Calculate the enthalpy of combustion of ethylene .
Given : `C_(2) H_(6) (g) + (7)/(2) O_(2) (g) to 2 CO_(2) + 3 H_(2)O (l) , Delta H = = 1562 kJ* mol^(-1)`
`H_(2) (g) + (1)/(2) O_(2) (g) to H_(2) O (l) , Delta H = -286 kJ * mol^(-1)`
`C_(2) H_(4) (g) + H_(2) (g) to C_(2) H_(6) (g) , Delta H = -32 kJ * mol^(-1)`

Explanation

`C_(2)H_(6) (g) + (7)/(2) O_(2) (g) to 2 CO_(2) (g) + 3 H_(2) O(l) , Delta H = 1562 kJ * mol^(-1) … (i)`
`H_(2) (g) + (1)/(2) O_(2) (g) to H_(2)O(l) , Delta H = -286 kJ* mol^(-1) … (ii)`
`C_(2)H_(4) (g) + H_(2) (g) + (1)/(2) O_(2) (g) , Delta H = -32kJ * mol^(-1) ... (iii)`
Reversing the equation (ii) , we obtain , `H_(2)O (l) to H_(2) (g) + (1)/(2) O_(2) (g) , Delta H = + 286 kJ * mol^(-1) ... (iv) `
Adding equation (i) , (ii) , (iii) and (iv) , we obtain ,
`C_(2) H_(4) (g) + 3O_(2) (g) to 2 CO_(2) (g) + 2H_(2)O (l) , Delta H = -1562 - 32 + 286 = - 1300 k J * mol^(-1)`
`therefore` The enthalpy of combustion of ethylene = `1816k J * mol^(-1)`