Question

A). For the reaction 3N0(g) → N20(g) + NO2(g) S[N201 = 219.7 JK find the value of ASrxn s[NOJ-210.65 JK-1 SINO1-239.9 JK Calculate ΔG for the reaction: 2CH,0H(g) + 302(g) → 2CO2(g) + 4H20(g) CH30H 02 CO2 H20 -201.2 238 205 213.7 -393.5 -241.826 188.72
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Answer #1

A)

DS0rxn = (1*S0,NO2 + 1*S0,N2O)-(3*S0,NO)

         = (239.9+210.65)-(219.7)

         = 230.85 j/mol.k

B)

DH0rxn = (2*DH0f,H2O + 2*DH0f,CO2)-(2*DH0f,CH3OH+3*DH0f,O2)

         = (2*-241.826+2*-393.5)-(2*-201.2+3*0)

         = -868.25 Kj/mol

dS0rxn = (2*S0f,H2O + 2*S0f,CO2)-(2*S0f,CH3OH+3*S0f,O2)

         = (2*188.72+2*213.7)-(2*238+3*205)

         = -286.16 j/mol.k

dg0 = DH0-TDS0

      = (-868.25)-(273.15*-286.16)

      = 77.3 kj/mol

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A). For the reaction 3N0(g) → N20(g) + NO2(g) S[N201 = 219.7 JK" find the value...
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