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Question: Calculate the standard enthalpy of formation of \( C{H_3}OH \) from the following data: \( C{H_3}...

Calculate the standard enthalpy of formation of CH3OHC{H_3}OH from the following data:
CH3OH(I)+32O2(g)→CO2(g)+2H2O(I);ΔrHθ=−726kJmpl−1C{H_3}OH(I) + \dfrac{3}{2}{O_2}(g) \to C{O_2}(g) + 2{H_2}O(I);{\Delta _r}{H^\theta } = - 726kJmp{l^{ - 1}}
C(graphite)+O2(g)→CO2(g);ΔCHθ=−393kJmol−1{C_{(graphite)}} + {O_2}(g) \to C{O_2}(g);{\Delta _C}{H^\theta } = - 393kJmo{l^{ - 1}}
H2(g)+12O2(g)→H2O(I);ΔfHθ=−286kJmol−1{H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(I);{\Delta _f}{H^\theta } = - 286kJmo{l^{ - 1}}

Explanation

Solution

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of one mole of the substance from its constituent elements, with all substances in their standard states. The standard enthalpy of formation is measured in units of energy per amount of substance, usually stated in kilojoule per mole, but also in kilocalorie per mole, joule per mole or kilocalorie per gram any combination of these units conforming to the energy per mass or amount guideline. All elements in their standard states oxygen gas, solid carbon in the form of graphite, etc. have a standard enthalpy of formation of zero, as there is no change involved in their formation.

Complete answer:
Formation enthalpy is the normal reaction enthalpy for the formation of the compound from its elements (atoms or molecules) at the chosen temperature (298.15K) and at 1 bar pressure in their most stable reference states.
ΔfHθ=∑vΔfHθ(products)−∑vΔfHθ(reactan⁡ts){\Delta _f}{H^\theta } = \sum\limits_{}^{} {v{\Delta _f}} {H^\theta }(products) - \sum\limits_{}^{} {v{\Delta _f}} {H^\theta }(reac\tan ts)
On reversing the first equation
CH3OH(I)+32O2(g)→CO2(g)+2H2O(I);ΔrHθ=+726kJmpl−1C{H_3}OH(I) + \dfrac{3}{2}{O_2}(g) \to C{O_2}(g) + 2{H_2}O(I);{\Delta _r}{H^\theta } = + 726kJmp{l^{ - 1}} (On reversing the reaction sign of ΔH\Delta H also reverse)
The equation remains same
C(graphite)+O2(g)→CO2(g);ΔCHθ=−393kJmol−1{C_{(graphite)}} + {O_2}(g) \to C{O_2}(g);{\Delta _C}{H^\theta } = - 393kJmo{l^{ - 1}}
And, on multiplying equation three by 22 , we get
H2(g)+12O2(g)→H2O(I);ΔfHθ=2×−286kJmol−1=−572kJmol−1{H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(I);{\Delta _f}{H^\theta } = 2 \times - 286kJmo{l^{ - 1}} = - 572kJmo{l^{ - 1}}
On adding these three equations we get,
C(graphite)+2H2(g)+12O2(g)→CH3OH(I){C_{(graphite)}} + 2{H_2}(g) + \dfrac{1}{2}{O_2}(g) \to C{H_3}OH(I)
Enthalpy of formation of CH3OH=ΔfH=ΔrHθ+ΔcHθ+2ΔfHθC{H_3}OH = {\Delta _f}H = {\Delta _r}{H^\theta } + {\Delta _c}{H^\theta } + 2{\Delta _f}{H^\theta }
ΔfH=+726−393−572kJmol−1=−239kJmol−1{\Delta _f}H = + 726 - 393 - 572kJmo{l^{ - 1}} = - 239kJmo{l^{ - 1}}
So, the enthalpy of formation of CH3OHC{H_3}OH is −239kJmol−1- 239kJmo{l^{ - 1}} .

Note:
In the estimation of reaction enthalpies, enthalpies (or heats) of formation are incredibly useful. This is because it is possible to imagine any reaction as happening along a path through which all reactant compounds are first converted to elements and then all elements are converted into compounds of the substance