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Using the bond energies in Table 7.2, determine the approximate enthalpy change for each of the following reactions:(a) $\mathrm{Cl}_{2}(g)+3 \mathrm{F}_{2}(g) \rightarrow 2 \mathrm{CIF}_{3}(g)$(b) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(g)+\mathrm{H}_{2}(g) \rightarrow \mathrm{H}_{3} \mathrm{CCH}_{3}(g)$(c) $2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$
a) $-807 \mathrm{kJ}$b) $-128 \mathrm{kJ}$c) $-2340 \mathrm{kJ}$
01:44
Aadit S.
Chemistry 101
Chapter 7
Chemical Bonding and Molecular Geometry
Chemical Bonding
Molecular Geometry
University of Central Florida
Rice University
Drexel University
University of Kentucky
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toe estimate the change in and thall P using bond energies, we need Thio. Sum up all the bonds of the reactant that need to be broken and then subtract off all of the bonds of the products that will be formed. We're subtracting off the bonds being formed because their energy will be released. An energy released has a negative sign. So for the first one, we have one CL bond and three FF bonds. CL Bond has a bond energy of 2 43 killer jewels. F F is 1 61 killer jewels and then subtract off all the bonds being formed. Because this is C L F. Three, there are actually a total of three bonds per molecule. And with two molecules, there's six C l F bonds with a bond energy of 2 55 killer jewels. This then gives us negative eight oh four killer jewels as an estimate for the anti LP of this reaction. The next one, I'm going to draw all the bonds so that it is obvious, except in the case of hydrogen, which is just in HH bond. We have four C H bonds at 4 15. Kill the jewels. One C double bond. See bond at 6 11 killer jewels and one H single bond. H bond at 4 36. Killer jewels. Then we have six C H bonds at 4 15 killer jewels and one see single bond. See bond at 3 45 which estimates 1 28 killer jewels. Negative. 1 28 Killer jewels for the Antheil. P Then for the next one, we have two of these molecules reacting with seven oxygen molecules. We draw the Louis structure for oxygen. We do see that it is a double bond producing four carbon dioxide that has each carbon dioxide to see double bond. Oh, bonds. And then we have six h two OS each model each h 20 having to h o bonds. Single bonded. So we have to of what we had appear. But now it is a reactant. So we will sum them up and keep them positive, too. Of the total energies to break all the bonds of the C two h six plus seven times the bond energy for O double Bondo minus. We've got two per carbon dioxide times four. So that's eight. See double Bondo and then we've got six times two. 12 h single Bondo and this estimates then tell P of negative 2340 killer jewels.
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