Question

A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by the reaction ATP(aq) + H2O(1) ADP(aq) + HPO- (aq) - for which AG x = -30.5 kJ/mol at 37.0 °C and pH 7.0. Calculate the value of AG in a biological cell in which (ATP) = 5.0 mm, (ADP) = 0.90 mm, and [HPO-) = 5.0 mM. AG = -8507 kJ/mol Is the hydrolysis of ATP spontaneous under these conditions? yes no Question Source: MRG-General Chemistry | Publisher: Un about us Careers privacy policy terms of use contact us help

Answer 1

Answer:

$\small \Delta$G_rxn = - 30.772 KJ/mol.

We know for spontaneous process $\small \Delta$ G_rxn is negative; hence this reaction must be spontaneous.

Yes, the hydrolysis of ATP is spontaneous under these condition.

Calculation:

The given reaction is,

ATP(aq) +H₂O(l) $\small \rightarrow$ ADP(aq) + HPO₄^2-(aq)

Equillibrium constant K, can be written as,

[ADP] x [HP02] KE [ATP]

Given concentrations are,

[ATP] = 5.0 mM; [ADP] = 0.90 mM; and [HPO₄^2-] = 5.0 mM.

Now, $\small K = \frac{(0.90 mM)\times (5.0mM)}{(5.0 mM)}$

Or, K = 0.90

And we know,

$\small \Delta G_{rxn} = \Delta G^{0}_{rxn} +RTln K$

Where, K is equillibrium constant; R is gas constant and T is temperature.

Given data:

$\small \Delta$G⁰_rxn = -30.5 KJ/mol; R = 8.314 J/mol = 8.314/1000 KJ/mol = 0.008314 KJ/mol; T = 37.0 ⁰C = (273.15+37.0) K = 310.15 K; and calculated K = 0.90.

Now, putting the all values in the above equation we get,

$\small \Delta G_{rxn} = \Delta G^{0}_{rxn} +RTln K$

$\small Or, \Delta G_{rxn} = -30.5 KJ/mol + (0.008314 KJ/mol)\times (310.15K)\times ln(0.90)$

$\small Or, \Delta G_{rxn} = -30.5 KJ/mol + (-0.272)KJ/mol$

$\small Or, \Delta G_{rxn} = (-30.5 -0.272) KJ/mol$

$\small Or, \Delta G_{rxn} = -30.772 KJ/mol$.