Subtopic: Fundamentals of Downstream processing (chromatography, filtration)
Question Statement:
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
Q.1 M, respectively. If the value of the ideal gas constant R is
111.9872 . .cal mol K--
, the free-energy change (in
1.kcal mol - ) for this passive
diffusion of A (rounded off to two decimal places) is ___________.
Official Answer Key:
-2.74 to -2.71
Step-by-Step Solution & Explanation
### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2023 Q.50)**
**1. Governing Concept & Definition**:
- **Topic**: Bioenergetics & Membrane Transport Thermodynamics
- For passive uncharged molecular diffusion from extracellular compartment to intracellular compartment:
$\Delta G = R T \ln\left(\frac{C_{\text{in}}}{C_{\text{out}}}\right)$
**2. Detailed Step-by-Step Calculation**:
- **Given**:
- Temperature $T = 298\text{ K}$
- Gas constant $R = 1.9872\text{ cal mol}^{-1}\text{ K}^{-1} = 1.9872 \times 10^{-3}\text{ kcal mol}^{-1}\text{ K}^{-1}$
- Intracellular concentration $C_{\text{in}} = 0.001\text{ M}$
- Extracellular concentration $C_{\text{out}} = 0.1\text{ M}$
- **Concentration Ratio**:
$\frac{C_{\text{in}}}{C_{\text{out}}} = \frac{0.001}{0.1} = 0.01 = 10^{-2}$
- **Natural Logarithm**:
$\ln(0.01) = \ln(10^{-2}) = -2 \ln(10) \approx -2 \times 2.302585 \approx -4.60517$
- **Free Energy Calculation**:
$\Delta G = 1.9872 \times 298 \times (-4.60517) \approx -2727.1\text{ cal mol}^{-1}$
$\Delta G = -2.727\text{ kcal mol}^{-1} \approx -2.73\text{ kcal mol}^{-1}$
**3. Final Answer**:
- **Calculated Value**: **-2.73 kcal/mol**
- **Official GATE Accepted Range**: **-2.74 to -2.71**
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