posted by BOB on .
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar for all species. For the reaction
the standard change in Gibbs free energy is ΔG° = -32.8 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are
P(C2H6)=0.500bar, P(H2)=0.500bar, and P(CH4)=0.850bar
dG = dGo + RTlnQ
how do you find lnQ?
Q is the same form as K; i.e., concentrations in M of products over concentrations in M of reactants and raise each reagent to the power shown by the coefficients. In this case, you're dealing with pressures so it's partial pressure instead of concentrations. In other words make it look like a Kp expression. For example,
A(g) + 2B(g) ==> 3C(g) you would have
Q = pC^3/pA*p^B^2
Ok thanks a bunch!