energy available from the combustion of acetylene for the given volume is -1299.16 kJ/mol 脳 (26.04 / 58.12), while for butane it is -2873.94 kJ/mol. Comparing these values, we can determine which gaseous fuel provides more energy per unit volume under the same conditions.

Step by step solution

01

Calculate energy released per mole of each gas

Using the molar mass of acetylene (C2H2) and butane (C4H10), determine the energy released per mole when each gas combusts. The molar mass of Acetylene (C2H2) is given by: Molar mass of (C) = 12.01 g/mol Molar mass of (H) = 1.01 g/mol Molar mass of Acetylene (C2H2) = 2 脳 12.01 + 2 脳 1.01 = 26.04 g/mol Energy released per mole of Acetylene = -49.9 kJ/g 脳 26.04 g/mol = -1299.16 kJ/mol The molar mass of Butane (C4H10) is given by: Molar mass of Butane (C4H10) = 4 脳 12.01 + 10 脳 1.01 = 58.12 g/mol Energy released per mole of Butane = -49.5 kJ/g 脳 58.12 g/mol = -2873.94 kJ/mol
02

Determine the number of moles in a given volume

Given the volume (V) of both gases, use the ideal gas law to find the number of moles (n) for each gas at the same temperature (T) and pressure (P). The ideal gas law is given by: PV = nRT Where R is the ideal gas constant (8.314 J/mol路K) and P, V, T, and n represent pressure, volume, temperature, and the number of moles, respectively. For both gases, P, V, and T are the same. Therefore, we can say: n_acetylene / n_butane = M_acetylene / M_butane Where M_acetylene and M_butane are the molar masses of acetylene and butane, respectively. Substitute the values for molar masses: n_acetylene / n_butane = 26.04 / 58.12
03

Compare combustion energy for the same volume of both gaseous fuels

Now, we calculate the energy available from the combustion of a given volume of acetylene and butane. Multiplying the energy per mole by the ratio of moles for a given volume, we get: Energy_acetylene = Energy per mole of Acetylene 脳 (n_acetylene / n_butane) Energy_acetylene = -1299.16 kJ/mol 脳 (26.04 / 58.12) Energy_butane = Energy per mole of Butane 脳 (n_butane / n_butane) Energy_butane = -2873.94 kJ/mol From the values of Energy_acetylene and Energy_butane, we can compare the energy available from the combustion of a given volume of acetylene to the combustion energy from the same volume of butane at the same temperature and pressure.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Using the following data, calculate the standard heat of formation of \(\operatorname{ICl}(g)\) in \(\mathrm{kJ} / \mathrm{mol}\) : $$ \begin{aligned} \mathrm{Cl}_{2}(g) & \longrightarrow 2 \mathrm{Cl}(g) & \Delta H^{\circ} &=242.3 \mathrm{~kJ} \\ \mathrm{I}_{2}(g) & \longrightarrow 2 \mathrm{I}(g) & \Delta H^{\circ} &=151.0 \mathrm{~kJ} \\ \mathrm{ICl}(g) & \longrightarrow \mathrm{I}(g)+\mathrm{Cl}(g) & \Delta H^{\circ} &=211.3 \mathrm{~kJ} \\ \mathrm{I}_{2}(s) & \Delta H^{\circ}=62.8 \mathrm{~kJ} \end{aligned} $$

One way to lose weight is to exercise! Walking briskly at 4.0 miles per hour for an hour consumes about \(400 \mathrm{kcal}\) of energy. How many hours would you have to walk at \(4.0\) miles per hour to lose one pound of body fat? One gram of body fat is equivalent to \(7.7 \mathrm{kcal}\) of energy. There are \(454 \mathrm{~g}\) in \(1 \mathrm{lb}\).

Calculate \(\Delta E\) for each of the following. a. \(q=-47 \mathrm{~kJ}, w=+88 \mathrm{~kJ}\) b. \(q=+82 \mathrm{~kJ}, w=-47 \mathrm{~kJ}\) c. \(q=+47 \mathrm{~kJ}, w=0\) d. In which of these cases do the surroundings do work on the system?

When \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid \(\left(\mathrm{BaSO}_{4}\right)\) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{~g} / \mathrm{mL}\), calculate the enthalpy change per mole of \(\mathrm{BaSO}_{4}\) formed.

A gas absorbs \(45 \mathrm{~kJ}\) of heat and does \(29 \mathrm{~kJ}\) of work. Calculate \(\Delta E\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.