Answer:
ΔG° of reaction = -47.3 x [tex]10^{3}[/tex] J/mol
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 = [tex]\frac{[HPO4-2] x [ADP]}{ATP}[/tex]
And we have values given for these quantities in the biological cell:
[HP04-2] = 2.1 x [tex]10^{-3}[/tex] M
[ATP] = 1.2 x [tex]10^{-2}[/tex] M
[ADP] = 8.4 x [tex]10^{-3}[/tex] M
Let's plug in these values in the above equation for equilibrium constant:
K = [tex]\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }[/tex]
K = 1.47 x [tex]10^{-3}[/tex] 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 [tex]10^{3}[/tex]) + (8.314)x (310K)xln(0.00147)
Where 8.314 = value of Gas Constant
ΔG° of reaction = (-30.5 x [tex]10^{3}[/tex]) + (-16810.68)
ΔG° of reaction = -47.3 x [tex]10^{3}[/tex] J/mol