For the following reaction, the value of G ° is618.8 kJ/mol at 25°C. HCHO( g ) +

Question

For the following reaction, the value of G° is618.8 kJ/mol at 25°C. HCHO(g) + 2/3O3(g) CO2(g) +H2O(g) Other data are below. H°f(kJ/mol) at25°C S°(J/mol · K) at25°C HCHO 117.0 219.0 H2O(g) 241.8 188.7 CO2(g) 393.5 213.7 O3(g) 142.7 ? Calculate the absolute entropy, S°, per mole ofO3(g).
J/mol·K
HCHO(g) + 2/3O3(g) CO2(g) +H2O(g) H°f(kJ/mol) at25°C S°(J/mol · K) at25°C HCHO 117.0 219.0 H2O(g) 241.8 188.7 CO2(g) 393.5 213.7 O3(g) 142.7 ? H°f(kJ/mol) at25°C S°(J/mol · K) at25°C HCHO 117.0 219.0 H2O(g) 241.8 188.7 CO2(g) 393.5 213.7 O3(g) 142.7 ?

Explanation / Answer

delH_rxn = [-241.8 + -393.5] - [-117.0 + 2/3 *142.7] = -613.433kJ delG_rxn = delH_rxn - T*delS_rxn delS_rxn = (delH_rxn - delG_rxn)/T = (-613.433 kJ - (-618.8kJ/mol))/298K = 0.018 kJ/mol/K = 18 J/mol/K delS_rxn = (188.7 + 213.7) - [2/3 *S_O3 + 219.0] = 18 183.4 - 2/3*S_O3 = 18 S_O3 = 248 J/mol/K

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