Question

9. Use the following experimentally derived combustion data to calculate the standard molar enthalpy of formation (ΔH°_f ) of liquid methanol (CH₃OH) from its elements.

2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(l)     ΔH°_rxn = −1452.8 kJ

C(graphite) + O₂(g) → CO₂(g)                               ΔH°_rxn = −393.5 kJ

2 H₂(g) + O₂(g) → 2 H₂O(l)                                     ΔH°_rxn = −571.6 kJ

(1) −238.7 kJ/mol    (2) 487.7 kJ/mol       (3) −548.3 kJ/mol    (4) 20.1 kJ/mol         (5) 47.1 kJ/mol

Answer 1

we need

CH3OH --> C(s) + 2H2(g) + 1/2O2(g) = CH3OH(l)

then:

get rxn 2 as it is

C(s) + O2(g) = CO2(g) H = -393.5

we need; 2H2 mol so add rxn 2

2H2(g) + O2(g) C(s) + O2(g) = CO2(g) + 2H2O(l) H = -393.5 + -571.6= -965.1

2H2(g) + C(s) + 2O2(g) = CO2(g) + 2H2O(l) H = -965.1

divide rxn 1 by 2 and invert it

CO2(g) + 2H2O(l) = CH3OH + 3/2O2(g) H = -1/2*(-1452.8) =726.4

add all

CO2(g) + 2H2O(l) + 2H2(g) + C(s) + 2O2(g) = CO2(g) + 2H2O(l) + CH3OH + 3/2O2(g) H = -965.1+726.4 = -238.7

simplify

2H2(g) + C(s) + 1/2O2(g) =CH3OH(l) H = -238.7 kJ/mol

choose option (1)