# Chem

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Consider the reaction
H2S(g) <==> H2(g) + 1/2S2(g)
At 1065*C the partial pressure at equilibrium of hydrogen sulfied, hydrogen and sulfur are found to be .670 atm, .221 atm, and .110 atm, respectively. Find Kp and delta-G* for the reaction. Find delta-G for the reaction if the partial pressure of each of the gases is 3.00 atm.

I know how to solve for Kp and delta-G*, but I'm confused about two things. The first is, is the partial pressure given for sulfur the pressure of 1/2S2, or the pressure of S2?

And two, how does delta-G differ from delta-G* in this problem? Thank you for the help.

Plug in 0.110 for partial pressure of S2 and calculate K. Don't forget to raise S (0.110) to the 1/2 power.
Delta G = Go + RT ln Q.
At equilibrium, delta G = 0 and Q = K.

After you know delta Go and K, then you put in the new the new conditions and calculate delta G. Delta G does not represent equilibrium conditions when it is other than zero.

Under the new conditions other than equilibrium, technically, K becomes Q when you substitute 3.00 atm for each of the gases.

I have a follow-up question concerning this. Here is my work done out-- my problem is, my units seem to be... odd. Are they correct or have I made an error?

Kp= ((PH2)(PS2)^1/2)/(PH2S)
Kp= ((.221 atm)(.110 atm)^1/2)/(.670 atm)= .109 atm
delta G*= -RT lnK
delta G*= -(8.314 x 10^-3 kJ/mol K)(1338.2 K) ln (.109 atm)
delta G*= 24.62 kJ atm/mol

Q= ((PH2)(PS2)^1/2)/(PH2S)
Q= ((.300 atm)(.300 atm)^1/2)/(.300 atm)
Q= .5477 atm

delta-G= delta-G* + RT ln Q
delta-G= 24.62 kJ atm/mol + (8.314 x10^-3 kJ/mol K)(1338.2K)ln .5477 atm
delta-G= 17.92 kJ atm/mol

kJ atm/mol? Is that odd for delta-G related answers? I tried multiplying this by (8.314 x 10-3 kJ/mol K)/(.08206 L atm/mol K) but I ended up with L in my denominator. Any help would be appreciated.