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: $=176.6kJmol^{-1}$

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)}

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

$=176.6kJmol^{-1}$