Test02SOL-docA

Combustion of ethanol with an excess oxygen at 29815

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: change in entropy for the system is written as ΔS and is obtained as the sum of the entropy changes associated with each of the three steps that must be taken to make the process reversible ((1) slow cooling o o under 1.00 atm of the gas phase from 130. C to 100. C, (2) reversible condensation of water vapor at o o 100 C and 1.00 atm, and (3) slow cooling under 1.00 atm of the resulting liquid phase from 100. C to o 30. C. ΔS(1) the change in entropy associated with the reversible isobaric cooling of 1.50 mol of H2O vapor from o o 130. C to 100. C. ΔS(1) = n CPm(H2O(g)) ln (373.15K/403.15K) = 1.50 mol x 33.6 J/K.mol x ln(373.15K/403.15K) ΔS(1) = - 3.897 J/K ΔS(2) the change in entropy associated with the reversible isobaric condensation of 1.50 mol of H2O vapor o at 100. C. ΔS(2) = n ΔcondHo (H2O(g)) / 373.15K = - n ΔvapHo (H2O(g)) / 373.15K = - 1.50 mol x 40.7 kJ/mol / 373.15K ΔS(2) = - 163.607 J/K ΔS(3) the change in entropy associated with the reversible isobaric cooling of 1.50 mol of H2O liquid from o o 100. C to 30. C. ΔS(3) = n CPm(H2O(l)) ln (303.15K/373.15K) = 1.50 mol x 75.3 J/K.mol x ln (303.15K/373.15K) ΔS(3) = - 23.466 J/K ΔS = ΔS(1) + ΔS(2) +ΔS(3) = - 190.970 J/K Δ S = - 191 J/K Question 9: (5 points) o o Given the value for the standard vaporization enthalpy of H2O at 100. C and 1 atm (ΔvapH = 40.7 kJ/mol), calculate the change in entropy for a system composed of 1.50 mol of water as a result of the ir...
View Full Document

Ask a homework question - tutors are online