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 \(R=1.9872\ \mathrm{cal\ mol^{-1}\ K^{-1}}\), the free-energy change (in \(\mathrm{kcal\ mol^{-1}}\)) for this passive diffusion of A (rounded off to two decimal places) is ________________.

Hint:

Use \(\Delta G=RT\ln\left(\frac{C_{\text{final}}}{C_{\text{initial}}}\right)\) for neutral solutes; a downhill (spontaneous) influx gives a negative \(\Delta G\).

Answer 1

Correct Answer: -2.74

Step 1: For an uncharged solute, \(\Delta G = RT\ln\!\left(\dfrac{C_{\text{in}}}{C_{\text{out}}}\right)\).
Step 2: Insert values: \[ \Delta G=(1.9872)(298)\ln\!\left(\frac{0.001}{0.1}\right)\ \mathrm{cal\ mol^{-1}} =592.1856\times \ln(0.01). \] Step 3: Since \(\ln(0.01)=-4.60517\), \[ \Delta G=-2727.12\ \mathrm{cal\ mol^{-1}}=-2.727\ \mathrm{kcal\ mol^{-1}} \approx \boxed{-2.73\ \mathrm{kcal\ mol^{-1}}}. \]