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
(a) The change in Gibbs free energy \((\Delta G)\) for the transfer of \(1 mol\) of potassium-ion from blood plasma to muscle cell fluid at body temperature \(37^{\circ} C\) is approximately \(-10,300 J/mol\). (b) The minimum amount of work that must be used to transfer this potassium-ion is approximately \(10,300 J/mol\).

Step by step solution

01

Identify the known and unknown values

We are given: - Potassium-ion concentration in blood plasma: \(5.0 \times 10^{-3} M\) - Potassium-ion concentration in muscle cell fluid: \(0.15 M\) - Temperature: \(37^\circ C\) We need to find: - Change in Gibbs free energy \((\Delta G)\) - Minimum amount of work needed to transfer \(1mol\) potassium-ion
02

Convert the temperature to Kelvin

To do calculations involving temperature, we must use Kelvin as the unit. Therefore, we need to convert the given Celsius temperature to Kelvin. The conversion is as follows: \(T(K) = T(^{\circ} C) + 273.15\) \(T(K) = 37 + 273.15\) \(T(K) = 310.15 K\)
03

Calculate the Gibbs free energy change using the Nernst equation

The Nernst equation is used to determine the change in Gibbs free energy \((\Delta G)\) when transferring ions between environments with different ion concentrations. It can be expressed as: \(\Delta G = -RT\ln\frac{[K^{+}]_{cell}}{[K^{+}]_{plasma}}\) where, - \(\Delta G\) = change in Gibbs free energy - \(R\) = universal gas constant (\(8.314 J/(mol\cdot K)\)) - \(T\) = temperature in Kelvin - \([K^{+}]_{cell}\) = potassium-ion concentration in muscle cell fluid - \([K^{+}]_{plasma}\) = potassium-ion concentration in blood plasma Now, we will plug in the given values and calculate the Gibbs free energy change: \(\Delta G = -(8.314 J/(mol\cdot K)) \cdot (310.15 K) \cdot \ln \frac{0.15}{5.0\times10^{-3}}\) \(\Delta G \approx -10,300 J/mol\)
04

Calculate the minimum amount of work needed to transfer 1 mol potassium-ion

Since the change in Gibbs free energy \((\Delta G)\) represents the maximum energy that can be converted into useful work, the minimum work that needs to be done is equal to the negative of the Gibbs free energy change. Therefore, Minimum work = \(- \Delta G\) Minimum work = \(10,300 J/mol\)
05

Summary of Results

(a) The change in Gibbs free energy \((\Delta G)\) for the transfer of \(1 mol\) of potassium-ion from blood plasma to muscle cell fluid at body temperature \(37^{\circ} C\) is approximately \(-10,300 J/mol\). (b) The minimum amount of work that must be used to transfer this potassium-ion is approximately \(10,300 J/mol\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nernst Equation
Understanding the Nernst equation is crucial for anyone studying biochemistry or physiology, especially when considering the movement of ions across a membrane. This equation provides a quantitative means to predict the equilibrium potential for a particular ion based on its concentration gradient across the membrane.

The Nernst equation is expressed as \(E = \frac{RT}{nF}\ln\frac{[\text{ion}]_{inside}}{[\text{ion}]_{outside}}\), where E is the equilibrium potential, R is the gas constant, T is the temperature in Kelvin, n is the charge of the ion, F is the Faraday constant, and the ion concentrations inside and outside the membrane are denoted by [ion]_{inside} and [ion]_{outside}, respectively.

Real-Life Application

In the given exercise, the Nernst equation has been adapted to calculate the Gibbs free energy change (\(\Delta G\)) during the transfer of potassium ions across a cell membrane. By plugging in the given potassium-ion concentrations and temperature into the Nernst equation, students can understand how a concentration gradient drives the energy cost for moving ions, which is essential in biological systems like nerve cells and muscle contraction.
Potassium-Ion Concentration
The concentration of potassium ions (\(K^+\)) is fundamentally important in cellular processes. For example, in neurons, the difference in \(K^+\) concentration inside and outside the cell creates an electrochemical gradient that facilitates nerve transmission.

In muscle cells, as in our exercise, potassium plays a significant role in maintaining cell volume and the resting membrane potential. This is why understanding the concentration gradients of potassium between different biological compartments, like blood plasma and muscle-cell fluid, is so essential. The concentrations given in the exercise—\(5.0 \times 10^{-3} M\) in plasma and \(0.15 M\) in muscle cells—highlight a sizeable gradient that translates to potential energy for physiological functions.

Practical Implications

By calculating the Gibbs free energy change for transferring \(K^+\) ions from an area of low concentration to high concentration, students grasp the thermodynamics behind cellular operations, like maintaining electrical neutrality and regulating ions to support cell functions.
Thermodynamics
Thermodynamics, particularly the concept of Gibbs free energy, plays a central role in understanding chemical and biochemical reactions. Gibbs free energy (\(\Delta G\)) is a thermodynamic potential that measures the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.

In biological systems, \(\Delta G\) denotes whether a reaction or process can occur spontaneously. A negative value of \(\Delta G\), as calculated in our exercise for potassium ion transfer, implies that the process is exergonic and spontaneous under the given conditions. Conversely, a positive value suggests that additional work or energy input is needed.

The Essence of Gibbs Free Energy in Biology

When considering cellular functions, such as ion transport against a concentration gradient, Gibbs free energy quantifies the energy required for these non-spontaneous processes. It establishes a connection between the seemingly abstract concepts of thermodynamics and the tangible physiological events in living organisms.

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Most popular questions from this chapter

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g)\) (d) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

A certain reaction has \(\Delta H^{\circ}=+23.7 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+52.4 \mathrm{~J} / \mathrm{K} .\) (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K} .(\mathbf{d})\) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Propanol \(\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)\) melts at \(-126.5^{\circ} \mathrm{C}\) and boils at \(97.4{ }^{\circ} \mathrm{C}\). Draw a qualitative sketch of how the entropy changes as propanol vapor at \(150^{\circ} \mathrm{C}\) and 1 atm is cooled to solid propanol at \(-150^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

For the majority of the compounds listed in Appendix \(\mathrm{C},\) the value of \(\Delta G_{f}^{\circ}\) is more positive (or less negative) than the value of \(\Delta H_{f}^{\circ} .\) (a) Explain this observation, using \(\mathrm{NH}_{3}(g), \mathrm{CCl}_{4}(l)\), and \(\mathrm{KNO}_{3}(s)\) as examples. (b) An exception to this observation is \(\mathrm{CO}(g)\). Explain the trend in the \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) values for this molecule.

The \(K_{b}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{b} .\) (b) By using the value of \(K_{b},\) calculate \(\Delta G^{\circ}\) for the equilibrium in part (a). (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=6.7 \times 10^{-9} \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]=2.4 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.098 \mathrm{M} ?\)

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