The heat of combustion of C_(2)H_(6) is -368.4 kcal. Calculate the heat of combustion of C_(2)H_(2), when the heat of combustion of H_(2) is 68.32 kcal mol^(-1). C_(2)H_(4)(g) + H_(2)(g) rarr C_(2)H_(6)(g) , Delta H = 37.1 kcal.

The heat of combustion of C_(2)H_(6) is -368.4 kcal. Calculate the hea

1.	The heat of combustion of C_(2)H_(6) is -368.4 kcal. Calculate the heat of combustion of C_(2)H_(2), when the heat of combustion of H_(2) is 68.32 kcal mol^(-1). C_(2)H_(4)(g) + H_(2)(g) rarr C_(2)H_(6)(g) , Delta H = 37.1 kcal.
Answer» Solution :`C_(2)H_(4)(G) + H_(2)(g) RARR C_(2)H_(6)(g) , Delta H = -37.1 kcl` `Delta H_(c )C_(2)H_(6) = -368.4 kcal, Delta H_(c )C_(2)H_(4) = ? , Delta H_(c )H_(2(g)) = -68.32 kcal` `Delta H = SigmaDelta H_("c(reactants)") = -Sigma Delta H_("c(products)")` `Delta H = Delta H_(c )C_(2)H_(4) + Delta H_(c )H_(2(g)) - Delta H_(c )C_(2)H_(6(c ))` `-37.1 kcal = Delta H_(c )C_(2)H_(4) - 68.32 - (-368.4)` `Delta H_(c )C_(2)H_(4) = -337.18 K cal`.