The equilibrium constant K for the hydrolysis of adenosine triphosphate (ATP) to adenosine diphosphate (ADP) and phosphate is 1.66 X 10^5 at 37o C, and ÄHè is -20.1 kJ/mol. Calculate ÄSè for the hydrolysis at 37o C. On the assumption that ÄHè and ÄSè are temperature independent, calculate K at 25o C.
To calculate the change in entropy (ΔS) for the hydrolysis of ATP at 37°C, we need to use the relationship between ΔS, ΔH (enthalpy change), and temperature (T).
The equation is as follows:
ΔG = ΔH - TΔS
Where ΔG is the Gibbs free energy change. In this case, we can assume that ΔG is zero because the reaction is at equilibrium.
So, we have:
0 = -20.1 kJ/mol - (37°C + 273.15 K)ΔS
Solving for ΔS:
(37°C + 273.15 K)ΔS = -20.1 kJ/mol
ΔS = -20.1 kJ/mol / (37°C + 273.15 K)
Calculating ΔS:
ΔS = -20.1 kJ/mol / 310.15 K
ΔS = -0.0648 kJ/(mol·K)
Now, to calculate the equilibrium constant, K, at 25°C, we can use the equation:
ΔG = -RTlnK
Where R is the gas constant (8.314 J/(mol·K)), T is the temperature in kelvin (298.15 K), and ln is the natural logarithm.
From the equation, we can rearrange it to solve for K:
K = e^(-ΔG/RT)
Substituting the given values:
K = e^(-(-20.1 kJ/mol) / (8.314 J/(mol·K) * 298.15 K))
K = e^(2.43)
Calculating K:
K ≈ 11.42
Therefore, the equilibrium constant, K, at 25°C, is approximately 11.42.