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Free Energy of Reaction

  1. Nov 3, 2007 #1
    2NH3(g) + 3O2(g) + 2CH4(g) [​IMG] 2HCN(g) + 6H2O(g)

    ΔH° for this equation
    -939.8 kJ

    ΔS° for this equation
    165 J/K

    ΔG° at 232°C, for this equation
    -1023.1 kJ

    Are the following statements about this process True or False?

    True: The high temperature required for this process is needed for thermodynamic reasons.
    True: The equilibrium position for this reaction is further to the left at lower temperatures.
    True: Thermodynamically, this reaction is spontaneous at any temperature.
    False: At temperatures significantly lower than 1000°C this reaction is spontaneous.
    True: This reaction is exothermic at room temperature.

    I dont really understand what im doing wrong for the TRUE AND FALSE, can someone help
  2. jcsd
  3. Nov 3, 2007 #2
  4. Nov 3, 2007 #3
    Well, just eyeballing it, it looks like you've got a couple things wrong there.

    This is an extremely exothermic reaction with a positive change in entropy. This means the product of this reaction is drastically lower in energy, and thus, more stable, than the reactant. I'll elaborate on this shortly. Generally, chemistry will do whatever it can to get to the lowest energy configuration that it can unless an outside force coerces it to do otherwise.

    In any chemical reaction that occurs, bonds are broken and formed. Energy is required to break these bonds in all cases. However, the formation of bonds adds stability to the system, and not as much energy is required to maintain it. The formation of bonds releases this now-unneeded energy. When the formation of product bonds releases more energy from the system than was required to break reactant bonds, the system experiences a net loss in energy. This net loss is symbolized by a negative ΔH°.

    Entropy is the amount of disorder in a system -- a system will try to get a state of higher entropy. Why does more disorder require less energy to maintain? Think about it -- it's much easier to keep a random pile of toothpicks the way it is than it is a carefully balanced model of the Eiffel Tower built out of them. More disorder leads to more 'freedom' for the molecules in the product.

    The high temperature isn't actually required for this process, as it's very spontaneous and will occur by itself at any temperature, however, it will take a prohibitively long time to reach equilibrium at lower temperatures.

    As discussed, this is an exothermic reaction. This means that at lower temperatures, the equilibrium state of the system will contain more product than at higher temperatures, hence, equilibrium will be further to the right.

    It's true that this reaction is spontaneous at any temperature.

    As far as the reaction being exothermic at room temperature, yes, it is. Whether or not a reaction is exothermic has nothing to do with the temperature at which it occurs, though temperature can slow down, speed up, or shift the equilibrium position of the reaction. Whether a reaction is exothermic or endothermic is based solely upon the configurations of the products and reactants.

    I hope this helped -- I'd be happy to discuss further, though this was posted in hurry as I'm on way out. I'll check back here later if you have any questions.
  5. Nov 4, 2007 #4
    Thanks i got the right answers :D ur the best but i still dont get why

    "At temperatures significantly lower than 1000°C this reaction is spontaneous. "

    this is true becuase significantly lower or higher this is spotanteous isnt it?? since

    DG = Dh -TDS

    dh is negative ds is positive so no matter what the magnitude of the temperature dG will always be NEGATIVE meaning it is spontaneous???

    so why are they asking that question??
  6. Nov 4, 2007 #5
    That question is trying to trick you into second-guessing yourself. Your reasoning that ΔG° will be negative for this reaction regardless of temperature is correct. Since -ΔG° indicates a spontaneous reaction, it does follow that this reaction is spontaneous at any given temperature.

    This may be going deeper than you cared to, but it's worth taking a look at a similar situation. If you have an endothermic reaction (+ΔH°), it will not be spontaneous unless the product of the temperature and ΔS° is a positive value in excess of the value of ΔH° (this is the only way to get a negative free energy change). It would also then be correct to reason that an endothermic reaction with a negative change in entropy (the products are more orderly than the reactants) will not be spontaneous at any temperature.

    In summary, the only time you need to watch for temperature affecting spontaneity in this sort of problem is when ΔH° and ΔS° have the same sign -- this is the only situation where a different temperature determines whether or not TΔS is greater or less than ΔH°.

    Think about it (keep in mind that T, in Kelvins, is always a positive value):
    +ΔH° - T(-ΔS°) always = positive + positive = +ΔG° regardless of T

    -ΔH° - T(+ΔS°) always = negative - negative = -ΔG° regardless of T

    It is only when ΔH° and ΔS° are both negative that T governs the outcome of ΔG°'s sign.

    Let me know if you have any other questions.
  7. Nov 4, 2007 #6
    thanks that makes sense :D

    anways apart from that i think i should open this ina new thread but ill try here anyways

    i had to do another q

    Calculate ΔG for the dilution of aqueous HCl from 0.89 M to 0.253 M at 25°C.

    the hint was

    The process you are considering is:
    H+(aq) + Cl-(aq)® H+(aq) + Cl-(aq) (for which ΔG°=0 i.e. standard states all round--both sides are the same).
    Now use ΔG=ΔG° + RTln(Q). (Use the actual concentrations to determine Q).

    i did this

    ΔG=RTln(Q) = (8.314)(273.15+25)(ln (0.253/0.89)) = -3117 J/mol

    but thats wrong according to my online assignment and i dont understand how else i can do it :S

    any help?
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