Friday
April 18, 2014

Homework Help: Chem

Posted by Jessica on Monday, April 16, 2007 at 1:59pm.

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.

I didn't check your numbers but the following may be helpful.
K, whether Kc or Kp, has no units (just as K for ANY reaction as no units) because the values that go into Kp and Kc or activities and not concentrations. (Activities don't have units.) MOST OF THE TIME, however, we assume concentrations and pressures are the same as activities and use those values, minus the unit, for activity. Is that the usual case. Not really, but it makes the calculations simpler. In fact, many times the activities are not even close to concentration. (I have seen some internet sites in which K is given "provisional units" but that is artificial.) As to R, you should use 8.314 J/mol*K. And G or Gf come out in terms of kJ/mol or J/mol. Does that help?

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

Chemistry - Consider the following reaction: H2+I2<->2HI A reaction ...
chem - Consider the decomposition of ammonium hydrogen sulfide: NH4Hs(s) <--&...
Chemistry - Nitrogen gas (N2) reacts with hydrogen gas (H2) to form ammonia (NH3...
chem - The following equilibrium pressures were observed at a certain ...
Chemistry - Consider the reaction: 2 H2S(g)--><-- 2H2(g) + S2 Kp= 2.4 x ...
Chemistry - Consider the reaction: 2 H2S(g)--><-- 2H2(g) + S2 Kp= 2.4 x ...
Chemistry - Consider the following equilibrium at 395K: NH4HS(s)<-> NH3(g...
Chemistry - Consider the following equilibrium at 395K: NH4HS(s) <--> NH3(...
Chemistry - Hydrazine, N2H4 can be made (on paper) by reaction of molecular ...
Chemistry - The reaction for my lab is Mg(s)+2HCl(aq)--> MgCl2(aq)+H2(g) I ...

Search
Members