Question

Question #57ef8

Answer 1

`W=8.42"kJ"`

Explanation:

The situation here is like a concentration cell. The natural tendency is that potassium ions of a high concentration will move across the cell membrane to the region of low concentration.

To get the potassium ions to move in the opposite direction will require energy to move them against the concentration gradient.

The potential difference `E` across the membrane is given by the Nernst Equation:

`E_(cell)=E_(cell)^(@)-(RT)/(zF)lnQ`

Because we are dealing with a `"K"^(+)"/""K"` system on both sides of the membrane `E_(cell)^(@)` is zero.

The Nernst Equation at 298K therefore simplifies down to:

`E=(0.05916)/(z)log([[K_("blood")^(+)]]/[[K_("muscle")^(+)]])`

`:.E=0.05916log((5xx10^(-3))/(0.15))`

`E=0.05916xx(-1.477)=-0.0873"V"`

This is the potential difference, in volts, across the membrane.

If I Coulomb of electric charge is moved across a potential difference of 1 Volt then 1 Joule of work is done.

In this case we are moving 1 mole of `K^(+)` ions.

The charge on 1 mole of `K^+` ions is `9.65xx10^(4)"C"`. This is known as the Faraday Constant `F`.

So to move 1 mole of `K^(+)` ions through a potential difference of `E` volts will require `FE` Joules of work.

So the work done `W` is given by:

`W=FE=9.65xx10^(4)xx(-0.0873)=-0.842xx10^(4)"J"`

`W=-8.42"kJ"`

The minus sign means work has to be done to move the ions across the membrane.

You may be familiar with the expression for the Gibbs Free energy change for a cell reaction:

`DeltaG^(@)=-nFE_(cell)^(@)`

This is the same thing. `-DeltaG^(@)` is the maximum amount of work you can get from a cell when it is operating under standard conditions.