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The free energy generated by ion movement in the cell is used by the cell to do

ID: 1073871 • Letter: T

Question

The free energy generated by ion movement in the cell is used by the cell to do work. Compute the minimum number of millimoles of sodium that would be needed to pump one millimole of calcium ions if the following concentrations were present in the cell. Assume the temperature to be 37 degree C and recall that R=0.008314 kJ/mol-K. Na^+ (aq, 36m M) rightarrow Na^+ (aq, 5.25 mM) Ca^2+ (aq, 39 mM) rightarrow Ca^z+ (aq, 1.50 times 10^2 mM) The equation for the conversion of graphite to carbon dioxide is C(s) + O_2 (g) CO_2(g) and the change of Gibbs free energy of the reaction can be found by summing the Gibbs energies of the products less the Gibbs energies of the reactants. Standard Gibbs free energies of formation can be found here. To find the expression for the net reaction, add the two chemical equations together. To find the net change in free energy, add the free energy of each reaction together. How is the equilibrium constant related to the net change in Gibbs free energy?

Explanation / Answer

Equilibrium constant for the reaction, Na+--à Na+

Is K= [Na+] /[Na+] = 5.25/36 =0.1458

Gibbs free energy change =-RTlnK= -8.314*(37+273)*ln(0.1458) =4962.72 joules/mole

Gibbs free energy per nmiilmole = n* moles is 4962.72*n Jou,les

For the reaction Ca+2 -----àCa+2

K= 1.5*102/39= 3.85

deltaG=-8.314*310*ln(3.85)= 3472 Joules/mole

Millimoles of Na = n*4962 = 3472*1millimole

There fore,n = 3472/4962 =0.70 millimoles

2. since standard gibbs free energy change of C and O2 =0 while that of CO2= -394.38 Kj/mole

standard gibbs free energy change= gibbs free energy change of products- gibbs free energy change of reactants.

Standard gibbs free energy change, deltaGO= -394.38 Kj/mole

deltaGo= -RTlnK

lnK= -deltaGO/RT= 394.38*1000/(310*8.314), K= 2.85*1066

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