1
$\begingroup$

So in my book I have the reaction

$$ATP \rightarrow ADP +P_i$$

and hence $$\Delta G^0 = G^0_{ADP} + G^0_{P_i} - G^0_{ATP}$$

This is easily understood, but I have not been given any actual concentrations of values to use to calculate the result, which is given as

$$\Delta G^0 = -31.0kJ\text{mol}^{-1}$$

Where do I find the values of $G^0_{ADP}, G^0_{P_i} \text{ and } G^0_{ATP}$?

I have the formula $$\Delta G = \Delta G^0 + RT\ln \bigg( \frac{AB}{CD} \bigg)$$

where $A,B$ are the concentrations of the products and $CD$ are the concentrations of the reactants.

$\endgroup$
  • $\begingroup$ There seemed to be a typo in your title, which I've corrected. $\endgroup$ – David Oct 10 '16 at 11:17
0
$\begingroup$

You can't measure the energy of ATP itself, but you can measure the amount of energy that is released every time a P group is removed from the molecule, that is about 30 kj/mol.

Here some more details on how the deltaG is experimentally determined.

Fomr Wikipedia:

https://en.wikipedia.org/wiki/Adenosine_triphosphate

https://en.wikipedia.org/wiki/ATP_hydrolysis

Hydrolysis of the terminal phosphoanhydridic bond is a highly exergonic process, releasing 30.5 kJ mol−1 energy.

... The actual value of ΔG for ATP hydrolysis varies, primarily depending on Mg2+ concentration, and under normal physiologic conditions is actually closer to -50 kJ mol−1.

and

The standard amount of energy released from hydrolysis of ATP can be calculated from the changes in energy under non-natural (standard) conditions, then correcting to biological concentrations. The net change in heat energy (enthalpy) at standard temperature and pressure of the decomposition of ATP into hydrated ADP and hydrated inorganic phosphate is −30.5 kJ/mol, with a change in free energy of 3.4 kJ/mol.[17] The energy released by cleaving either a phosphate (Pi) or pyrophosphate (PPi) unit from ATP at standard state of 1 M are:[18]

ATP + H 2O → ADP + Pi ΔG° = −30.5 kJ/mol (−7.3 kcal/mol)

ATP + H 2O → AMP + PPi ΔG° = −45.6 kJ/mol (−10.9 kcal/mol)

... The values given for the Gibbs free energy for this reaction are dependent on a number of factors, including overall ionic strength and the presence of alkaline earth metal ions such as Mg2+ and Ca2+ . Under typical cellular conditions, ΔG is approximately −57 kJ/mol (−14 kcal/mol).

$\endgroup$
0
$\begingroup$

I think the answer to your question of where you find standard free energy values is ‘nowhere’. Free energy is a concept that you can expres in equations but that you cannot measure. You can only measure the change in free energy of a process.

I cannot provide definitive support for this opinion — a chemistry list may be a better bet.

$\endgroup$
  • $\begingroup$ I guess you could use the boltzmann distribution or something - if you know the partition function. $\endgroup$ – Inhibitor Apr 16 '18 at 1:56
0
$\begingroup$

You can use tables of standard transformed Gibbs energies of formation, $\ce{\Delta _fG^{'o}}$ , such as those found in Table 1 of the following reference, to calculate apparent equilibrium constants of biochemical reactions, and to calculate standard transformed Gibbs energies of formation, $\ce{\Delta _rG^{'o}}$ (at pH 7), which I think is exactly what you require.

The first thing to say it that you need to include water as a reactant.

$$\ce{ATP + H2O = ADP + P_i}\tag{1}$$

Secondly, the reaction shown in Eqn(1) is used as an example by Alberty in the paper cited above, where he calculates both $\ce{\Delta _rG^{'o}}$ (at pH 7) and the apparent equilibrium constant (at pH 7).

 


The following values were taken from Table 1 of the Alberty reference cited above. They refer to pH 7, 25oC, and an ionic strength of 0.1.


              $\ce{\Delta _fG^{'o}/{kJ mol^{-1}}}$

ATP -2276.43

ADP -1409.95

Pi -1059.17

H2O -156.05


Knowing these values, it is now very easy to calculate the standard transformed Gibbs energies of formation, $\ce{\Delta _rG^{'o}}$, for the reaction shown in Eqn (1) (it is merely a case of 'pluggin in' the values).

$$\Delta _rG^{\ce{'o}} = \sum v_i^{'} \Delta_f G_i{^o} = (2276.43 + 156.05 - 1409.95 - 1059.17)\tag{2}$$

$$\ce{\Delta _rG^{'o} = -36.68 kJ mol^{-1}\tag{3}}$$

In Eqn (2), $v_i^{'}$ is the stoichiometric number (positive for reactants and negative for products)

The criterion which determines whether a reaction will 'go' in the forward direction or the reverse direction is the transformed Gibbs energy of reaction, $\Delta_rG^{'}$, (not $\Delta_rG^{'o}$).

If $\Delta_rG^{'}$ is negative, the reaction goes in the forward direction.

We can say that under standard conditions, Eqn (1) goes to the right, and as $\Delta_rG^{'o}$ (not $\Delta_rG^{'}$) is directly related to the apparent equilibrium constant (Eqn 4), we can say that the equilibrium constant for the rection of Eqn (1) is very much to the right.

$$\Delta _rG^{\ce{'o}} = \ce{-RT ln K^{'}}\tag{4}$$

We can go a step further (Alberty,2000) and calculate the apparent equilibrium constant, $K^{'}$, for the reaction shown in Eqn (1).

$$K^{'} = e^{\frac{-\Delta _rG^{'o}}{RT}}=e^{36680/(8.3145 \times 298.15)}\tag{5}$$

$$K^{'} = 2.67 \times 10^6\tag{6}$$

The great value of tables of $\ce{\Delta _fG^{'o}}$ it that they allow $\Delta _rG^{\ce{'o}}$ and $K^{'}$ to be calculated for many reactions, even those for which there is no known enzyme, or where the equilibrium constant has not been determined.

Notes

In Eqn (5) above, $R$ is the gas constant ($\ce{8.3145 J K^{-1} mol^{-1}}$ ) and $T$ is the temperature (in Kelvin)

IMO, the Alberty reference cited above is a model of clarity that makes these type of calculations very easy. It is also an 'easy read' with many great examples.

Two other good refs, if you are 'into' such things, are

Other tables of standard transformed Gibbs energies of formation, $\ce{\Delta _fG^{'o}}$, may be found in

  • Calculation of Standard Transformed Gibbs Energies and Standard Transformed Enthalpies of Biochemical Reactants (by R.A. Alberty)

  • Calculation of Standard Transformed Formation Properties of Biochemical Reactants and Standard Apparent Reduction Potentials of Half Reactions (by R.A. Alberty)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy