Question

23. Question (1 point) W Calculate the free-energy change of the following reaction at 221°C: C2H(g) +302(g) → 2002(g) + 2H20(g) A Hºt, kJ/mol 52.3 Sº, J/mol K 219.5 C2H4(g) O₂(g) CO2(g) H2O(g) -393.5 205.0 213.6 188.7 -241.8 1st attempt

Answer 1

C2H4(g) + 3 O2(g) ------------------ 2 CO2(g) + 2 H2O(g)

$\Delta$H0rxn =[ 2 x $\Delta$ H0f of CO2 + 2 x $\Delta$ H0f of H2O] - [ $\Delta$ H0f of C2H4 + 3 x $\Delta$ H0f of O2]

$\Delta$H0rxn =[ 2 x ( -393.5) + 2 x ( - 241.8)] - [ 52.3 + 3x 0.0]

$\Delta$H0rxn = -1322.9 KJ = - 1322900 J

$\Delta$S0rxn =[ 2 x $\Delta$ S0f of CO2 + 2 x $\Delta$ S0f of H2O] - [ $\Delta$ S0f of C2H4 + 3 x $\Delta$ S0f of O2]

$\Delta$S0rxn = 2 x 213.6 + 2 x 188.7 - [ 219.5 + 3 x 205.0]

$\Delta$S0rxn = - 29.9 J

T= 221C = 221+273 = 494K

$\Delta$G0rxn = $\Delta$ H0rxn - T $\Delta$ S0rxn

$\Delta$G0rxn = - 1322900 - [ 494x( -29.9)]

$\Delta$G0rxn = - 1308129.4 J

$\Delta$G0rxn = - 1308.13 KJ