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Calculate the enthalypy of combustion of ethylene (gas) to form CO_(2) (gas ) and H_(2) O (gas) at 298 K and 1 atmospheric pressure. The enthalpies of formation ofCO_(2), H_(2)O and C_(2)H_(4) are - 393.7, -241.8 ,+52.3kJ per mole respectively. |
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Answer» Solution :We are given `:` (i) `C(s) +O_(2)(g) rarr CO_(2)(g), DELTAH^(@) = - 393.5kJ mol^(-1)` (ii)`H_(2)(g) + (1)/(2) O_(2)(g)rarr H_(2)O(g), Delta H^(@) = - 241.8 kJ mol^(-1)` (iii)`2C(s) + 2H_(2)(g) rarr C_(2)H_(4)(g), Delta H^(@)= + 52.3 kJ mol^(-1)` We aim at `:``C_(2)H_(4)(g) + 3O_(2)(g)rarr 2CO_(2)(g) + 2H_(2)O(g) , Delta _(c ) H^(@) = ?` ` 2 xx` Eqn.`(i) + 2 xx ` Eqn. (ii) - Eqn. (iii) gives `ul{(" "2C(s) ,+2O_(2)(g)rarr2CO_(2)(g)+O_(2)(g),+2H_(2)(g)+2H_(2)O(g)),(-2C(s),,-2H_(2)(g)-C_(2)H_(4)(g)):}` `3O_(2)(g)rarr 2CO_(2) +2H_(2)(g)-C_(2)H_(4)(g)` or `C_(2)H_(4)(g)+3O_(2)(g) rarr 2CO_(2)(g) + 2H_(2)O(g)` `Delta_(c) H^(@)= 2 ( -393.5) + (-241.8) - (52.3) = -1322.9 kJ mol^(-1)` Alternative Method ` :` We aim at `:``C_(2)H_(4)(g)+3O_(2)(g) rarr2CO_(2)(g)+2H_(2)O(g)` We are given `:``Delta_(f) H_((CO_(2)))^(@) = - 393.5 kJ mol^(-1)` `Delta_(f) H_((H_(2)O))^(@)= - 241.8 kJ mol^(-1)` `Delta _(f) H_((C_(2)H_(4)))^(@)= + 52.3 kJ mol^(_1)` `Delta_(R)H^(@) = ("Sum of " Delta_(f)H^(@)" valuesof Products")-("Sum of" Delta_(f)H^(@)" values of Reactants")` `=[ 2 xx Delta_(f) H^(@) (CO_(2))+2xxDelta_(f)H^(@)(H_(2)O)]-[Delta_(f)H^(@)(C_(2)H_(4))+3xxDelta_(f)H^(@)(O_(2))]` `=[ 2 xx ( -393.5) + 2 xx ( -241.8) ] - [ ( 52.3) + 0]( :' Delta_(f) H^(@)` for elementary subtance `=0 )` `=[ -787 .0 -483.6 ] - 52.3 = - 1322.9 kJ mol^(-1)` |
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