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Chapter 8: Problem 75

Use bond energies to estimate \(\Delta H\) for the combustion of one mole of acetylene: $$\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$

Short Answer

Expert verified
To estimate the change in enthalpy (∆H) for the combustion of one mole of acetylene, we first identify the types and quantities of bonds in the reactants and products: 1 C≡C triple bond, 2 C-H single bonds, and 5 O=O double bonds in reactants; 4 C=O double bonds and 2 O-H single bonds in the products. Using the given bond energies, we calculate the energy used to break bonds in reactants (4151 kJ/mol) and the energy released from the formation of bonds in products (4124 kJ/mol). Subtracting these values, we find ∆H to be -27 kJ/mol.

Step by step solution

01

Identify the bonds in the reactants and products and their quantities

The balanced equation for the combustion reaction of acetylene is given below: \[\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\] In the reactants, we have: - \(1\) C≡C triple bond in acetylene molecule - \(2\) C-H single bonds in acetylene molecule - \(5\) O=O double bonds in oxygen molecule In the products, we have: - \(4\) C=O double bonds in carbon dioxide molecules - \(2\) O-H single bonds in water molecule
02

Obtain bond energy values

Use the provided bond energy values or look them up if needed. For this example, we will use the following bond energy values: C≡C: 835 kJ/mol C-H: 413 kJ/mol O=O: 498 kJ/mol C=O: 799 kJ/mol O-H: 464 kJ/mol Note: Bond energies may vary slightly between sources.
03

Calculate the energy required to break bonds in reactants

Multiply the bond energy value of each bond by the number of that bond type in the reactants and then sum up all the values: Energy used to break bonds = (1 × C≡C bond energy) + (2 × C-H bond energy) + (5 × O=O bond energy) = (1 × 835 kJ/mol) + (2 × 413 kJ/mol) + (5 × 498 kJ/mol) = 835 kJ/mol + 826 kJ/mol + 2490 kJ/mol = 4151 kJ/mol
04

Calculate the energy released from the formation of bonds in products

Multiply the bond energy value of each bond by the number of that bond type in the products and then sum up all the values: Energy released from the formation of bonds = (4 × C=O bond energy) + (2 × O-H bond energy) = (4 × 799 kJ/mol) + (2 × 464 kJ/mol) = 3196 kJ/mol + 928 kJ/mol = 4124 kJ/mol
05

Estimate the change in enthalpy (∆H)

Subtract the energy released from the formation of bonds in products from the energy used to break bonds in reactants: ∆H = Energy released - Energy used = 4124 kJ/mol - 4151 kJ/mol = -27 kJ/mol The estimated change in enthalpy for the combustion of one mole of acetylene is -27 kJ/mol.

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Most popular questions from this chapter

Consider the following reaction: $$A_{2}+B_{2} \longrightarrow 2 A B \quad \Delta H=-285 \mathrm{kJ}$$ The bond energy for \(\mathrm{A}_{2}\) is one-half the amount of the AB bond energy. The bond energy of \(\mathrm{B}_{2}=432 \mathrm{kJ} / \mathrm{mol}\) . What is the bond energy of \(\mathrm{A}_{2}\) ?

Predict the molecular structure (including bond angles) for each of the following. (See Exercises 115 and 116.) a. \(\mathrm{XeCl}_{2}\) b. \(\mathrm{ICl}_{3}\) c. \(\mathrm{TeF}_{4}\) d. \(\mathrm{PCl}_{5}\)

Two different compounds have the formula \(\mathrm{XeF}_{2} \mathrm{Cl}_{2}\) . Write Lewis structures for these two compounds, and describe how measurement of dipole moments might be used to distinguish between them.

An alternative definition of electronegativity is $$\text { Electronegativity } = \text { constant } (\mathrm{I.E.}-\mathrm{E.A.})$$ where I.E. is the ionization energy and E.A. is the electron affinity using the sign conventions of this book. Use data in Chapter 7 to calculate the \((\mathrm{I} . \mathrm{E} .-\mathrm{E} \cdot \mathrm{A} .)\) term for \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\) and \(\mathrm{I}\). Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are 1678, 1255, 1138, and 1007 kJ/mol, respectively. (Hint: Choose a constant so that the electronegativity of fluorine equals 4.0. Using this constant, calculate relative electronegativities for the other halogens and compare to values given in the text.)

The lattice energies of \(\mathrm{FeCl}_{3}, \mathrm{FeCl}_{2},\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) are (in no particular order) \(-2631,-5359,\) and \(-14,774 \mathrm{kJ} / \mathrm{mol}\) . Match the appropriate formula to each lattice energy. Explain.
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