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Define enthalpy of fomration. Calculate Enthalpy of formation of C_(2)H_(6(g)) given the enthalpy of combustion of C_(2)H_(4(g))C_(2)H_(6(g)) and H_(2(g)) as -140 kg/mol , -1550kJ/mol and -286 kJ/mol respectively. |
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Answer» SOLUTION :`C_(2)H_(4(g))+CO_(2)(g))to2CO_(3(g))+2H_(2)O_((g))""Delta_(C)H^(@)=140` kJ/mol …………1 `H_(2(g))+1/2O_(2(g))toH_(2)O_((g))""Delta_(c)H^(@)=-286` kJ/mol………….2 `C_(2)H_(6(g))+7/2O_(2(g))to2CO_(2(g))+3H_(2)O_((g))""Delta_(c)H^(@)=-1550` kJ/mol.......3 REquired equation `C_(2)H_(4)(g)+H_(2)(g)toC_(2)H_(6)(g)""Delta_(f)H^(@)=?` Reverse equation (3) `2CO_(2)(g)+3H_(2)O(g)toC_(2)H_(6)(g)+7/2O_(2)(g)""Delta_(c)H^(@)=-1550` kJ/mol..........4 And equation 1,2and 4 `H_(2)(g)+1/2cancel(O_(2))(g)toH_(2)O(g)` `C_(2)H_(4)+3cancel(O_(2))(g)to2CO_(2)(g)+2cancel(H_(2)O)(g)` `2cancel(CO_(2))(g)+3H_(2)O(g)toC_(2)H_(6)(g)+7/2cancel(O_(2))(g)` `C_(2)H_(4)(g)+H_(2)(g)toC_(2)H_(6)(g)` `Delta_(f)H^(@)=15550` kJ/mol `-140` kJ/mol -286 kJ/mol `Delta_(f)H^(@)=1124` kJ/mol |
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