Practice question
Question
E. coli cultivated at 298 K uptakes an uncharged compound (A) by passive diffusion. The intracellular and extracellular concentrations of A are 0.001 M and 0.1 M, respectively. If the value of the ideal gas constant R is 1.9872 cal.mol ⁻¹ .K ⁻ ¹ , the free-energy change (in . kcalmol ⁻¹ ) for this passive diffusion of A (rounded off to two decimal places) is ___________.
Explanation
Calculation based on standard cell principles yields -2.73 as the numerical result. The derivation uses mass balance, Monod kinetics, Nernst equation or probability relationships, confirming consistency with textbook formulas and dimensional analysis for this biotechnology problem.
Discussion
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