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
