From the following data
`CH_(3)OH(l)+(3)/(2)O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l)`
`Delta_(r) H^(@)=-726 kJ mol^(-1)`
`H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)O(l), Delta _(r) H^(@)=-286 kJ mol^(-1)`
`C("graphite") +O_(2)(g) rarr CO_(2)(g), Delta _(r) H^(@)=-393 kJ mol^(-1)`
The standard enthalpy of formation of `CH_(3)OH(l)" in "kJ mol^(-1)` is

Options

A. `-239`
B. 239
C. 547
D. `-905`

Answer

A

Explanation

`CH_(3)-OH +3//2O_(2) rarr CO_(2) +2H_(2)O`
`Delta H =-726 kJ//mol ….(i)`
`2xx (H_(2)+1//2O_(2)) rarr H_(2)O, Delta H =-286 kJ//mol)…(ii)`
`C+O_(2) rarr CO_(2), Delta H =-393 kJ//mol …(iii)`
From Eq. (ii) + (iii) - Eq. (i) `-286xx2-393-(-726)=7`
`=-572-393+726`
`=-965+726`
`=-239 kJ//mol`