/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is 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 ky of heat emitted.

Short Answer

Expert verified
The balanced equation for the combustion of ethanol is \(C_2H_5OH (l) + 3O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (g)\). The standard enthalpy change for the reaction is -1367 kJ/mol. The heat produced per liter of ethanol by combustion under constant pressure is -23389 kJ. The mass of COâ‚‚ produced per kJ of heat emitted is 0.064 g/kJ.

Step by step solution

01

(a) Balanced equation for combustion of ethanol

We can represent ethanol as \(C_2H_5OH\) and oxygen gas as \(O_2\). When ethanol combusts in the presence of oxygen, it produces carbon dioxide (\(CO_2\)) and water (\(H_2O\)). The balanced equation is: \(C_2H_5OH (l) + 3O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (g)\)
02

(b) Standard enthalpy change for the reaction

To calculate the standard enthalpy change, we'll use the following equation: \(\Delta H_{rxn}^\circ = \Sigma n_i\Delta H_{f,i}^\circ(products) - \Sigma n_j\Delta H_{f,j}^\circ(reactants)\) We'll consider the following standard enthalpy of formation (\(\Delta H_f^\circ\)) values: \(C_2H_5OH (l):\ -277.7\ \mathrm{kJ/mol}\) \(O_2 (g)\): 0 (since it is an element in its standard state) \(CO_2 (g):\ -393.5\ \mathrm{kJ/mol}\) \(H_2O (g):\ -241.8\ \mathrm{kJ/mol}\) Now, plug in the values: \(\Delta H_{rxn}^\circ\ = [2(-393.5) + 3(-241.8)] - [(-277.7) + 3(0)]\) \(\Delta H_{rxn}^\circ = -1367\ \mathrm{kJ/mol}\)
03

(c) Heat produced per liter of ethanol by combustion under constant pressure

We first need to calculate the mass of 1L of ethanol: Mass of ethanol = Volume × Density = 1L × (0.789 \(\mathrm{g/mL}\)) = 789 g Since 1 mol ethanol = 46.07 g (i.e. molecular weight of ethanol), moles of ethanol in 1L =84 (i.e. 789g ÷ 46.07g = 17.11 mol) Heat produced for 17.11 mol = -1367kJ/mol × 17.11 mol = -23389 kJ Thus, the heat produced per liter of ethanol by combustion under constant pressure is -23389 kJ.
04

(d) Mass of COâ‚‚ produced per kJ of heat emitted

1 mol of ethanol combustion produces 2 mol of CO₂. Then, 17.11 mol of ethanol combustion produces 34.22 mol of CO₂. Now, we can calculate the mass of CO₂ produced: Mass of CO₂ = 34.22 mol × 44.01 g/mol (i.e. molecular weight of CO₂) = 1505.52 g Mass of CO₂ per kJ of heat emitted is: 1505.52 g ÷ 23389 kJ = 0.064 g/kJ

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change in Combustion
Understanding the concept of enthalpy change is crucial when dealing with reactions like the combustion of ethanol. Enthalpy change, represented as \(\Delta H\text{rxn}\), is the amount of heat released or absorbed during a chemical reaction at constant pressure.
To comprehend the enthalpy change for the combustion of ethanol, you consider both the reactants and products. According to the law of conservation of energy, the energy required to break the chemical bonds in the reactants is different from the energy released when new bonds form in the products.
The standard enthalpy change for a reaction is calculated using standard enthalpies of formation \(\Delta H_f^\circ\), which represent the heat change when one mole of a compound is formed from its elements in their standard states.

Calculating Standard Enthalpy Change

The standard enthalpy change for the combustion of ethanol is found by subtracting the enthalpy of the reactants from that of the products. The formula is: \[\Delta H_{\text{rxn}}^\circ = \Sigma n_i\Delta H_{f,i}^\circ(\text{products}) - \Sigma n_j\Delta H_{f,j}^\circ(\text{reactants})\]Here, \(n_i\) and \(n_j\) are the stoichiometric coefficients from the balanced chemical equation. For the combustion of ethanol, this process typically shows energy is released, indicating that the reaction is exothermic.
Stoichiometry of Ethanol Combustion
Stoichiometry is the mathematical relationship between the amounts of reactants and products in a chemical reaction. It is a key concept in ensuring chemical equations are balanced with respect to mass and charge.
Stoichiometry involves calculations based on the molar ratios depicted in a balanced chemical equation. For the combustion of ethanol, we use stoichiometry to reveal that one mole of ethanol reacts with three moles of oxygen to produce two moles of carbon dioxide and three moles of water.

Applying Stoichiometry to Calculate Heat Production

The problem at hand requires calculating the heat produced per liter of ethanol. Using its density, you first determine the mass and then convert this into moles. And since the enthalpy change of the reaction is known, stoichiometry allows you to calculate the heat produced from a known quantity of ethanol.
Issues like density conversions, molar mass calculations, and understanding the combustion process become less daunting when you get a solid grip on stoichiometry. This pillar of chemistry doesn't just apply to ethanol combustion; it's the backbone for all quantitative analysis in chemical reactions.
Writing Chemical Equations for Combustion
Chemical equations are symbolic representations of chemical reactions, showcasing the reactants and products along with their respective quantities. Getting familiar with writing chemical equations involves understanding the reactants involved, predicting the products, and balancing the equation.
For the combustion of ethanol, the equation begins with ethanol \(C_2H_5OH\) and oxygen gas \(O_2\). The products of this combustion are carbon dioxide \(CO_2\) and water \(H_2O\), which for our purposes is considered in its gaseous state.

Balancing the Combustion Equation

The chemical equation for the combustion of ethanol is balanced by adjusting the coefficients to ensure an equal number of atoms of each element on both sides of the equation:\[C_2H_5OH (l) + 3O_2 (g) \rightarrow 2CO_2 (g) + 3H_2O (g)\]Balancing chemical equations is a foundational skill in chemistry that confirms the law of conservation of mass. It ensures that the quantity of each element does not change in the reaction. Learning to write and balance equations like this one for ethanol's combustion aids in understanding the reaction process and is essential for thorough chemical analysis.

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

(a) Under what condition will the enthalpy change of a process equal the amount of heat transferred into or out of the system? (b) During a constant- pressure process, the system releases heat to the surroundings. Does the enthalpy of the system increase or decrease during the process? (c) In a constant-pressure process, \(\Delta H=0 .\) What can you conclude about \(\Delta E, q,\) and \(w ?\)

A 2.200 -g sample of quinone \(\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)\) is burned in a bomb calorimeter whose total heat capacity is 7.854 \(\mathrm{kJ} / \mathrm{c}\) . The temperature of the calorimeter increases from 23.44 to \(30.57^{\circ} \mathrm{C}\) . What is the heat of combustion per gram of quinone? Per mole of quinone?

Consider the decomposition of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) to gaseous acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g) :\) $$\mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \quad \Delta H=+630 \mathrm{kJ}$$ (a) What is the enthalpy change for the reverse reaction? (b) What is \(\Delta H\) for the formation of 1 mol of acetylene? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) were consumed instead of \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

Consider the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g) .(\mathbf{a})\) Use the bond enthalpies in Table 5.4 to estimate \(\Delta H\) for this reaction, ignoring the fact that iodine is in the solid state. (b) Without doing a calculation, predict whether your estimate in part (a) is more negative or less negative than the true reaction enthalpy. (c) Use the enthalpies of formation in Appendix \(C\) to determine the true reaction enthalpy.

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 \(\mathrm{H}_{2},(\mathbf{c}) 1 \mathrm{mol} \mathrm{H}_{2}(g)\) 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 \(25^{\circ} \mathrm{C},(\mathbf{d}) 1 \mathrm{mol} \mathrm{N}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) or 1 \(\mathrm{mol} \mathrm{N}_{2}(g)\) at \(300^{\circ} \mathrm{C}\) .

See all solutions

Recommended explanations on Chemistry Textbooks

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.