Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be written as

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Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be written asATP(aq)+H2O(l)⟶ADP(aq)+H2PO−4 (aq)where ADP represents adenosine diphosphate. For this reaction, ΔG∘=−30.5kJ/mol.a. Calculate K at 25∘C .b. If all the free energy from the metabolism of glucoseC6H12O6(s)+6O2(g)⟶6CO2(g)+6H2O(l)goes into forming ATP from ADP, how many ATP molecules can be produced for every molecule of glucose?

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  1. Answer:

    Step-by-step explanation:

    From the given information:

    ΔG° = -30.5 kJ/mol

    By applying the following equation to calculate the value of K.

    ΔG° =-RT㏑K

    making ㏑ K  the subject of the formula:

    [tex]\mathtt{ In \ K} = \dfrac{\Delta G^0}{-RT}[/tex]

    where;

    Temperature at 25° C = (25 + 273)K

    = 298K

    R = 8.3145 J/mol.K (gas cosntant)

    [tex]\mathtt{ In \ K} = \dfrac{-30.5 \times 10^{3}\ J /mol} {-(8.3145 \ J/mol. K \times 298 \ K}[/tex]

    [tex]\mathtt{ In \ K} = \dfrac{-30.5 \times 10^{3}\ J /mol} {-2477.721 J/mol }[/tex]

    ㏑K = 12.309

    [tex]K = e^{12.309}[/tex]

    K = 221682.17

    K = 2.22 × 10⁵

    b) The reaction for the metabolism of glucose is given as:

    [tex]C_6H_{12} O_6 + 6O_{2(g)} \to + 6CO_{2(g)} + 6H_2O_{(l)}[/tex]

    From the above expression, let calculate the Gibbs free energy by using the formula:

    [tex]\Delta G^0_{rx n }= \Delta G^0_{product}- \Delta G^0_{reactant}[/tex]

    [tex]\Delta G^0_{rx n }= [6 \times \Delta G^0_{f}(CO_2) + 6 \times \Delta G^0_{f}(H_2O)] – [1 \times \Delta G^0_{f}(C_6H_{12}O_6) + 6 \times \Delta G^0_{f}(O_2)][/tex]

    At standard conditions;

    The values of corresponding compounds are substituted into the equation above:

    Thus,

    [tex]\Delta G^0_{rx n }= [6 \times (-394) + 6 \times (-237)] – [1 \times (-911) + 6 \times (0)] \ kJ/mol[/tex]

    [tex]\Delta G^0_{rx n }= [-2364-1422] – [-911+0] \ kJ/mol[/tex]

    [tex]\Delta G^0_{rx n }= -3786 +911 \ kJ/mol[/tex]

    [tex]\Delta G^0_{rx n }= -2875 \ kJ/mol[/tex]

    [tex]\Delta G^0_{rx n }= -2875000 \ J/mol[/tex]

    Now, the no of ATP molecules generated = [tex]\dfrac{\Delta G^0 \text{of metabolism for glucose}}{\Delta G^0 \text{of hydrolysis for ATP}}[/tex]

    = (-2875000 J/mol ) / -30500 J/mol

    = 94.26

    ≅ 94 ATP molecules generated

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