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Chemical thermodynamics

Bond energy

Exercise 1

     

Use the →   Table of bond energies to recalculate the enthalpy of formation of $HCl(g)$.

$H-H(g)$ $+$ $Cl-Cl(g)$ $\longrightarrow$ $2 H-Cl (g)$ $\Delta H$ $= $ $E_{bond}(H-H)$ $ +$ $ E_{bond}(Cl-Cl)$ $-$ $ 2E_{bond}(H-Cl)$ $=$ $ -183\;kJ$ Enthalpy of formation (one mole !) of $HCl(g$): $\Delta H_f(HCl(g))$ $=$ $\frac{-183}{2}=-91.5kJ$ , an approximate value! (see the →   Table of formation enthalpies!)