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As per your syllabus, given ΔG° = 13.8 kJ/mol at 298 K for glucose phosphorylation, the value of Kc is closest to which?
A5.6 × 10²
B3.8 × 10⁻³
C1.0
D3.8 × 10³
Answer & Solution
Correct answer: B. 3.8 × 10⁻³
syllabus Problem 6.10: ln Kc = −13800 / (8.314 × 298) = −5.57; Kc = e^−5.57 = 3.81 × 10⁻³.
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