Question

The typical potassium ion concentration in the fluid outside a cell is$\mathbf{0}\mathbf{.}\mathbf{0050}\mathbf{\text{M}}$, whereas that inside a muscle cell is$\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{\text{M}}$.

(a) What is the spontaneous direction of motion of ions the through the cell wall?

(b) In active transport, cells use free energy stored in ATP (see Problem 34) to move ions in the direction opposite their spontaneous direction of flow. Calculate the cost in free energy to move$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{mol}}\mathbf{}{\mathbf{\text{K}}}^{\mathbf{+}}$through cell wall by active transport. Assume no change in${\mathbf{\text{K}}}^{\mathbf{+}}$concentrations during this process

Short answer

The potassium ions move from muscle cells to outside of the cell and the energy for $1\text{molK}$is $2.9\times {10}^{24}\text{kJ}$

Step-by-step solution

Given information

a) The potassium ions move from muscle cells to the outside of the cell

potassium ion outside of the cell)$=0.0050\text{M}$

potassium ion inside of the cell)$=0.15\text{M}$

It is known that ions move in the direction from higher concentration tolower concentration

Concept of spontaneous reaction

A spontaneous process is irreversible and can only be stopped by external forces. The degree of randomness in each system is defined by its entropy. Total entropy change is the most important parameter in determining the spontaneity of any process.

Calculate the number of ATP required

For the reaction to occur the number of ATP molecules is required for transport as:

$$\begin{gathered} 1\text{molK}=1\text{molK}\times \frac{6.022\times {10}^{23}\text{ions}}{1\text{molK}}\times \frac{1\text{ATP molecule}}{8\text{potassium ions}} \\ =8.278\times {10}^{22}moleculeofATP \end{gathered}$$

Calculate the energy needed for transport

The energy needed to transport$1mol$of potassium ion:

Energy for $1\text{molK}=1\text{mol}$potassium is$34.5\text{kJ}$

$\times \frac{8.278\times {10}^{22}\text{molecules ATP}}{1\text{mol potassium}}\times \frac{34.5\text{kJ}}{1\text{molecules ATP}}=2.9\times {10}^{24}\text{kJ}$