GATE 2023 BT – Question 50
E. coli cultivated at 298 K uptakes an uncharged compound A by passive diffusion. Intracellular and extracellular concentrations are 0.001 M and 0.1 M. With $R=1.9872\ \mathrm{cal\ mol^{-1}K^{-1}}$, the free-energy change in kcal/mol (two decimal places) is ___________.
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Correct answer: -2.74 to -2.72
Explanation
For passive diffusion of an uncharged compound, the free-energy change when it moves from outside to inside is
$$\Delta G=RT\ln\frac{C_{in}}{C_{out}} .$$
**Numbers.** $R=1.9872$ cal mol⁻¹ K⁻¹, $T=298$ K, $C_{in}/C_{out}=0.001/0.1=0.01$:
$$RT=1.9872\times298=592.2\text{ cal/mol}$$
$$\Delta G=592.2\times\ln(0.01)=592.2\times(-4.605)=-2727\text{ cal/mol}.$$
$$\Delta G=\mathbf{-2.73\ kcal/mol}.$$
The sign is negative, so the movement of A down its concentration gradient into the cell is spontaneous.