Calcutate enthalpy for formation of ethylene form the following data
(I)` C_((graphtie))+O _(2)(g)toCO_(2)(g), """""Delta=-393.5kJ`
(II)`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l), """" DeltaH=-286.2kJ`(III)`C_(2)H_(4)(g)+3O_(2)(g)to2CO_(2)(g)+2H_(2)O(l)"""""DeltaH=-1410.8kJ`

Options

A. 54.1kJ
B. 44.8kJ
C. 51.4kL
D. 48.4kJ

Answer

C

Explanation

(i) `C_(graphite)+O_(2)(g)toCO_(2)(g),""""Delta=-393.5kJ`
(ii) `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),"""'DeltaH=-286.2kJ`
(iii) `C_(2)H_(4)(g)+3O_(2)(g)to2CO_(2)(g)+2h_(2)O(l)""""DeltaH=-1410.5kJ`
By multiplying the equations (i)and (ii) by 2, and then adding them `2C+2O_(2)to2CO_(2),""""DeltaH=-787.0kJ`
underline `(2H_(2)+O_(2)to2H_(2)O,""""DeltaH=-572.4kJ)`(iv) `2C+2H_(2)+3O_(2)to2CO_(2)+2H_(2)O,DeltaH=-1359.4kJ`
Substracting equation (iii) from (iv)
`2C+2H_(2)+3O_(2)to2CO_(2)+2H_(2)O,DeltaH=-1359.4kJ`

underline`(C_(2)+2H_(4)(g)+3O_(2)(g)to2CO_(2)(g)+2H_(2)O(l),DeltaH=-1410.8kJ`)`2C+2H_(2)"""""toC_(2)H_(4),"""""DeltaH=51.4kJ`