Fundamentals of Biochemistry: Life at the Molecular Level 5th Edition

Published by Wiley
ISBN 10: 1118918401
ISBN 13: 978-1-11891-840-1

Chapter 1 - Introduction to the Chemistry of Life - Exercises - Page 21: 14


$K_{eq}= 4.6×10^3$

Work Step by Step

Consider the formula for calculating the equilibrium constant : $K_{eq}=\frac{([C]_eq^c [B]_eq^a)}{([A]_eq^a [B]_eq^b )}=e^{(-∆°G')/RT}$ We know the following values : $∆G°^{'}=-20.9 kJ∙mol^{-1}$ $R= 8.314 J∙ mol^{-1}∙K^{-1}$ $T=25℃ $ Before we can proceed with the calculation, we need to first make sure that the units match up. $T=25℃$ needs to be converted to Kelvin $→$ $T (K)= T(℃)+273$ $→$ $ T(K)=25+273=298$ $∆G°^{'}= -20.9 kJ∙mol^{-1}$ needs to be converted to $kJ∙mol^{-1}$ $\frac{-20.9 kJ}{mol} ∙\frac{1000 J}{1 kJ}=-20.9 J∙ mol^{-1} $ Now $ T= -248 K$ $R= 0.008314 kJ∙ mol^{-1}∙K^{-1}$ $∆G°^{'}=-20.9 kJ∙mol^{-1}$ and $K_{eq}=e^{(-∆°G)/RT}$ $=e^{\frac{-(-20.9 ×10^3 J∙mol^{-1})}{8.314 J∙ mol^{-1}∙K^{-1} ∙ (298 K)}}$ $=e^{8.4357} = 4.6 \times 10^{3}$ Thus, $K_{eq}=4.6 \times 10^{3}$
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