Calculate the enthalpy of formation of methanol `(DeltaH_(f) (CH_(3) OH) (I) )` in KJ `mol^(-1)` using the data :
`DeltaH_("vap") (CH_(3)OH)(I) ` is
38 KJ `mol^(-1)`
The heat of formation of gaseous atoms from elements in their standard states are
`H = 218 KJ mol^(-1) C ` = 715 KJ
`mol^(-1)`O = 249 KJ` mol^(-1)`
Average bond energy data is
C- H= 415 KJ ` mol^(-1) ` C-O
= 356 KJ `mol^(-1) ` O -H = 463
KJ `mol^(-1)`

Options

A. `-266`
B. 266
C. 313
D. `-313`

Answer

A

Explanation

Formation reaction is
`4H_((g)) + C_((g)) +O_((g)) rarr H - underset(H)underset(|)overset(H)overset(|)(C) -O -H_((g))`……. (i)
`DeltaH_(f) =sum_("react") BE - sum_("prod") BE`
`DeltaH_(i) = - 228 KJ m^(-1)`
`= [ 4xx 218 + 715 + 249 ]`
`- [ 3xx 415 + 356 + 463 ]`
`= [872 + 715 + 249] `
`- [ 1245 + 356 + 463 ]`
`= 1836 - 2064 = - 228 KJm^(-1)`
`CH_(3) OH_((g)) rarr CH_(3) OH_((I)) DeltaH_((II)) -38 KJm^(-1)`
`:. DeltaH_(f) (CH_(3) OH) l = - 228 - 38 = -266 kj m^(-1)`