InterviewSolution
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InterviewSolution
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Calculate the ethalpy of combustion of ethylene (g) to form `CO_(2)` (gas) and `H_(2)O` (gas) at 298 K and 1 atmospheric pressure. The enthalpies of formation of `CO_(2),H_(2)O` and `C_(2)H_(4)` are `-393.7, -241.8+52.3` kJ per mole respectively. |
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Answer» Given : `{:((i),C(s)+O_(2)(g)rarr CO_(2)(g),,Delta H=-393.7 kJ 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)):}` `"Objective reaction : " C_(2)H_(4)(g)+3O_(2)(g)rarr 2CO_(2)(g)+2H_(2)O(g)," "Delta H = ?` `2xx` Equation `(i)+2xx` Equation (ii) - Equation (iii) gives `{:(2C(s)+2O_(2)(g)" "rarr " "2CO_(2)(g)),(+O_(2)(g)+2H_(2)(g)" "rarr " "2H_(2)O(g)),(ul(-2C(s)-2H_(2)(g)" "rarr " "-C_(2)H_(4)(g))),(" "3O_(2)(g)" "rarr " "2CO_(2)(g)+2H_(2)O(g)-C_(2)H_(4)(g)),(or C_(2)H_(4)(g)+3O_(2)(g)rarr 2CO_(2)(g)+2H_(2)O(g)):}` Alternative Method : We aim t : `C_(2)H_()(g)+3O_(2)(g)rarr 2CO_(2)(g)+2H_(2)O(g)` We are given : `Delta H_(f(CO_(2))=-393.7 kJ mol^(-1), Delta H_(f(H_(1)O))=-21.8 kJ mol^(-1), Delta H_(f(C_(2)H_(4))=+52.3 kJ mol^(-1)` `Delta H_("Reaction")` = (Sum of `Delta H_(f)^(@)` values of Products) - (Sum of `Delta H_(f)^(@)` values of Reactants) `=[2xxDelta H_(f(CO_(2)))^(@)+2xxDelta H_(f(H_(2)O))^(@)]-[Delta H_(f(C_(2)H_(4)))^(@)+3xxDelta H_(f(CO_(2)))^(@)]` `=[2xx(-393.7)+2xx(-241.8)]-[(-52.3)+0]` `(because Delta H_(f)^(@) "for elementary substance" = 0)` `=[-787.4-483.6]-52.3=-1323.3 kJ mol^(-1)` |
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