Question

Use bond energies from Table 10.3 in the textbook to estimate the enthalpy change (ΔH) for the following reaction. C2H2(g)+H2(g)→C2H4(g) ΔH=?

Answer 1

Answer:
`=176.6kJmol^(-1)`

Step-by-step explanation:Bond energy of H-H is 436.4 kJ/mole

Bond energy of C-H is 414 kJ/mol

Bond energy of C=C is 620 kJ/mol

Bond energy of C≡C is 835 kJ/mol

`\Delta H= {\text {sum of bond energies of reactants}}-  {\text {sum of bond energies of products}}`

`\Delta H`= {1B.E(C≡C)+2B.E(C-H) +1B.E(H-H)} - {1B.E(C=C)+4B.E(C-H)}

`\Delta H= {1B.E(835kJmole^(-1))+2B.E(414kJmole^(-1)) +1B.E(436.4kJmole^(-1))} -  {1B.E(620kJmole^(-1))+4B.E(414kjmole^(-1))}`

`=176.6kJmol^(-1)`