a.) The distribution of sodium ions across a certain biological membrane is 11.0

Question

a.) The distribution of sodium ions across a certain biological membrane is 11.0 mmol/L inside the cell and 146 mmol/L outside the cell. At equilibrium the concentrations would be equal (and the cell would be dead!!), so the cell must expend metabolic energy to pump ions across the membrane and maintain a concentration gradient. What is the reaction Gibbs free energy in kJ/mol for a mole of ions going from the inside to the outside under these conditions? Na+(in) ----> Na+(out)  The temperature is 294 K.

b.) If the equilibrium constant for the reaction:

2NO2 --> N2O4

at a certain temperature is K = 6.51 , what would be the equilibrium constant of the related reaction:

2N2O4 --> 4NO2

Explanation / Answer

a) ?G can be calculated by using the expression: ?G= -RT lnQc

R = gas constant= 8.314 J/(K.mol), T = 294 K

Qc is the reaction quotient,it gives the value at a state other than state of equilibrium.

The given reaction is : Na+(in)--------------->Na+(out)

So, Qc = [Na+(out)]/[Na+(in)]

the concentration should be in terms of mol/L

So, [Na+(in)] = 11. 0 mmol/L = 0.011 mol/L

[Na+(out)] = 146 mmol/L = 0.146 mol/L

So, Qc = 0.146/0.011 = 13.27

Now, ?G = -8.314 J/(K.mol) × 294 K × ln(13.27)

= - 6319.79 J/mol = -6.320 kJ/mol

b). The given reaction is: 2NO2 ------------->N2O4

for this reaction K = [N2O4]/([NO2)^2 =6.52

The second given reaction is: 2N2O4 ---------------------> 4NO2

For this reaction

K' = ([NO2])^4/([N2O4])^2 = ([NO2])^2/[N2O4])^2

Since the expression ([NO2])^2/[N2O4]) is just inverse of expression in reaction first , so it will be equal to 1/6.51

Now substituting this value

K' = (1/6.51)^2 = 0.0236

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.