The enthalpy of formation `(Delta_rH)` of methanol, formaldehyde and water -239, -116 and -286KJ`mol^-1` respectively. The enthalpy change for the oxidation of emthanol to formaldehyde and water in KJ is

Options

A. `-136`
B. `-173`
C. `163`
D. `-163`

Answer

D

Explanation

Given `H_(f^(@)) for CH_(3)OH = - 239 kJ mol^(-1)`
`H.CHO = - 116 kJ mol^(-1)`
`H_(2) O=- 286 kJ mol^(-1) 0`
Required relation :
`2CH_(3) OH + O_(2) to 2 HCHO + 2H_(2)O`
`Delta H_(R) = (1)/(2) H_(f^(@)) of H.CHO . and H_(2)O`
`Delta H_(R) = (1)/(2) H_(f^(@)) [(2x - 116)+ (2x - 286)]`
`- 1 [(2 x 239)]`
`= (1)/(2) [(-232 )+ (-572)] - [ - 478]`
` = - (1)/(2) [ - 804 + 478]`
`rArr (1)/(2) [-326 ] = - 163 kJ mol^(-1)`