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 reactionATP(aq)+ H2O(l) → ADP(aq)+ HPO4^-2 (aq)for which ΔGrxn = -30.5 kj/mol at 37.0C and pH 7.0. Required:a. Calculate the value of ΔGrxn in a biological cell in which [ATP] = 5.0 mM, [ADP] = 0.30 mM, and HPO4^-2= 5.0mMb. Is the hydrolysis of ATP spontaneous under these conditions?

Answer 1

Δ
`G_(rxn)` = −51. 4 kJ/mol

However, since Δ
`G_(rxn)` is negative. The hydrolysis of ATP for this reaction is said to be spontaneous

Step-by-step explanation:

From the question; The equation for this reaction can be represented as :

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

where:

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

= -30.5 kJ/mol × 1000 J/ 1 kJ

= -30.5 × 10 ⁻³ J/mol

Temperature T = 37 ° C

= (37+273)

= 310 K

pH = 7.0

[ATP] = 5.0 mM

= 5.0mM × 1M/1000mM

= 0.005 M

[ADP] = 0.30 mM

= 0.30 mM × 1M/1000mM

= 0.0003 M

`[HPO_4^(2-)}]` = 5.0 mM

= 5.0mM × 1M/1000mM

= 0.005 M

The objective is to calculate the value for Δ
`G_(rxn)` in the biological cell and to determine if the hydrolysis of ATP is spontaneous under these conditions.

Now;

From the equation given; the equilibrium constant
`K_(eq)` can be expressed as:

`K_(eq) = \frac{[ADP][ HPO_4^(2-)]} {[ATP]}`

`K_(eq) = \frac{(0.0003 \ M)(0.005 \ M)} {(0.005 \ M)}`

`K_(eq) = 3*10^(-4)`

The Δ
`G_(rxn)` in the biological cell can now be calculated as:

Δ
`G_(rxn)` =
`(-30.5 * 10 ^3 \ J/mol) + (8.314 \ J/mol.K)(310 K ) In ( 3*10^(-4))`

Δ
`G_(rxn)` =
`(-30.5 * 10 ^3 \ J/mol) + (-20906.68126)`

Δ
`G_(rxn)` = −51406.68 J/mol

Δ
`G_(rxn)` = −51. 4 × 10³ J/mol

Δ
`G_(rxn)` = −51. 4 kJ/mol

Thus since Δ
`G_(rxn)` is negative. The hydrolysis for this reaction is said to be spontaneous