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Archy [21]
2 years ago
12

Using the data: C2H4(g), = +51.9 kJ mol-1, S° = 219.8 J mol-1 K-1 CO2(g), = ‑394 kJ mol-1, S° = 213.6 J mol-1 K-1 H2O(l), = ‑286

.0 kJ mol-1, S° = 69.96 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = 205 J mol-1 K-1 calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)
Chemistry
1 answer:
ra1l [238]2 years ago
6 0

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.  

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At this point there are two answers:  one says that there is no difference, except for the units;  the molecular weight (MW) is the mass of one molecule, while a molar mass is the mass of a mole (6.02 x 10 ^ 23) of molecules.

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So, we'll look at your question from a different perspective.   Take a real example, something simple like water.  Water has a nominal MW of 18 (so we'll use nominal precision for simplicity, instead of the 4-places often used for these types of calculations).

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From the CRC reference book we find that one amu weighs 1.66 x 10 ^--24 grams.  Multiplying the two and in keeping with the two units of precision we're working with, one molecule of water has a mass of  29.8 x 10^-24g, or [3.0 x 10 ^ -23 g] per molecule of water.

A Mole is simply Avagadro's number (6.02 x 10 ^ 23) of anything... protons, baseballs, whatever.  The term Molar Mass in chemistry refers to the mass of a mole of molecules.  So in this case a molar mass of water molecules is Avagadro's number of them, the mass therefore being [6.02 x 10 ^ 23] x 3.0 x 10 ^ -23 g/ molecule] =  18.0 g

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