Question

# 9. Use the following experimentally derived combustion data to calculate the standard molar enthalpy of formation...

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)                                     Δrxn = −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

#### Homework Answers

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)

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