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5 Activity: expressing the dependence of Gibbs free energy on concentration

# 5 Activity: expressing the dependence of Gibbs free energy on concentration

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7.5 Activity and Gibbs free energy

149

so that

dH = dU + p dV + V dp

By definition,

dU = dq + dw

If we take a reversible path and only pressure-volume work is possible,

dU = T dS − p dV

We now substitute this expression for dU into dH , and dH into dG to obtain

dG = V dp − S dT

(7.3)

Equation (7.3) is a very important equation: it tells us how the Gibbs free energy depends

on p and T . It is very general, and can be applied to any system.

For an isothermal process, dG = V dp. We can calculate (2) Gm directly from this

equation:

(2) Gm

=

p

p◦

dGm =

p

p◦

Vm dp = RT

p

p◦

dp

p

= RT ln ◦

p

p

In this derivation, we used the ideal gas law in the form pVm = RT to eliminate Vm .

We can now put the pieces together. The free energy of formation of an ideal gaseous

compound at pressure p is

p

(7.4)

f Gm =

f Gm + RT ln ◦

p

The fraction p/p◦ measures how far from the standard state the gas is. We call this quantity

the activity a. When a gas is in the standard state, a = 1 and the logarithm in Equation (7.4)

is zero. If a = 1, the gas is out of the standard state, and we have the logarithmic correction

to the free energy of formation shown above. By analogy, for any substance, we write

f Gm

=

f Gm

+ RT ln a

(7.5)

The activity is a dimensionless quantity that gives the extent of the deviation from the

standard state.

For ideally behaving substances, the activity is defined as follows:

Solids: a = 1.

Liquids: a = X, the mole fraction. The mole fraction is the number of moles of a

particular substance (in this case, a liquid, often the solvent) divided by the total

number of moles of all species in a system.

Gases: a = p/p◦ where p◦ is the standard pressure.

Solutes: a = c/c◦ where c◦ is the standard concentration for that species. In the chemists’

standard state, this would be 1 mol L−1 if our standard state is defined on the molarity scale, and 1 mol kg−1 for the molality-scale standard state. In the biochemists’

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150

Free Energy

standard state, the standard concentration would be 1 mol L−1 for most substances

(with species differing only in their protonation state treated as one substance), but

10−7 mol L−1 for hydrogen and hydroxide ions.

We have already seen why p/p◦ is a sensible definition for the activity of an ideal gas.

Solids don’t have a property corresponding to concentration; they are either there or they

are not, which is why the activity of a pure solid is just 1. We will see later why the other

choices are correct for dilute solutions.

For non-ideal substances, Equation (7.5) will remain valid, but we will have to modify

our definitions of the activities. This important topic will be discussed in Chapter 9.

To get the dependence of the free energy change during a reaction on the activities of

the participating species, we simply add the free energies of formation of the products and

subtract the free energies of formation of reactants in the usual way. Using the logarithm

manipulation rules, we get, for any reaction,

r Gm

=

r Gm

ai

⎜ i ∈ products ⎟

+ RT ln ⎜

aj ⎠

(7.6)

j ∈ reactants

(The symbol

represents a product, in the same way that

represents a sum of terms.

N

Thus, i=1 xi = x1 x2 x3 . . . xN .) The quantity following the logarithm is the reaction

quotient, usually denoted Q:

r Gm

=

r Gm

+ RT ln Q

(7.7)

Example 7.3 Calculating the free energy of reaction under non-standard conditions In Example 7.1 we computed the standard change in Gibbs free energy for the

reaction

C6 H12 O6(aq) + 6O2(aq) → 6CO2(aq) + 6H2 O(l)

Now suppose that we want to calculate the actual change in free energy when the concentrations of glucose, oxygen and carbon dioxide are, respectively, 0.005, 0.001 and

0.03 mol L−1 at 298 K. For this reaction,

Q=

(aCO2 )6 (XH2 O )6

(aC6 H12 O6 )(aO2 )6

The activities of the solutes are easy to calculate. For example, aCO2 = [CO2 ]/c◦ =

(0.03 mol L−1 )/(1 mol L−1 ) = 0.03. We need the mole fraction of water to compute Q. The

mole density of water at 298 K is 55.33 mol L−1 . If there are no other solutes than those listed

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7.5 Activity and Gibbs free energy

151

above, the total number of moles of all species in one liter of water is [H2 O] + [C6 H12 O6 ] +

[O2 ] + [CO2 ] = 55.33 + 0.005 + 0.001 + 0.03 mol L−1 = 55.37 mol L−1 . Thus,

55.33 mol L−1

= 0.999

55.37 mol L−1

r Gm =

r Gm + RT ln Q

XH2 O =

= −2922.99 kJ mol−1

+ (8.314 472 × 10−3 kJ K−1 mol−1 )(298 K) ln

(0.03)6 (0.999)6

(0.005)(0.001)6

= −2859.31 kJ mol−1

Since the mole fraction of the solvent is usually very close to 1, we generally leave it out. If

we had done this here, we would have obtained −2859.30 kJ mol−1 , a negligible difference

from the value computed above.

Exercise group 7.5

(1) Consider the following reaction occurring in aqueous solution:

H2 O(l) + A(aq) → P(aq) .

Suppose that this reaction is not allowed under some particular conditions, but only just.

In other words, suppose that the free energy change is just slightly positive. Holding

all other conditions constant, we add an inert salt to the solution. Can this make the

reaction thermodynamically allowed? Why or why not?

(2) Is the reaction

2HgS(s) + Cl2(g) → Hg2 Cl2(s) + 2S(s)

thermodynamically allowed if the pressure of chlorine gas is 2 bar and the temperature

is 25 ◦ C?

(3) 20 mL of a 0.04 mol/L solution of lead (II) nitrate is mixed with 15 mL of a 0.003 mol/L

solution of ammonium sulfate at 25 ◦ C. Is a lead (II) sulfate precipitate formed?

(4) The overall reaction in an alkaline battery is

Zn(s) + 2MnO2(s) → ZnO(s) + Mn2 O3(s) .

(a) Calculate the maximum electrical work that an alkaline battery can perform per

kilogram of zinc at 25 ◦ C.

(b) If an alkaline battery initially contains 10 g of zinc and 15 g of manganese

(IV) oxide, what is the maximum total electrical work that can be produced at

25 ◦ C?

(5) A fuel cell is a device that generates electricity from an oxidation reaction at constant

temperature and pressure (i.e. the net reaction is combustion but no flame or explosion

is involved; this is similar to biological oxidations). Calculate the maximum electrical

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152

Free Energy

work that can be obtained from the oxidation of 1 kg of hydrogen by gaseous oxygen

in a hydrogen fuel cell under the following conditions: pH2 = 1 bar, pO2 = 0.2 bar and

pure liquid water is produced at 298 K.

G to different temperatures

If we want to calculate r G at a non-standard temperature, we typically just apply Equation

(7.2), i.e. r G = r H − T r S. We can use the methods described in previous chapters to

first adjust r H and r S to the desired temperature. More commonly, however, we simply

assume that r H and r S are temperature independent. We can treat these two quantities

as being independent of temperature over a significant range of temperatures, because they

both depend on Cp,m , the difference in heat capacities of reactants and products, which

is typically small, and certainly much smaller than r S ◦ for most reactions. We are also

helped by the fact that r H ◦ and r S ◦ change in the same direction (either both increase

or both decrease, depending on the sign of Cp,m ) as we change the temperature, so the

negative sign in Equation (7.2) tends to result in some cancellation of the error we introduce

by neglecting the temperature variation of these quantities.

Example 7.4 ATP hydrolysis at 37 ◦ C Muscle contraction is powered by the hydrolysis

of ATP:

ATP + H2 O → ADP + Pi

2−

where Pi represents any phosphate (PO3−

4 , HPO4 , etc.), as discussed on page 146. A

muscle is an isothermal free energy machine. It does not produce work from heat, but by

harnessing the free energy change of a chemical reaction.

The standard Gibbs free energy change for this hydrolysis in a physiological biochemical

standard state1 is r G◦ = −32.49 kJ mol−1 , and the standard enthalpy change is r H ◦ =

−30.88 kJ mol−1 . These data are at 298.15 K, but a muscle works at 37 ◦ C, i.e. 310 K. From

the basic definition of G, we have

− r G◦

= 5.40 J K−1 mol−1

298.15 K

Neither r H ◦ nor r S ◦ vary dramatically with temperature so their values at 310 K are

approximately the same as their values at 298 K. Thus

rS

r G310

=

rH

= −30.88 kJ mol−1 − (310 K)(5.40 × 10−3 kJ K−1 mol−1 ) = −32.55 kJ mol−1 .

Note that this value isn’t terribly different from the value at 25 ◦ C. Of course, 37 ◦ C isn’t

very far removed from 25 ◦ C, either.

The approximate physiological concentrations of phosphate, ATP and ADP are, respectively, 2 mmol L−1 , 4 mmol L−1 and 0.3 mmol L−1 .2 Thus, under physiological conditions,

1

2

T = 298.15 K, pH = 7, pMg = 3, I = 0.25 mol L−1 . Data from Alberty and Goldberg, Biochemistry 31, 10610 (1992).

Bernard L. Oser (Ed.), Hawk’s Physiological Chemistry, 14th ed.; McGraw-Hill: New York, 1965, pp. 217–218.

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G to different temperatures

153

the change in free energy during hydrolysis is

r Gm

=

r G310

+ RT ln Q

= −32.55 kJ mol−1

+ (8.314 472 × 10−3 kJ K−1 mol−1 )(310 K) ln

(0.3 × 10−3 )(2 × 10−3 )

4 × 10−3

= −55.25 kJ mol−1

This is the maximum amount of non-pressure–volume work available. Of course, muscles

mainly do non-pV work, since their volume is relatively constant, being made primarily

of incompressible components. Therefore, this is the maximum work that can be done by

muscle, per mole of ATP hydrolyzed.

Note that in the last calculation of the example, we must use a standard free energy of

reaction adjusted to the temperature of the calculation. We can’t use r G◦ (or r G◦ ) from

25 ◦ C and a different T in this equation.

Exercise group 7.6

(1) Calculate the maximum work available from the oxidation of glucose at 37 ◦ C and

solute concentrations of 0.005 mol L−1 of glucose, 0.001 mol L−1 of dissolved oxygen

and 0.03 mol L−1 of dissolved carbon dioxide. Compare your answer to the calculation

of Example 7.3.

(2) (a) Based on thermodynamic considerations alone, can liquid boron trichloride be

made by reacting diborane (B2 H6(g) ) with chlorine gas if the pressures of the two

reactants are both 0.5 bar and the pressure of hydrogen gas is 0.03 bar at 25 ◦ C?

(b) What if the reaction was run at −10 ◦ C instead of 25 ◦ C? Assume that r Hm◦ and

r Sm are independent of temperature.

(3) Creatine phosphate is an alternative energy storage compound found in many cells.

Just as in the case of ATP, considerable free energy can be released by the hydrolysis

of creatine phosphate:

creatine phosphate + H2 O → creatine + Pi

with a r G◦ of −37.7 kJ mol−1 at 310 K. Suppose that, in a certain muscle with a

volume of 1 L, the concentrations of creatine and creatine phosphate are both approximately 1 mmol L−1 and that the concentration of phosphate is 2 mmol L−1 . What is

the maximum work that can be done by such a muscle if all the concentrations are

held constant by homeostatic mechanisms when the entire store of creatine phosphate

is turned over once? How far would this amount of work lift a 1 kg mass?

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