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Ch.9 - Thermochemistry: Chemical Energy
Chapter 9, Problem 105

Calculate ∆H°f in kJ/mol for benzene, C6H6, from the following data: 2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(l) ∆H°=-6534 kJ ∆H°f (CO2) = -393.5 kJ/mol ∆H°f(H2O) = -285.8 kJ/mol

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Write the balanced chemical equation for the formation of benzene, C6H6, from its elements in their standard states: C(s, graphite) + H2(g) → C6H6(l).
Use Hess's Law to relate the given reaction to the formation reactions of CO2, H2O, and C6H6. Hess's Law states that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in.
Calculate the total enthalpy change for the formation of the products in the given reaction using their formation enthalpies: 12 moles of CO2 and 6 moles of H2O. Multiply the given ∆H°f values by the respective moles and sum them.
Set up the equation for the enthalpy change of the reaction using the enthalpy of formation of the products and the given overall reaction enthalpy. Include the unknown ∆H°f for benzene in this equation.
Solve for ∆H°f of benzene, C6H6, by rearranging the equation to isolate ∆H°f (C6H6) on one side.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Enthalpy of Formation (∆H°f)

The enthalpy of formation (∆H°f) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It is a crucial concept in thermodynamics, as it allows for the calculation of the energy changes associated with chemical reactions. Standard enthalpy values are typically given in kJ/mol and are used to determine the overall energy change in a reaction.
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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 achieve the reaction. This principle allows chemists to calculate the enthalpy change of a reaction by using known enthalpy values of other reactions. It is particularly useful when direct measurement of a reaction's enthalpy change is difficult or impossible.
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Stoichiometry in Chemical Reactions

Stoichiometry involves the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It provides the mole ratios needed to relate the quantities of substances involved in the reaction. Understanding stoichiometry is essential for accurately applying thermodynamic data, such as enthalpy changes, to calculate the energy changes in a given reaction.
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Related Practice
Textbook Question

Write balanced equations for the formation of the following compounds from their elements. (b) Sodium sulfate

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Textbook Question

Write balanced equations for the formation of the following compounds from their elements. (c) Dichloromethane (a liquid, CH2Cl2)

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Textbook Question
Sulfuric acid (H2SO4), the most widely produced chemical in the world, is amde yb a two-step oxidaiton of sulfur to sulfur trioxide, SO3, followed by reaciton with water. Calculate ΔH°f for SO3 in kJ/mol, given the following data: S(s) + O2(g) → SO2(g) ΔH° = -296.8 kJ SO2(g) + 1/2 O2(g) → SO3(g) ΔH° = -98.9 kJ
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Textbook Question
The standard enthalpy change for the reaciton of SO3(g) with H2O(l) to yield H2SO4(aq) is ΔH° = -227.8 kJ. Use the information in Problem 9.104 to calculate ΔH°f for H2SO4(aq) in kJ/mol. [For H2O(l), ΔH°f = -285.88 kJ/mol.]
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Textbook Question
Acetic acid (CH3CO2H), whose aqueous solutions are known as vinegar, is prepared by reaction of ethyl alcohol (CH3CH2OH) with oxygen: CH3CH2OH(l) + O2(g) → CH3CO2H(l) + H2O(l) Use the following data to calculate ∆H° in kilojoules for the reaction: ∆H°f [CH3CH2OH(l)] = -277.7 kJ/mol ∆H°f [CH3CO2H(l)] = -484.5 kJ/mol ∆H°f [H2O(l)] = -285.8 kJ/mol
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Textbook Question
Styrene (C8H8), the precursor of polystyrene polymers, has a standard heat of combustion of -4395 kJ/mol. Write a balanced equation for the combustion reaction, and calculate ΔH°f for styrene in kJ/mol. ΔH°f [CO2(g)] = -393.5 kJ/mol; ΔH°f [H2O(l)] = -285.8 kJ/mol
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