One enzyme- catalyzed reaction in a biochemical cycle has an equilibrium constant (K1) that is 10 times the equilibrium constant (K2) of a second reaction. If the standard Gibbs energy of the former reaction is - 300 kJ mol-1, what is the standard reaction Gibbs energy of the second reaction?
There may be an easier way to do it but I would use dGo = -RTlnK and solve for K. Multiply that K by 10 and recalculate dGo.
Bun pengtieng
To find the standard reaction Gibbs energy of the second reaction, we can use the relationship between equilibrium constants and the standard Gibbs energy change (ΔG°) of a reaction.
The relationship is given by the equation: ΔG° = -RTln(K)
Given that the equilibrium constant for the first reaction (K1) is 10 times the equilibrium constant for the second reaction (K2), we have the following relationship:
K1 = 10 * K2
Taking the natural logarithm (ln) of both sides, we get:
ln(K1) = ln(10 * K2)
Using the property of logarithms that ln(a * b) = ln(a) + ln(b), the equation becomes:
ln(K1) = ln(10) + ln(K2)
Now, we are given that the standard Gibbs energy change (ΔG°) for the first reaction is -300 kJ/mol. Using the equation mentioned earlier, we can write:
ΔG°1 = -RTln(K1)
Substituting the values into the equation, we have:
-300 kJ/mol = -RTln(K1)
Since the temperature (T) is not given, we will assume it to be room temperature, 298 K.
-300 kJ/mol = -(8.314 J/mol·K) * 298 K * ln(K1)
Now, let's express the standard Gibbs energy change (ΔG°2) for the second reaction:
ΔG°2 = -RTln(K2)
Substituting the relationship between K1 and K2, we have:
ΔG°2 = -RTln(K1/10)
Using ln(a/b) = ln(a) - ln(b), the equation becomes:
ΔG°2 = -RTln(K1) + RTln(10)
Substituting the given values, we have:
ΔG°2 = -300 kJ/mol + (8.314 J/mol·K) * 298 K * ln(10)
Calculating this expression will give you the standard reaction Gibbs energy of the second reaction.
To determine the standard Gibbs energy of the second reaction, we need to use the relationship between equilibrium constants and the standard Gibbs energy change (ΔG°) of a reaction.
The relationship is given by the equation: ΔG° = -RT ln(K)
Where:
ΔG°: standard Gibbs energy change of the reaction (in J/mol)
R: gas constant (8.314 J/mol·K)
T: temperature (in K)
K: equilibrium constant of the reaction
Given that the equilibrium constant (K1) of the first reaction is 10 times the equilibrium constant (K2) of the second reaction, we can write the relationship as: K1 = 10K2
We are also given that the standard Gibbs energy change (ΔG°) of the first reaction is -300 kJ/mol. To convert it to J/mol, we multiply by 1000: ΔG° = -300,000 J/mol.
Now, let's solve for the equilibrium constant (K1) of the first reaction:
ΔG°1 = -RT ln(K1)
-300,000 = -8.314 * T * ln(K1)
Similarly, we can solve for the equilibrium constant (K2) of the second reaction:
ΔG°2 = -RT ln(K2)
ΔG°1 = ΔG°2 (since we are comparing the standard Gibbs energy changes of the reactions)
-300,000 = -8.314 * T * ln(K2)
Now, we can use the given relationship between the equilibrium constants:
K1 = 10K2
ln(K1) = ln(10K2)
ln(K1) = ln(10) + ln(K2)
-8.314 * T * ln(K2) = -8.314 * T * ln(10) -300,000
We can now cancel out the -8.314 * T term from both sides of the equation:
ln(K2) = ln(10) - 300,000 / (8.314 * T)
Finally, we can solve for the standard Gibbs energy change (ΔG°2) of the second reaction using the equation ΔG° = -RT ln(K):
ΔG°2 = -8.314 * T * ln(K2)
Now, you can substitute the value of ln(K2) that we found in the previous step to calculate ΔG°2.