/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 Many biochemical reactions that ... [FREE SOLUTION] | 91影视

91影视

Many biochemical reactions that occur in cells require relatively high concentrations of potassium ion \(\left(\mathrm{K}^{+}\right) .\) The concentration of \(\mathrm{K}^{+}\) in muscle cells is about \(0.15 \mathrm{M}\). The concentration of \(\mathrm{K}^{+}\) in blood plasma is about \(0.0050 \mathrm{M}\). The high internal concentration in cells is maintained by pumping \(\mathrm{K}^{+}\) from the plasma. How much work must be done to transport 1.0 mole of \(\mathrm{K}^{+}\) from the blood to the inside of a muscle cell at \(37^{\circ} \mathrm{C}\), normal body temperature? When 1.0 mole of \(\mathrm{K}^{+}\) is transferred from blood to the cells, do any other ions have to be transported? Why or why not?

Short Answer

Expert verified
The work required to transport 1.0 mole of K鈦 ions from blood plasma into the muscle cell is approximately -6860 J. Other ions, specifically sodium ions (Na鈦), need to be transported along with K鈦 ions to maintain electrical neutrality within the cell. This occurs through the sodium-potassium pump (Na鈦/K鈦-ATPase), an active transport system that helps regulate the balance of ions and maintain overall electrical neutrality in cells.

Step by step solution

01

Find the Gibbs free energy change

The Gibbs free energy change can be found using the formula: 螖G = -nRT ln(Q) where 螖G is the change in Gibbs free energy, n is the number of moles of K鈦, R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and Q is the reaction quotient. First, we need to convert the given temperature in Celsius to Kelvin: T(K) = 37掳C + 273.15 = 310.15 K
02

Calculate the reaction quotient (Q)

The reaction quotient (Q) can be calculated as: Q = [K鈦篯_inside / [K鈦篯_outside where [K鈦篯_inside is the concentration of K鈦 inside the muscle cell, and [K鈦篯_outside is the concentration of K鈦 in blood plasma. Given concentrations are: [K鈦篯_inside = 0.15 M [K鈦篯_outside = 0.0050 M We can now plug the values and find Q: Q = 0.15 / 0.0050 = 30
03

Calculate the work required

Now, we can use the given values to calculate the work required: 螖G = -nRT ln(Q) 螖G = -(1.0 mol)(8.314 J/mol K)(310.15 K) ln(30) 螖G 鈮 -6860 J The work required to transport 1.0 mole of K鈦 ions from blood plasma into the muscle cell is approximately -6860 J.
04

Discuss if any other ions need to be transported

Yes, other ions need to be transported along with K鈦 ions. The reason is that cells need to maintain electrical neutrality. When 1.0 mole of positively charged K鈦 ions is transported into the muscle cell, the same amount of a charge must be moved out of the cell to maintain the overall charge balance in the cell. Typically, sodium ions (Na鈦) are moved out of the cell in exchange for the K鈦 ions being pumped in. This process is commonly known as the sodium-potassium pump (Na鈦/K鈦-ATPase). This active transport system allows cells to maintain a proper balance of ions and overall electrical neutrality.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Potassium Ion Concentration
The potassium ion (K鈦) concentration within cells plays a critical role in maintaining cellular functions such as electrical signaling in neurons and muscle contraction. In muscle cells, a high concentration of K鈦 is essential, with typical values around 0.15 M compared to just 0.0050 M in blood plasma. This gradient is crucial for various physiological processes and is actively regulated by cellular mechanisms.

To understand the importance of maintaining this ion concentration gradient, let鈥檚 consider the cellular environment as a busy marketplace. In the market, certain items (or ions) need to be in greater abundance on one side to ensure the town (or cell) operates smoothly. The K鈦 ions, much like a popular product, are in higher demand within the cell than outside in the blood plasma, akin to a storage facility. This difference in concentration is not just a happenstance but a carefully regulated balance that allows cells to function optimally.
Sodium-Potassium Pump
The sodium-potassium pump, scientifically referred to as Na鈦/K鈦-ATPase, is a vital cellular mechanism that uses energy to move ions across the cell membrane against their concentration gradients. For every three sodium ions (Na鈦) pumped out, two potassium ions (K鈦) are pumped into the cell. This mechanism functions much like a gate that controls the entry and exit of these ions, and it's fueled by ATP, the cell鈥檚 energy currency.

Picture a bustling cafe where coffee (Na鈦) needs to leave the shop for every two cups of tea (K鈦) that enter. This exchange is not equal, reflecting the actual 3:2 ratio that the sodium-potassium pump works with to maintain cellular homeostasis. This pump is fundamental because it helps in setting the stage for electrical signal passage, nutrient transport, and overall cellular health.
Gibbs Free Energy Equation
The Gibbs free energy equation, given by 螖G = -nRT ln(Q), allows us to calculate the amount of energy change during a process. In this equation, 螖G represents the Gibbs free energy change, n is the number of moles, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient, which represents the ratio of the concentrations of products to reactants.

Imagine you are assessing whether a business venture (a biochemical reaction) is profitable. The Gibbs free energy is like the profit (or loss) from this venture. If 螖G is negative, the process (or reaction) can occur spontaneously, just as a business makes a profit without any extra investment. In the given exercise, the negative 螖G signifies that the work required to move K鈦 from the blood to the inside of a cell is energy favorable under standard conditions.
Cellular Ion Transport
Cellular ion transport involves moving ions across the cell membrane, requiring specialized proteins such as ion channels and pumps. This transport is essential for maintaining the electrolyte and fluid balance within cells and throughout the body.

Consider a diligent mailroom in a large office. The mailroom workers (ion channels and pumps) are responsible for sorting and delivering packages (ions) to various departments (cellular compartments). These workers need to recognize and move the packages to the right places, despite a constant influx of items coming in and going out. In biology, maintaining the proper distribution of ions like K鈦 and Na鈦 is paramount for physiological functions such as muscle contraction, nerve impulse generation, and osmoregulation. The transport process, especially when against the concentration gradient, consumes energy but ensures the cell has the right ingredients for optimal functionality.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?

The standard free energies of formation and the standard enthalpies of formation at \(298 \mathrm{K}\) for diffuoroacetylene \(\left(\mathrm{C}_{2} \mathrm{F}_{2}\right)\) and hexafluorobenzene \(\left(\mathrm{C}_{6} \mathrm{F}_{6}\right)\) are $$\begin{array}{ccc} & \Delta G_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & \Delta H_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) \\ \hline \mathrm{C}_{2} \mathrm{F}_{2}(g) & 191.2 & 241.3 \\\ \mathrm{C}_{6} \mathrm{F}_{6}(g) & 78.2 & 132.8 \end{array}$$ For the following reaction: $$\mathrm{C}_{6} \mathrm{F}_{6}(g) \rightleftharpoons 3 \mathrm{C}_{2} \mathrm{F}_{2}(g)$$ a. calculate \(\Delta S^{\circ}\) at \(298 \mathrm{K}\). b. calculate \(K\) at 298 K. c. estimate \(K\) at \(3000 .\) K, assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does O_2. Consider the following reactions and approximate standard free energy changes: $$\begin{array}{cl} \mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{kJ} \end{array}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

Consider the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$$ a. Use \(\Delta G_{\mathrm{f}}^{\circ}\) values in Appendix 4 to calculate \(\Delta G^{\circ}\) for this reaction. b. Is this reaction spontaneous under standard conditions at \(298 \mathrm{K} ?\) c. The value of \(\Delta H^{\circ}\) for this reaction is \(100 .\) kJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Gas \(\mathrm{A}_{2}\) reacts with gas \(\mathrm{B}_{2}\) to form gas \(\mathrm{AB}\) at a constant temperature. The bond energy of AB is much greater than that of either reactant. What can be said about the sign of \(\Delta H ? \Delta S_{\text {surr }}\) ? \(\Delta S ?\) Explain how potential energy changes for this process. Explain how random kinetic energy changes during the process.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.