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From the following heats of combustion, CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l)        ΔHorxn= –726.4 kJ/mol C(graphite) + O2(g)...

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Chemistry

From the following heats of combustion,

CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l)       ΔHorxn=–726.4 kJ/mol

C(graphite) + O2(g) →CO2(g)                             ΔHorxn = –393.5 kJ/mol

H2(g) + ½O2(g) →H2O(l)                                                ΔHorxn = –285.8 kJ/mol

Calculate the enthalpy offormation of methanol (CH3OH) from its elements.

C(graphite) + 2H2(g) +½O2(g) → CH3OH(l)

Answer & Explanation Solved by verified expert
4.2 Ratings (489 Votes)
Number equations 123 Now Invert EQN1 CO2g 2H2Ol CH3OHl 32O2g Horxn 7264 kJmol Multiply EQN3 by 2 2H2g O2g 2H2Ol Horxn 22858 kJmol 5716    See Answer
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