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Calculate the enthalpy of the following reaction: `H_(2)C =CH_(2)(g) +H_(2)(g) rarr CH_(3)-CH_(3)(g)` The bond enegries of `C-H, C-C,C=C`, and `H-H` are `99,83,147`,and `104kcal` respectively. |
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Answer» The reaction is: `underset(H)underset(|)overset(H)overset(|)(C)=underset(H)underset(|)overset(H)overset(|)C(g)+H-H(g)rarrH-underset(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(H)overset(|)(C)-H(g), DeltaH^(Theta) =` ? `DeltaH^(Theta) =` Sum of bond energies of reactants -Sum of the bond energies of products `[DeltaH_(C=C) +4 xx DeltaH_(C-H) + DeltaH^(H-H)] -[DeltaH_(C-C) +6 xx DeltaH_(C-H)]` `(147 +4 xx 99 +104)-(83+6 xx 99) =- 30 kcal` |
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