Textbook Question
From the following data for three prospective fuels, calculate which could provide the most energy per unit mass and per unit volume:
Ch.5 - Thermochemistry
Chapter 5, Problem 114
Three hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: Hydrocarbon Formula ΔHfº (kJ/mol) Butane C4H10(g) -125 1-Butene C4H8(g) -1 1-Butyne C4H6(g) 165. (a) For Butane, calculate the molar enthalpy of combustion to CO2(g) and H2O(l). (kJ/mol) (b) For 1-Butene, calculate the molar enthalpy of combustion to CO2(g) and H2O(l). (kJ/mol) (c) For 1-Butyne, calculate the molar enthalpy of combustion to CO2(g) and H2O(l).
Verified step by step guidance1
Step 1: Write the balanced chemical equation for the combustion of each hydrocarbon. For butane (C4H10), the combustion reaction is: C4H10(g) + 6.5O2(g) -> 4CO2(g) + 5H2O(l). For 1-butene (C4H8), the reaction is: C4H8(g) + 6O2(g) -> 4CO2(g) + 4H2O(l). For 1-butyne (C4H6), the reaction is: C4H6(g) + 5.5O2(g) -> 4CO2(g) + 3H2O(l).
Step 2: Use the standard enthalpies of formation (ΔHfº) to calculate the enthalpy change for the combustion reaction. The formula is: ΔHº_combustion = ΣΔHfº(products) - ΣΔHfº(reactants).
Step 3: Calculate the enthalpy of formation for the products. For CO2(g), ΔHfº = -393.5 kJ/mol, and for H2O(l), ΔHfº = -285.8 kJ/mol. Multiply these values by the number of moles of each product in the balanced equation.
Step 4: Calculate the enthalpy of formation for the reactants. Use the given ΔHfº values for each hydrocarbon and remember that the ΔHfº for O2(g) is zero because it is in its standard state.
Step 5: Substitute the calculated values into the formula from Step 2 to find the molar enthalpy of combustion for each hydrocarbon.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Enthalpy of Formation (ΔHfº)
The enthalpy of formation (ΔHfº) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It is a crucial value in thermodynamics, as it provides a reference point for calculating the energy changes in chemical reactions, particularly combustion. Negative values indicate that the formation of the compound is exothermic, while positive values suggest endothermic processes.
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Enthalpy of Formation
Combustion Reaction
A combustion reaction is a chemical process in which a hydrocarbon reacts with oxygen to produce carbon dioxide and water, releasing energy in the form of heat. The general equation for the combustion of a hydrocarbon can be represented as CxHy + O2 → CO2 + H2O. The enthalpy change associated with this reaction, known as the enthalpy of combustion, is essential for understanding the energy yield of fuels.
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Hess's Law
Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the number of steps taken to complete the reaction. This principle allows for the calculation of enthalpy changes for complex reactions by using known enthalpy values of formation and combustion. It is particularly useful in determining the enthalpy of combustion for hydrocarbons by combining the enthalpy of formation values of the reactants and products.
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Related Practice
Textbook Question
When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, MgO. The other is the product of the reaction of Mg with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (e) The standard enthalpy of formation of solid magnesium nitride is -461.08 kJ>mol. Calculate the standard enthalpy change for the reaction between magnesium metal and ammonia gas.
Textbook Question
A 201-lb man decides to add to his exercise routine by walking up three flights of stairs (45 ft) 20 times per day. Hefigures that theworkrequired to increasehis potential energy in this way will permit him to eat an extra order of French fries, at 245 Cal, without adding to his weight. Is he correct in this assumption?
Textbook Question
Sucrose (C12H22O11) is produced by plants as follows: 12 CO2(g) + 11 H2O(l) → C12H22O11 + 12 O2(g) H = 5645 kJ About 4.8 g of sucrose is produced per day per square meter of the earth's surface. The energy for this endothermic reaction is supplied by the sunlight. About 0.1 % of the sunlight that reaches the earth is used to produce sucrose. Calculate the total energy the sun supplies for each square meter of surface area. Give your answer in kilowatts per square meter 1kW/m2 where 1W = 1 J/s2.