When the reaction glucose-1-phosphate (aq) --> glucose-6-phosphate (aq) is at equilibrium at 25.0 degrees celcius, the amount of glucose-6-phosphate present is 95% of the total. A.) Calculate the Standard Delta G at 25.0 degrees celcius. B.) Calculate Delta G for the reaction in the presence of 10^(-2) M glucose-1-phosphate and 10^(-4) M glucose-6-phosphate. In which direction does the reaction proceed under these conditions?
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To calculate the standard Delta G at 25.0 degrees Celsius, you can use the equation:
ΔG° = -RTln(K)
where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (298.15 K), and K is the equilibrium constant.
In this case, the equilibrium constant (K) can be determined from the information given, which states that the amount of glucose-6-phosphate present is 95% of the total. The equilibrium constant (K) is defined as the ratio of the concentrations of products to the concentrations of reactants, each raised to their stoichiometric coefficients.
If x is the concentration of glucose-1-phosphate at equilibrium, then the concentration of glucose-6-phosphate is 0.95x.
Hence, K = [glucose-6-phosphate]/[glucose-1-phosphate]
= 0.95x / x
= 0.95
Now, we can plug the values into the equation for ΔG°:
ΔG° = -RTln(K)
= -(8.314 J/(mol·K))(298.15 K)ln(0.95)
= -8.314 Jln(0.95)
≈ -0.026 kJ/mol
Therefore, the standard Delta G at 25.0 degrees Celsius is approximately -0.026 kJ/mol.
To calculate Delta G for the reaction in the presence of specific concentrations of glucose-1-phosphate and glucose-6-phosphate (10^(-2) M and 10^(-4) M), you can use the equation:
ΔG = ΔG° + RTln(Q)
where ΔG is the Gibbs free energy change under non-standard conditions, ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (298.15 K), and Q is the reaction quotient.
The reaction quotient (Q) is a measure of the relative concentrations of reactants to products at any given point in a reaction. It is calculated in the same way as the equilibrium constant, but using the actual concentrations instead of the equilibrium concentrations.
In this case, Q can be determined as follows:
Q = [glucose-6-phosphate]/[glucose-1-phosphate]
= (10^(-4) M)/(10^(-2) M)
= 10^(-4)
Now, we can plug the values into the equation for ΔG:
ΔG = ΔG° + RTln(Q)
= -0.026 kJ/mol + (8.314 J/(mol·K))(298.15 K)ln(10^(-4))
≈ -0.026 kJ/mol + (8.314 J/(mol·K))(298.15 K)(ln(10^(-4)))
To determine the direction in which the reaction proceeds under these conditions, you need to consider the sign of ΔG. If ΔG is negative, the reaction is thermodynamically favorable and proceeds in the forward direction. If ΔG is positive, the reaction is thermodynamically unfavorable and proceeds in the reverse direction.
So, calculate the value of ΔG using the equation above, and then determine its sign to determine the direction of the reaction under the given conditions.