Answer :
Answer:
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
Explanation:
The formation reaction of CH_3OH will be,
[tex]C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?[/tex]
The intermediate balanced chemical reaction will be,
[tex]C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole[/tex]..[1]
[tex]H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole[/tex]..[2]
[tex]CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole[/tex]..[3]
Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:
We get :
[tex]C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole[/tex]..[1]
[tex]2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol[/tex]..[2]
[tex] CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole[/tex] [3]
The expression for enthalpy of formation of [tex]C_2H_4[/tex] will be,
[tex]\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3[/tex]
[tex]\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)[/tex]
[tex]\Delta H=-238.7kJ/mole[/tex]
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
The enthalpy of formation [tex]\Delta H_{f} ^{o}[/tex] of methanol (CH₃OH) from its elements is -238.7 kJ/mol
From the question,
We are to determine the enthalpy of formation of methanol (CH₃OH) from its elements
From the given information,
(1) CH₃OH(l) + ³/₂O₂(g) → CO₂(g) + 2H₂O(l) ΔH°rxn = -726.4 kJ
(2) C(graphite) + O₂(g) → CO₂(g) ΔH°rxn = -393.5 kJ
(3) H₂(g) + ¹/₂O₂(g) → H₂O(l) ΔH°rxn = -285.8 kJ
Since, the product we want to form is methanol CH₃OH,
Reverse equation (1) to give (4)
That is,
(4) CO₂(g) + 2H₂O(l) → CH₃OH(l) + ³/₂O₂(g) ΔH°rxn = 726.4 kJ
Now, multiply equation (3) by 2 to give (5)
(5) 2H₂(g) + O₂(g) → 2H₂O(l) ΔH°rxn = -571.6 kJ
Now, add equations (2), (4), and (5)
(2) C(graphite) + O₂(g) → CO₂(g) ΔH°rxn = -393.5 kJ
(4) CO₂(g) + 2H₂O(l) → CH₃OH(l) + ³/₂O₂(g) ΔH°rxn = 726.4 kJ
(5) 2H₂(g) + O₂(g) → 2H₂O(l) ΔH°rxn = -571.6 kJ
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C(graphite) + 2H₂(g) + ¹/₂O₂(g) → CH₃OH(l) ΔH°rxn = -238.7 kJ
Since, the above reaction represent the formation of 1 mole of methanol (CH₃OH) from its elements,
Then,
The enthalpy of formation [tex]\Delta H_{f} ^{o}[/tex] of methanol = -238.7 kJ/mol
Hence, the enthalpy of formation [tex]\Delta H_{f} ^{o}[/tex] of methanol (CH₃OH) from its elements is -238.7 kJ/mol
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