Answer to Question #337702 in General Chemistry for zaza

Question

Calculate the standard enthalpy change, ΔH, for the following reaction:

2CH3OH(l) + 3O3(g) —> 2CO2(g) + 2H2O(g)

Given:

CH3OH (l) ΔHf° = -238.7 kJ

CO2 (g) ΔHf° = -393.5 kJ

H2O (l) ΔHf° = -285.8 kJ

Answer 1

$2CH_3OH+3O_2 \rightarrow 2 CO_2+4H_2O$

$\Delta H=4\Delta H[H_2O]+ 2\Delta H [CO_2]- 3\Delta H[O_2]-2\Delta H [CH_3OH]=4(-285.8)+2(-393.5)-3(0)-2(-238.7)=-1143.2-787-0+477.2=-1453 kJ$