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Relationship between Free energy change and Equilibrium constant (K)

  1. Aug 11, 2014 #1
    1. I have a question regarding the equation ΔG° = -RT lnK. When solving for the equilibrium constant K, what is the relationship between the vapor pressure of each molecular compound in a reaction?

    ΔG° = Free energy change
    R = Universal gas constant
    T = Temperature
    K = equilibrium constant
    *K can be solved for in concentration or pressure or solubility


    2. Meaning, for the reaction aA(s) + bB(g) --> cC(g) + dD(g) where the lower case coefficients are the number of moles and the upper case coefficients are the molecular compounds. Assuming I solved for ΔG at a specific temperature and then solved for K If I wanted to find the the vapor pressure or the concentration of bB how would I relate the value of the equilibrium constant K to solve for B?

    My thought is that 1/Bb = K. Is this correct?

    There is a specific applied problem from where my question is derived, so if anyone would like me to post it to clarify areas where I may not be explicit enough, I can.
     
  2. jcsd
  3. Aug 11, 2014 #2

    Borek

    User Avatar

    Staff: Mentor

    C and D are gases as well, so they are part of K as well.

    Hard to say what you are expected to do not knowing the problem, but typically you should find C and D from the stoichiometry (for example using ICE table).
     
  4. Aug 11, 2014 #3
    Here's the problem:

    A humidity sensor consists of a cardboard square that is colored blue in dry weather and red in humid weather. The color change is due to the reaction:

    CoCl2(s) + 6 H2O(g) ⇌ [Co(H2O)6]Cl2(s)

    For this reaction at 25 °C, ΔH° = -352 kJ/mol and ΔS° = -899 J/(K mol). Assuming that ΔH and ΔS are independent of temperature, what is the vapor pressure of water (in mm Hg) at equilibrium for the above reaction at 35 °C on a hot summer day?

    I used the equation ΔG° = ΔH° - TΔS° at T = 35°C = 308.15 K. So, ΔG = -74973.15 J

    Then, I used ΔG° = - RTlnK and solved for K at T = 308.15 Kelvin. Thus, K = 5.5 x 1012 atm.

    This is where I am confused and had to look up the solution: 1/(H2O)6 = 5.5 x 1012. They then solved for H2O

    I think I actually just understood what they did. wow. So if there was more than one gas in the equation then I would need more information to solve for the pressure of water because there would be more than one variable?
     
  5. Aug 11, 2014 #4

    Borek

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    Staff: Mentor

    Yes.
     
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