Answer the following question. Calculate the standard enthalpy of formation of CH3OH(l) from the following data: - Chemistry

Sum

Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l); ΔrH° = - 726 kJ mol-1

"C"_("graphite") + "O"_(2("g")) -> "CO"_(2("g")); ΔrH° = - 393 kJ mol-1

H2(g) + 1/2 O2(g) → H2O(l); ΔrH° = - 286 kJ mol-1

Solution

Given: Given equations are,

CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l); ΔrH° = - 726 kJ mol-1      ....(i)

"C"_("graphite") + "O"_(2("g")) -> "CO"_(2("g")); ΔrH° = - 393 kJ mol-1      ....(ii)

H2(g) + 1/2 O2(g) → H2O(l); ΔrH° = - 286 kJ mol-1 ....(iii)

To find: The standard enthalpy of formation (ΔfH°) of CH3OH(l)

Calculation:

Required equation is, "C"_("graphite") + 2 "H"_(2("g")) + 1/2 "O"_(2("g")) -> "CH"_3"OH"_(("l"))

Multiply equation (iii) by 2 and add to equation (ii),

2"H"_(2("g")) + "O"_(2("g")) -> 2"H"_2"O"_(("l")), ΔrH° = - 575 kJ mol-1

"C"_("graphite") + "O"_(2("g")) -> "CO"_(2("g")), ΔcH° = - 393 kJ mol-1

"C"_("graphite") + 2"H"_(2("g")) + 2"O"_(2("g")) -> "CO"_(2("g")) + 2"H"_2"O"_(("l"))

ΔrH° = - 572 - 393 = - 965 kJ mol-1

Reverse equation (i) and add to equation (iv),

"CO"_(2("g")) + 2"H"_2"O"_(("l")) -> "CH"_3"OH"_(("l")) + 3/2 "O"_(2("g")), ΔrH° = 726 kJ mol-1

"C"_("graphite") + 2"H"_(2("g")) + 2"O"_(2("g")) -> "CO"_(2("g")) + 2"H"_2"O"_(("l")), ΔrH° = - 965 kJ mol-1

"C"_("graphite") + 2"H"_(2("g")) + 1/2 2"O"_(2("g")) -> "CH"_3"OH"_(("l"))

ΔfH° = ΔrH° = 726 - 965 = - 239 kJ mol-1

The standard enthalpy of formation (ΔfH°) of CH3OH(l) from the given data is – 239 kJ mol-1

Concept: Enthalpy (H)
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Chapter 4: Chemical Thermodynamics - Exercises [Page 89]

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Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercises | Q 4.16 | Page 89
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