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Calculate the bond energy of C-H bond,given that thee heat of formation ofCH_(4),heat of sublimation of carbon and heat of dissociation of H_(2) are-74.8, + 719.6 and 435.4 kJ mol^(-1) respectively. |
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Answer» SOLUTION :Here, we are GIVEN `C(s) +2H_(2)(g)rarr CH_(4) (g), Delta_(f)H^(@)= - 74.8 KJ`....(i) `C(s)rarr C(g), Delta_(r) H^(@) = +719.6kJ `....(ii) `H_(2)(g) rarr 2H(g), Delta_(r) H^(@) = + 435 .4 kJ`...(iii) We aim at `:` Eqn. (ii) `+ 2 xx`Eqn. (iii) - Eqn (i) gives `C(s) +2H_(2)(g) rarr C(g) + 4H(g)` `- C(s) -2H_(2)(g)rarr - CH_(4) (g) ` `0 =C(g)+4H(g)-CH_(4)(g), Delta_(r) H^(@)= 719.6+2 (435.4)- (- 74.8)` or ` CH_(4)(g) rarr C(g) + 4H(g), DELTA H = + 1665.2 kJ` This gives the enthalpy of dissociation of four moles of C-H bonds ( called enthalpy of atomization) . Hence, bond energy for C-H bond (average value ) i.e., `Delta_(C-H) H^(@) = (1665.2)/(4)=416.3 kJ mol^(_1)` |
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