Question

A critical reaction in the production of energy in biological systems is the hydrolysis of adenosine triphosphate (ATP) to adenosine diphosphate (ADP) as described by ATP(aq) + H2O(l)  ADP(aq) + HPO4-2(aq) for which ΔG°rxn = –30.5 kJ/mol at 37.0°C. Calculate the value of ΔGrxn in a biological cell in which [ATP] = 1.2 x10-2 M, [ADP] = 8.4 x10-3 M, and [HPO4-2] = 2.1x10-3 M.

Answer 1

ΔG° of reaction = -47.3 x
`10^(3)` J/mol

Step-by-step explanation:

As we can see, we have been a particular reaction and Energy values as well.

ΔG° of reaction = -30.5 kJ/mol

Temperature = 37°C.

And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:

The first step is to calculate the equilibrium constant for the reaction:

Equilibrium Constant K =
`([HPO4-2] x [ADP])/(ATP)`

And we have values given for these quantities in the biological cell:

[HP04-2] = 2.1 x
`10^(-3)` M

[ATP] = 1.2 x
`10^(-2)` M

[ADP] = 8.4 x
`10^(-3)` M

Let's plug in these values in the above equation for equilibrium constant:

K =
`([2.1x10^(-3)] x [8.4x10^(-3)] )/([1.2 x 10^(-2)] )`

K = 1.47 x
`10^(-3)` M

Now, we have to calculate the ΔG° of reaction for the biological cell:

But first we have to convert the temperature in Kelvin scale.

Temp = 37°C

Temp = 37 + 273

Temp = 310 K

ΔG° of reaction = (-30.5
`10^(3)`) + (8.314)x (310K)xln(0.00147)

Where 8.314 = value of Gas Constant

ΔG° of reaction = (-30.5 x
`10^(3)`) + (-16810.68)

ΔG° of reaction = -47.3 x
`10^(3)` J/mol