Question

Free energy change for transport of an uncharged solute across a membrane against a \(1 \times 10^3\) fold concentration gradient at 25°C is ............... kJ mol\(^{-1}\). (rounded off to 2 decimals) \[ R = 8.315 \, \text{J mol}^{-1} \, \text{K}^{-1} \]

Hint:

To calculate the free energy change for transport, use the formula \(\Delta G = - RT \ln (C_2/C_1)\) where \( R \) is the gas constant, \( T \) is temperature in Kelvin, and \( C_2/C_1 \) is the fold concentration gradient.

Answer 1

Step 1: Formula for Free Energy Change The formula for the free energy change (\(\Delta G\)) for the transport of an uncharged solute across a membrane is given by: \[ \Delta G = - RT \ln \left( \frac{C_2}{C_1} \right) \] Where:
- \( R \) = Gas constant = 8.315 J mol\(^{-1}\) K\(^{-1}\)
- \( T \) = Temperature in Kelvin = 25°C = 298 K
- \( C_2/C_1 \) = Fold concentration gradient = \(1 \times 10^3\)
Step 2: Substituting the given values
Now, substituting the known values into the equation: \[ \Delta G = - (8.315) \times (298) \times \ln (1 \times 10^3) \] Step 3: Calculation \[ \ln(1 \times 10^3) = \ln(1000) = 6.907 \] \[ \Delta G = - (8.315 \times 298 \times 6.907) \, \text{J mol}^{-1} \] \[ \Delta G = - 17160.23 \, \text{J mol}^{-1} = -17.16 \, \text{kJ mol}^{-1} \] Final Answer: \[ \boxed{-17.16 \, \text{kJ mol}^{-1}} \]