Problem Set 8 Solutions

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107. ContinuedTo determine the extent of reaction, we can use the ideal gas law,Then, in terms of the extent of reaction,n tot = PVRT( 29.33atm) ( 1.0 L)( )=( 0.08206 L atmmol −1 K −1) 523.15Kn tot = 0.683mol.€n tot = 0.574 + ξ0.683mol = 0.574 + ξξ = 0.109 mol.Finally, the equilibrium constant may be calculated (using P = 29.33 atm = 29.72 bar),€K eq ==K eq = 2.23.ξ 2 P( 0.574 +ξ) 0.341−ξ( ) P !( 0.109 mol) 2 ( 29.72 bar)( 0.574 + 0.109 mol) ( 0.341− 0.109 mol) ( 1bar)€
118. Determine the equilibrium constant and extent of reaction for the dissociation of hydrogen gas at 3000 Kand 1 atm,H 2 (g) 2 H (g) .The standard molar Gibbs free energies of formation,kJ/mol, respectively.ΔG f! , of H2 and H at 3000 K are 0 and 46.01The equilibrium constant can be expressed in terms of the extent of reaction,€The equilibrium constant isH 2 (g) 2 H (g) .moles init. 1 0moles equil. 1−ξ 2ξx i1−ξ1+ξ2ξ1+ξ€K eq =or K eq =2" P H%$# P ! '&" P H %$ 2# P ! '&2" x H P %$# P ! '&." x H 2P %$# P ! '&Writing the equilibrium constant in terms of the extent of reaction yields€K eq ==K eq =2" x H P %$# P ! '&" x H 2P %$# P ! '&2" 2ξ % " P %$ ' $# 1+ξ & # P ! '&" 1−ξ %$# 1+ξ &' " P %$# P ! '&24ξ 2 " P %( 1−ξ 2 $#) P ! ' .&€
Page 2 and 3: 22. The equilibrium constant for th
Page 6 and 7: 65 b.) ContinuedSolving for the ext
Page 8 and 9: 86. ContinuedSubstituting into the
Page 12: 128. ContinuedTo determine the exte

Problem Set 8 Solutions

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