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Chapter 19: Problem 102

The potassium-ion concentration in blood plasma is about \(5.0 \times 10^{-3} M,\) whereas the concentration in muscle-cell fluid is much greater \((0.15 \mathrm{M}) .\) The plasma and intracellular fluid are separated by the cell membrane, which we assume is permeable only to \(\mathrm{K}^{+} .\) (a) What is \(\Delta G\) for the transfer of 1 \(\mathrm{mol}\) of \(\mathrm{K}^{+}\) from blood plasma to the cellular fluid at body temperature \(37^{\circ} \mathrm{C} ?\) (b) What is the minimum amount of work that must be used to transfer this \(\mathrm{K}^{+} ?\)

Short Answer

Expert verified
The minimum work required to transfer 1 mol of K+ ions from blood plasma to muscle-cell fluid at body temperature \(37^{\circ}C\) is approximately 12000 Joules per mole.

Step by step solution

01

Write down the given information.

We have the following information: - Potassium-ion concentration in blood plasma: \(5.0 \times 10^{-3} M\) - Potassium-ion concentration in muscle-cell fluid: \(0.15 M\) - Body temperature: \(37°C\) We need to convert the temperature from Celsius to Kelvin by adding 273.15 to the Celsius value.
02

Convert the temperature to Kelvin.

To convert the temperature to Kelvin, we add 273.15 to 37: \[T = 37 + 273.15 = 310.15 K\]
03

Calculate the change in concentration.

The change in potassium-ion concentration can be calculated using the initial concentration in blood plasma and the final concentration in muscle-cell fluid: \[\Delta concentration = 0.15 M - 5.0 \times 10^{-3} M = 0.145 M\]
04

Calculate the Gibbs free energy change (ΔG).

We will use the following equation for calculating ΔG: \[\Delta G = RT \ln \frac{C_{2}}{C_{1}}\] where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, \(C_{1}\) is the initial concentration, and \(C_{2}\) is the final concentration. Substituting the values and solving for ΔG, we get: \[\Delta G = (8.314 \frac{J}{mol \cdot K})(310.15 K) \ln \frac{0.15 M}{5.0 \times 10^{-3} M}\] \[\Delta G \approx 12000 J/mol\] So, the Gibbs free energy change (ΔG) for transferring one mole of K+ from blood plasma to the muscle-cell fluid is approximately 12000 J/mol.
05

Calculate the minimum work required for transferring the ions.

The work to transfer one mole of K+ ions can be calculated using the ΔG value we found. As the minimum work required is equal to ΔG, we already have the answer: Minimum work required = ΔG Minimum work required ≈ 12000 J/mol The minimum work required to transfer 1 mol of K+ ions from blood plasma to muscle-cell fluid is approximately 12000 Joules per mole.

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false. If it is false, correct it. (a) The feasibility of manufacturing \(\mathrm{NH}_{3}\) from \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) depends entirely on the value of \(\Delta H\) for the process \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) The reaction of \(\mathrm{Na}(s)\) with \(\mathrm{Cl}_{2}(g)\) to form \(\mathrm{NaCl}(s)\) is a spontaneous process.(c) A spontaneous process can in principle be conducted reversibly. (d) Spontaneous processes in general require that work be done to force them to proceed. (e) Spontaneous processes are those that are exothermic and that lead to a higher degree of order in the system.

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