GATE 2023 BT Q.50: solution and answer

GATE 2023 · BTQ.50 · 2 marks · NAT

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 kcal.mol^-1) for this passive diffusion of A (rounded off to two decimal places) is ______.

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Answer and solution

ANSWER

The free-energy change for passive diffusion of A into the cell is about $- 2.73 kcal mol^{- 1}$ .

Official answer: -2.74 to -2.71Source: Official answer key, GATE 2023 BT (IIT Kanpur)

Passive diffusion of an uncharged solute is movement down its concentration gradient, so the free-energy change has only the concentration term: $Δ G = R T ln \frac{C _{in}}{C _{o u t}}$ . There is no membrane-potential term because A is uncharged.
Concentration ratio: $\frac{C _{in}}{C _{o u t}} = \frac{0.001}{0.1} = 0.01$ , so $ln 0.01 = - 4.6052$ .
$R T = 1.9872 \times 298 = 592.19 cal mol^{- 1} = 0.59219 kcal mol^{- 1}$ .
$Δ G = 0.59219 \times (- 4.6052) \approx - 2.727 kcal mol^{- 1}$ .
The sign is negative, so the process is spontaneous. This fits movement from the high (outside) concentration to the low (inside) concentration.
Rounded to two decimal places: $Δ G \approx - 2.73 kcal mol^{- 1}$ .

KEY IDEA

For an uncharged solute, the free-energy change of transport is $Δ G = R T ln (C_{in} / C_{o u t})$ . Take the ratio as destination over source and convert the units of R to kcal before multiplying.

WATCH OUT FOR

Using $C_{o u t} / C_{in}$ gives $+ 2.73$ , the wrong sign. Forgetting to convert cal to kcal gives $- 2727$ . Using $lo g_{10}$ in place of $ln$ gives a value about 2.3 times too small.

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GATE 2023 BT Q.50: solution and answer

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