Question

Calculate the enthalpy of formation of CH3OH from the following data:            (i) CH3OH + O2       - >    CO2 + 2H2O            H  = -726 KJ/mol          
(ii) C + O2          -   >       CO2 H = -393 KJ/mol          
  (iii) H2 + ½ O2       -  >    H2O                     H = -286 KJ/mol​

Answer 1

$\small\bold{Heat \: of \: formation \: of \: CH_3OH = 2× heat \: of}$

$\small \bold{combustion \: of \: H_2O + heat \: of \: combustion}$

$\small \bold{of \: CO_2 - heat \: of \: combustion \: of \: CH _3OH}$

$\small \bold{Heat \: of \: formation \: of \: CH_3OH }$

= 2×(−286)−(−394)−(−726)

$\bold{=−240KJmol^−1}$