If enthalpies of formation of `C_(2)H_(4)(g)1`,CO_(2) and`H_(2)O(l)` at`25(@)C` and`1`atm pressure be `52,-394`and`-286kJmol^(-1)`respectively ,the enthalpy of combustion of `C_(2)H_(4)(g)`will be

Options

A. `-141.2 KJ//mol`
B. `-1412` KJ/mol
C. `+14.2` KJ/mol
D. `+1412` KJ/mol

Answer

B

Explanation

Combustion of hydrocarbon,
`C_(2)H_(4)(g) + 3O_(2)(g) to 2CO_(2)(g) + 2H_(2)O(1)`
`Delta_(I)H=sumDelta_(f)H_(P)-sumDelta_(f)H_(I)`
`={2xxDelta_(f)H(CO_(2)) + 2 xx Delta_(f) H(O_(2))}-{Delta_(f)H(C_(2)H_(4))+ 3Delta_(f)H(O_(2))}`
`=(2xx-394 + 2 xx -286 )-(52 + 3xx0)`
`=-788 - 572 - 52 =-1412 KJ "mol"^(-1)`