Chemistry question involving k, ∆S˚, ∆H˚, and ∆G˚.?
Consider the following reaction: HI + F- > HF + I-
Predict whether k will be bigger or smaller than unity (1 unit). What is the sign of ∆S˚ and ∆H˚ and which one makes a greater contribution to ∆G˚?
Sorry dad for using your yahoo!
EDIT: This is all the information I was given, word for word. I also thought I would need at least one of the values, but I was assured that this information was sufficient. As for the phases, I think everything is aqueous.
Remember, I don't need any numbers, I just need to predict the signs and k.
- 8 years agoFavorite Answer
Could you please clarify the phase that each reagent is in. We need to know atleast ∆H˚ to go any further.
**Using an example analogous to your question:
Question) Consider the endothermic reaction: N2(g) + O2(g) →2 NO(g), ∆Hº = 192.5 kJ/mol. At 200 K the equilibrium constant is 5.0 x 10-4. At 2500 Kelvin, the value of the equilibrium constant is?
Answer) ∆Sº for the forward direction is difficult to determine without values, but one can apply Le Chatlier's Principles.
Since ∆Hº for the forward reaction is positive, this is endothermic thus..
N2(g) + O2(g) + Joules → 2 NO(g) ∆Hºrxn > 0
If the temperature increases (from 200K to 2500K), the "endothermic route" (forward reaction) will be favored, thus forming more products of the forward reaction (NO(g)) (the numerator in the reaction quotient expression). Thus, K2, after temperature raise, is > K1 (5E-4), because only temperature can affect the "value" of the equilibrium constant K.
**From here, we can state that since ∆Gº = -RTlnK,
(implying an inverse relationship between ∆Gº and K) that an increase in K (to the point where it's GREATER than one) makes ∆Gº lower thus the reaction will eventually become spontaneous at a higher temperature than 2500K (b/c the K will subsequently get "higher/further passed one" as well). Thus, this means that the spontaneity of the reaction is temperature dependent, indicating that ∆Hº and ∆Sº are of the same signs and since ∆Hº is positive, so is ∆Sº (thus forward reaction does increase in entropy, possibly b/c of microstates involved). This condition allows spontaneity to occur at high temperatures only.
**Nevertheless, combining two gases will always increase the entropy (while dissolving gas to a lower phase won't). Since ∆Gºsys is < 0, ∆Sºuniverse is > 0 and since ∆Sºuniverse = ∆Sºsystem + ∆Sºsurroundings, ∆Sºsystem must be a more positive value than ∆Sºsurrounding is a negative one and thus the "system" (or in this case, the forward reaction) DOES more work than the surrounding putting in thus w < 0 for all spontaneous reactions whereas w is > 0 for a nonspontaneous reaction (∆Gº > 0 thus ∆Sºuniverse < 0 thus ∆Sºsurroundings is more negative than ∆Sºsystem is positive) and work has to be done by the surrounding continually (such as an electrolytic cell) to allow for the forward reaction to continue.
**Also, since the equilibrium constant K increases from K1 to K2 (making the reaction more spontaneous towards the right) and is equal to the ratio of the forward over the reverse rate constants, the rate constant of the forward reaction is larger than the rate constant of the reverse reaction and thus the activation energy of the reverse reaction is higher than the forward reaction.
*******With all the species being aqueous, then there is nothing that can be determined about K or any other variable without further information.Source(s): U of H Brain