When dG is - (i.e., <0) the reaction is favored. Therefore, if dGo is >0 it must be + and that means the system is not favorable for a reaction. That is the equilibrium lies to the left. That takes care of 1 and 5.
2. Knowing that the equilibrium lies to the left means H2PO4^- is not likely to be transformed to HSO4^-.
4. If dG<0 (that is -) keq is to the right and >1/
If dG = 0, the Keq = 1
If dG > 0 (that is +) then Keq <1
So 4 can't be right.
3. Ka for HSO4^- is k2 and you can look that up. I think it's about 0.012 but you should confirm that.
Kb for SO4^2- = Kw/k2 = 1E-14/0.012 = about 3E-13 and this is < Ka (which equals 0.012) so 3 is OK.
Hope this helps.
Thanks to Dr. Bob for answering all these questions!
I always do poorly on conceptual problems.
Consider the following equilibrium system at 25°C and select the true statement below: H2PO4−(aq) + SO42−(aq) ↔ HSO4−(aq) + HPO42−(aq)
ΔG°rxn > 0 at 25°C
(1) this is a product-favored reaction
(2) H2PO4− is a stronger acid than HSO4−
(3) The Kb of SO42− is less than the Ka of HSO4−
(4) the Keq of this reaction is > 1.0
(5) if ΔG°rxn > 0, then ΔGrxn must also be > 0
I know that 3 is the correct answer but I was wondering why? How could you know that the Kb of SO42- was less if HSO4 is not a strong acid?
Also, how would you know it was product favored? How would you know if Keq is greater than 0?
1 answer
1 answer