Question

a) The metabolism of glucose can be written as C6H12O6(s) + 602() - → 6CO2(8) + 6H20m Calculate the value of AHⓇ and ASo for the reaction b) Calculate the value of AGº for the reaction at 298 K. c) Cells use energy derived from the metabolism of glucose to synthesize ATP from ADP and phosphate ion(written as P). ADP + P → ATP That reaction has AGⓇ = +30.5 kJ/mol. How many moles of ATP could theoretically be made from the metabolism of 1 mole of glucose?

Answer 1

Answer:

(a) -2802.74 kJ,  261.81 J·K⁻¹; (b) -2880.8 kJ; (c) 90.5 mol

Explanation:

a) ΔH° and ΔS°

                         C₆H₁₂O₆(s) + 6O₂(g) ⟶ 6CO₂(g) + 6H₂O(ℓ)

ΔHf°/kJ·mol⁻¹:     -1273.3                          -393.51   -285.830

Sf°/J·K⁻¹mol⁻¹:       209.19      205.0           213.6       69.9

ΔHr° = 6(-393.51) + 6(-285.530) - 1273.3 = -2361.06 - 1714.98 + 1273.3 = -2802.74 kJ

ΔSr° = 6(213.6) + 6(69.9) - 209.19 - 6(205.0) = 1281.6 + 419.4 - 209.19 - 1230.0 = 261.81 J/K

(b) ΔG°

ΔG° = ΔH° - TΔS°

$\begin{array}{rcl}\Delta G^{\circ} & = & -2 802 740 - 298.15 \times 261.81\\& = & -2 802 740 - 78 059\\& = & \text{-2 880 800 J}\\& = & \textbf{-2880.8 kJ}}\\\end{array}$

(c) No. of ATP molecules

ADP + P ⟶ ATP; ΔG = 30.5 kJ·mol⁻¹

$\text{Moles of ATP} = \text{2880.8 kJ} \times \dfrac{\text{1 mol ATP}}{\text{30.5 kJ}} = \textbf{94.5 mol ATP}$