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(l)⟶ADP(aq)+HPO2−4(aq) ATP(aq)+H2O(l)⟶ADP(aq)+HPO42−(aq) for which ΔG∘rxn=−30.5 kJ/molΔGrxn∘=−30.5 kJ/mol at 37.0 °C and pH 7.0. Calculate the value of ΔGrxnΔGrxn in a biological cell in which [ATP]=5.0 mM,[ATP]=5.0 mM, [ADP]=0.40 mM,[ADP]=0.40 mM, and [HPO2−4]=5.0 mM.

Answer 1

`\Delta G_(rxn) = -50.7 \ kJ/mol`

Step-by-step explanation:

The equation of the reaction is given as:

`ATP_((aq)) + H_2O_((l)) \to ADP_((aq)) + HPO_4^(-2)_((aq))`

Given that:

[ATP] = 5.00 mM

[ADP] = 0.40 mM

[HPO₄²⁻] = 5.00 mM

converting mM to M; we have:

5.00 mM = 5.00 mM × 1 M/ 1000 mM

= 0.005 M

0.40 mM = 0.40 mM × 1 M/ 1000 mM

= 0.0004 M

The equilibrium constant
`K_(eq)` is:

`K_(eq) = ((0.0004)(0.005))/((0.005))`

`K_(eq)` = 0.0004

Given that:

`\Delta G^ {^0} _(rxn) =-30.5 \ kJ/mol`

To Joule (J) ; we have

`\Delta G^ {^0} _(rxn) =-30.5 \ kJ/mol * (1000 \ J)/(1 \ kJ)`

`\Delta G^ {^0} _(rxn) =-30.5 *10^3 \ J/mol`

Temperature T = 37° C = (37+ 273)K = 310 K

`\Delta G_(rxn)` is then calculated by using the equation:

`\Delta G_(rxn) = \Delta G^0 _(rxn)+ RT In \ K_(eq)`

`\Delta G_(rxn) = (-30.5*10^3 \ J/mol})+ (8.314 \ J/mol.K)(310) \ In \ (0.0004)`

`\Delta G_(rxn) = -50.665.23 \ J/mol`

`\Delta G_(rxn) = -50.7 \ kJ/mol`

Since
`\Delta G_(rxn)` is negative; the hydrolysis of ATP is spontaneous .