calculate the enthalpy change for the reaction C + O2 ---> CO2 Given that energy required for sublimation of one mole of carbon is 716.7kJ. Also the energy required to break one mole of O=O & C=O BONDS IS 494 kJ and 707kJ respectively. (please explain every step, I'm reporting irrelevant answers :D)
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Answer:
Your answer is in the above attachment.
Answer:
a Hess' law discussion and then solve example #1 again.
However, there is a difficulty: we wind up with a Hess' Law formulation that is slightly different than we use when we manipulate chemical equations with their associated enthalpies.
One final note before solving some problems: the ΔH values determined via this technique are only approximations. This is because the bond enthalpy values used are averages. Bond enthalpies actually differ slightly from substance to substance.
Here is what I mean: take a carbon-carbon single bond (C−C).
Example #4 has a little trick in it. Just so you know!
Example #1: Hydrogenation of double and triple bonds is an important industrial process. Calculate (in kJ) the standard enthalpy change ΔH for the hydrogenation of ethyne (acetylene) to ethane:
H−C≡C−H(g) + 2H2(g) ---> H3C−CH3(g)
Bond enthalpies (in kJ/mol): C−C (347); C≡C (839); C−H (413); H−H (432)
Solution:
1) You have to put energy into a bond (any bond) to break it. Bond breaking is endothermic. Let's break all the bonds of the reactants:
one C≡C ⇒ +839 kJ
two C−H ⇒ 413 x 2 = +826 kJ
two H−H ⇒ 432 x 2 = +864 kJ
The sum is +2529 kJ
Note there are two C−H bonds in one molecule of C2H2 and there is one H−H bond in each of two H2 molecules. Two different types of reasons for multiplying by two.
2) You get energy out when a bond (any bond) forms. Bond making is exothermic. Let's make all the bonds of the one product:
one C−C ⇒ −347 kJ
six C−H ⇒ −413 x 6 = −2478
The sum is −2826 kJ
3) ΔH = the energies required to break bonds (positive sign) + the energies required to make bonds (negative sign):
+2529 + (−2825) = −296 kJ/mol
In step 3 just above, I wrote the ΔH calculation in the form of Hess' Law, but with words. Let's try some symbols:
ΔH = Σ Ebonds broken + Σ Ebonds formed
I'm using E to represent the bond energy per mole of bonds (for example, E for the C≡C bond is 839 kJ/mol). Also, a reminder:
Σ Ebonds broken ⇒ always a positive value
Σ Ebonds formed ⇒ always a negative value
Now, I want to rearrange the negative sign on the second value in step three above. Here is what I wrote above:
+2529 + (−2825) = −296 kJ/mol
and in writing the Hess' law formulation, I want to do this:
+2529 − (+2825) = −296 kJ/mol
Here is how it affects the Hess' Law formulation :
Example #1 (again):
H−C≡C−H(g) + 2H2(g) ---> H3C−CH3(g)
Bond enthalpies (in kJ/mol): C−C (347); C≡C (839); C−H (413); H−H (432)
Solution:
1) Hess' Law for bond enthalpies is:
ΔH = Σ Ereactant bonds broken − Σ Eproduct bonds broken
2) On the reactant side, we have these bonds broken:
Σ [two C−H bonds + one C≡C bond + two H−H bonds]
Σ [(2 x 413) + 839 + (2 x 432)] = 2529 kJ
3) On the product side, we have these bonds broken:
Σ [one C−C bond + six C−H bonds]
Σ [347 + (6 x 413)] = 2825 kJ
4) Using Hess' Law, we have:
ΔH = 2529 − 2825 = −296 kJ
Example #2: Using bond enthalpies, calculate the reaction enthalpy (ΔH) for:
CH4(g) + Cl2(g) ---> CH3Cl(g) + HCl(g)
Bond enthalpies (in kJ/mol): C−H (413); Cl−Cl (239); C−Cl (339); H−Cl (427)
Solution:
1) Hess' Law for bond enthalpies is:
ΔH = Σ Ereactant bonds broken − Σ Eproduct bonds broken
2) On the reactant side, we have these bonds broken:
Σ [four C−H bonds + one Cl−Cl bond]
Σ [(4 x 413) + 239] = 1891 kJ
3) On the product side, we have these bonds broken:
Σ [three C−H bonds + one C−Cl bond + one H−Cl bond]
Σ [(3 x 413) + 339 + 427] = 2005 kJ
4) Using Hess' Law, we have:
ΔH = 1891 − 2005 = −114 kJ
reactant side: [one C−H bond + one Cl−Cl bond] = 413 + 239 = 652
product side: [one C−Cl bond + one H−Cl bond] = 339 + 427 = 766
ΔH = 652 − 766 = −114 kJ
Example #3: What is the enthalpy of reaction for the following equation:
2CH3OH(ℓ) + 3O2(g) ---> 2CO2(g) + 4H2O(g)