Thermodynamics (IIT question)

CH₃OH (l)--->CH₃OH ΔH = 38 kJ/mol

1/2H₂(g)------------->H(g) ΔH = 218 kJ/mol

C(graphite)-->C(g) ΔH = 715 kj/mol

1/2 O₂---------->O(g) ΔH = 249 kJ/mol

CH(g)---------------> C(g) +H(g) ΔH= 415kJ/mol

C-O(g)---------->C(g) + O(g) ΔH= 356kJ/mol

O-H(g)--------------->O(g) + H(g) ΔH=463kJ/mol

For the calculation of ΔH_f of CH₃OH, the thermochemical equation is written as

C(s) + 2H₂ +1/2 O₂(g)----->CH₃OH, ΔH=?-----(a)

On th basis of bond enthalpy concept,

ΔH = - {sum of bond enthalpy of all bonds of products - sum of bond enthalpies of all bonds of reactants}

ΔH = {(3 x E_C-H + E_C-O + E_O-H) - (E_{C(s)----->C(g)}+ E_{1/2 H-H(g)} + E_1/2O2(g)}

= - {(3 x 415 +356 +463) - (715 + 4 x 218 +249)}

ΔH = -228 kJ/mol -------- (answer of equation (a))

CH₃OH (g) ------> CH₃OH(l)--------- (b)

On adding two thermochemical equation (a) and (b),

ΔH_f = -266kJ/mol,

Hence the answer.

Oye Talwar Machhli [3][3], Tell here whether the answer is correct or not.