Delta G = G not + RTlnQ
==69.0 +(0.008314*298K*(ln((0.8^2)/((0.25^2)*0.15))
=-69.0 +10.46383111
=-58.53616889
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction
2NO(g) + O2(g) --> 2NO2(g)
the standard change in Gibbs free energy is ΔG° = -69.0 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are
P[NO] = 0.25 atm
P[O2] = 0.15 atm
P[NO2] = 0.8 atm
2 answers
dG = dGo + RTlnQ
where Q is set up the same way as Keq.
where Q is set up the same way as Keq.