Infatuation Rules
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Bond Strength: Covalent Bonds The stronger a bond, the greater the energy required to break it. The average C–H bond energy, DC – H, is 1660/4 = 415 kJ/mol because there are four moles of C–H bonds broken per mole of the reaction.
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Read More »Average Bond Lengths and Bond Energies for Some Common Bonds Bond Bond Length (Å) Bond Energy (kJ/mol) C–C 1.54 345 C = C C = C 1.34 611 C ≡ C C ≡ C 1.20 837 C–N 1.43 290 C = N C = N 1.38 615 C ≡ N C ≡ N 1.16 891 C–O 1.43 350 C = O C = O 1.23 741 C ≡ O C ≡ O 1.13 1080 Table 7.3 We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic. An exothermic reaction (ΔH negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants. An endothermic reaction (ΔH positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants. The enthalpy change, ΔH, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy “in”, positive sign) plus the energy released when all bonds are formed in the products (energy “out,” negative sign). This can be expressed mathematically in the following way: Δ H = ƩD bonds broken − ƩD bonds formed Δ H = ƩD bonds broken − ƩD bonds formed In this expression, the symbol Ʃ means “the sum of” and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table (like Table 7.3) and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction.
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Read More »H 2 ( g ) + Cl 2 ( g ) ⟶ 2 HCl ( g ) H 2 ( g ) + Cl 2 ( g ) ⟶ 2 HCl ( g )
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