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Calculate the enthalpy of hydrogenation of ethylene, given that the enthalpy of combustion of ethylene, hydrogen and ethane are -1410.0,-286.2 and -1560.6 kJmol^(-1) respectively at 298 K |
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Answer» Solution :We aregiven (i)`C_(2)H_(4)(G) +3O_(2)(g)rarr 2CO_(2)(g) +2H_(2)O(l), DeltaH= - 1410kJ mol^(-1)` (II) `H_(2)(g) +(1)/(2) O_(2)(g) rarr H_(2)O(l), DeltaH = - 286.2 kJ mol^(-1)` (iii)`C_(2)H_(6)(g) + 3(1)/(2) O (g) rarr 2CO_(2)(g) + 3H_(2)O(l), Delta H = - 1560.6 kJ mol^(_1)` We aim at `:``C_(2)H_(4)+ H_(2)(g) rarr C_(2)H_(6)(g), DeltaH = ?` Equation (i) `+` Equation (ii) - Equation (iii) gives `C_(2)H_(4)(g) +H_(2)(g) rarr C_92)H_(6)(g), Delta H = - 1410.0 + ( - 286.2) - ( 1560.6)` `= - 135.6 kJ mol^(-1)` |
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