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Posted by on Monday, February 11, 2013 at 10:45am.

1. Given the following equation:
Glycerol-3-phosphate glycerol + Pi Go’ = -9.7 kJ/mol
At equilibrium the concentrations of both glycerol and inorganic phosphate are 1 mM. Under these conditions, calculate the final concentration of glycerol-3-phosphate. Remember to convert kJ to J.

2. Using the data for the reaction in question 1, calculate the G value (Go’ = -13.8 kJ/mol).

  • Biochemistry - , Monday, February 11, 2013 at 11:49am

    I'm not sure if you typed in the reaction correctly. If the reaction is as followed:

    Glycerol-3-phosphate----> Glycerol + Pi
    ∆G0'= -9.7kj


    Then this what you I believe you should do:

    The removal of the phosphate is the only way that I can see the reaction being energetically favorable, and at equilibrium, ∆G =0.


    R=0.0083145 kJ/mol
    T=273.15+25=298.15K
    ∆G=0
    ∆Go'= -9.7kj/mol
    Pi=1mM
    Glycerol= 1mM
    Q=reactants/products
    Glycerol-3-phosphate=G3P=?
    Q=reactants/products=[1mM][1mM]/G3P

    Plug in your values and solve for Glycerol 3-phosphate

    ∆G=∆G0'+RTlnQ

    0=∆G0'+RTlnQ

    -∆G0'=RTlnQ

    -∆G0'/RT=lnQ

    10^(-∆G0'/RT)=Reactants/Products

    Reactants/10^(-∆G0'/RT)=Products

    [1mM][1mM]/10^(-∆G0'/RT)=G3P


    [1mM][1mM]/(9.7kj/mol)/[(0.0083145 kJ/mol)(298.15K)]=G3P

    For the second part, use the value that you obtained for G3P, and the values that are given to you for Glycerol, ∆G0' (-13.8 kJ/mol), and Pi and plug into
    the equation below and solve.

    ∆G=∆G0'+RTlnQ

  • Biochemistry - , Monday, February 11, 2013 at 11:53am

    Last part of the equation for the first step should be,


    [1mM][1mM]/10^[(9.7kj/mol)/[(0.0083145 kJ/mol)(298.15K)]=G3P

  • Biochemistry - , Monday, February 11, 2013 at 12:58pm

    I have a bad typo in my original post that may lead to confusion: replace the words reactants with products and vis versa. Everything else is okay except for that.

  • Biochemistry - , Monday, February 11, 2013 at 1:16pm

    Meaning, Q=products/reactants, not the other way around. for 10^(-∆G0'/RT)=Reactants/Products part of the equation manipulation.

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