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For 7.81,(3C2H2 --> C6H6) I subtracted 3 C double bonds + 3 C single bonds + 6 C-H bonds from 3 double C bonds + 6 C-H bonds. However, in the solutions manual, the C-H bonds do not come into play when calculating the bond enthalpies, only the C double bonds do. I don't understand why that is? Thanks!
6 mol of C-H bonds are already present on the reactants' side, and we want to form 6 mol of C-H bonds in the end. The reason they're left out is because if you included them, they would end up canceling out since the positive value is used for the reactants, where bonds are broken, and the negative bond enthalpy value is used for the products, where bonds are formed. You would essentially be breaking the same 6 mol C-H bonds to form them again, so that's why you don't include bonds that are already present on both sides of the equation.
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