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01/10/2021 12:55 pm
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What is the equilibrium constant if ATP hydrolysis by water produces standard free energy of -50 kJ/mol under normal body conditions?
(a) 2.66 × 108
(b) 5.81 × 108
(c) 1.18 × 107
(d) 1.98 × 108
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01/10/2021 1:25 pm
Correct answer: (a) 2.66 × 108
Explanation:
ΔG = -RT ln Keq:
Normal body temperature = 37°C
⇒ -50 \(\frac{1}{2}\) = 8.314 \(\frac{J}{K\;mol}\) x 310 lnKeq
⇒ 19.39 = lnKeq
⇒ Keq = 2.66 × 108