For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 bar for all species. For the reaction

C2H6(g)+H2(g)---->2CH4
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=____kj/mol

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!

To calculate ΔG for the reaction at the given partial pressures, we can use the equation:

ΔG = ΔG° + RTln(Q)

Where:
ΔG is the change in Gibbs free energy
ΔG° is the standard change in Gibbs free energy
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (298 K)
ln is the natural logarithm
Q is the reaction quotient

First, we need to calculate the reaction quotient (Q) using the partial pressures provided. Since the reaction involves gases, we can use the partial pressures directly as the concentrations in Q. Hence:

Q = [CH4]^2 / ([C2H6] * [H2])

Now, we can substitute the values into the equation and solve for ΔG:

ΔG = ΔG° + RTln(Q)
ΔG = -32.8 kJ/mol + (8.314 J/(mol·K) * 298 K * ln(0.850^2 / (0.500 * 0.500)))

Calculating this expression will give you the value of ΔG in kilojoules per mole.