12-16 PROBLEMS 1. Calculate the entropy change at 25 °C and one bar pressure for
ID: 718031 • Letter: 1
Question
12-16 PROBLEMS 1. Calculate the entropy change at 25 °C and one bar pressure for the reaction THERMODYNAMICS (VII) aas-15 K using the following data at one bar pressure: Ahyus (kJ mol) 226.73 g 52.26 (kJ mol) 209.20 68.15 r(298.15) 2. Calculate the maximum useful work that may be obtained from the process at 298.15K and one bar pressure, using the following data, at 25°C and one bar co CO (si (kJ mol -110.525 -393.509 S JK mol) 197.674 213.74574 3. Calculate the maximum "other" work that could be obtained from the process: 1 .00 kg water vapor 1 .00 kg liquid water at 80°C and one atm. Heat of vaponization of water at the normal boiling point 2257 J g heat capacity of liquid water = 4 20 J g"K": heat capacity of water vapor-210 J g-1 K" I mol liquid toluene 70.0°C, 1 atn I mol toluene vapo 70.0°C, 1 ann 4. For the process calculate: (a) AS (b) AG. The normal boiling point of toluene (methyl benzene) is 111°C;its molar massis 92 15 g mol. For toluene at 111 C and one atm pressure: Ah 33.18 kJ moliquid toluene) 1.67 J g- K-": c, ltoluene vapor] = 1.13 J g-1 K-1 Is this a spontaneous process?Explanation / Answer
Ans 1
Enthalpy change for the reaction = sum of Enthalpy of formation of products - sum of Enthalpy of formation of reactants
H = Hf(C2H4) - Hf(H2) - Hf(C2H2)
= 52.26 - 0 - 226.73
= - 174.47 kJ/mol
Free energy change
G = Gf(C2H4) - Gf(H2) - Gf(C2H2)
= 68.15 - 0 - 209.20
= - 141.05 kJ/mol
We know that
G = H - TS
- 141.05 kJ/mol = - 174.47 kJ/mol - 298.15 K x S
33.42 = - 298.15 x S
S = - 0.11209 kJ/mol·K x 1000J/kJ
Entropy change of reaction S = - 112.09 J/mol·K
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