Question

Calculate the enthalpies of methanol if the enthalpies of formation of CO2,H2O and CH3OH are respectively -393.5kj/mol,-285.8kj/mol and -239.0kj/mol.

Answer 1

$\Delta_сH^0=\displaystyle\sum_{i=1}v_i\Delta_fH^0(B_i)-\displaystyle\sum_{i=1}v_i\Delta_fH^0(A_i)$

Where B are products and A are initial reagents

Equation for combustion of methanol:

CH₃OH + 1.5O₂ = CO₂ + 2H₂O

Then

$\Delta_сH^0(CH_3OH)=\Delta_fH^0(CO_2)+2\Delta_fH^0(H_2O)$ $-1.5\Delta_fH^0(O_2)-\Delta_fH^0(CH_3OH)$

For all simple substances (including O₂) under normal conditions enthalpies of formation are taken as 0. So $\Delta_fH^0(O_2) = 0$ and

$\Delta_сH^0(CH_3OH)= -393.5 + 2*(-285.8)$$-1.5*0-(-239.0)=-726.1$ kj/mol