Answer to Question #155146 in General Chemistry for Alexis

Question #155146

Calculate the Gibbs energy for the next reaction 2NH3 (g) + 2.5O2 (g) = 2NO (g) + 3H2O (g)

Expert's answer

at 298K,

"\\Delta G^\\circ = \\sum {G^\\circ _f {\\rm{products}}} - \\sum {G^\\circ _f {\\rm{reactants}}}"

ENTHALPY OF REACTION

"{[}4\u0394Hf(NO (g)) + 6\u0394Hf(H2O (g)){]} - {[}4\u0394Hf(NH3\n(g)) + 5\u0394Hf(O2 (g)){]}\n\\\\{[}4(90.25) + 6(-241.82){]} - {[}4(-46.11) +\n5(0){]}\\\\ = -905.48 kJ"

ENTROPY CHANGE

"{[}4\u0394Sf(NO (g)) + 6\u0394Sf(H2O (g)){]} - {[}4\u0394Sf(NH3 (g)) +\n5\u0394Sf(O2 (g)){]}\\\\{[}4(210.65) + 6(188.72){]} - {[}4(192.34) + 5(205.03){]}\n\\\\= 180.41 J\/K"

FREE ENERGY OF REACTION (AT 298.15 K)

From ΔGf° values:

"{[}4\u0394Gf(NO (g))\n+ 6\u0394Gf(H2O (g)){]} - {[}4\u0394Gf(NH3 (g)) + 5\u0394Gf(O2 (g)){]}\\\\{[}4(86.57) +\n6(-228.59){]} - {[}4(-16.48) + 5(0){]}\\\\ = -959.34 kJ \n(spontaneous)"

FROM "\\Delta G^\\circ = \\sum {G^\\circ _f {\\rm{products}}} - \\sum {G^\\circ _f {\\rm{reactants}}}" ,

=-959.27 kJ (spontaneous)