1. The equilibrium constant for a certain system is 50 at 25 celsius. Calculate the free energy change
2. When the fee energy change:
a) =0, the position of the equilibrium, temperature * change in entropy, free energy change at standard states respectively are???
b) =1, the position of the equilibrium, temperature * change in entropy, free energy change at standard states respectively are???
c) >1, the position of the equilibrium, temperature * change in entropy, free energy change at standard states respectively are?
To calculate the free energy change (ΔG) for a reaction using the equilibrium constant (K), you can use the formula:
ΔG = -RT ln(K)
where ΔG is the free energy change, R is the gas constant (8.314 J/mol*K), T is the temperature in Kelvin, and ln denotes the natural logarithm.
1. Given that the equilibrium constant (K) is 50 at 25 degrees Celsius, we need to convert the temperature to Kelvin before proceeding with the calculation.
T = 25°C + 273.15 = 298.15 K
Using the equation:
ΔG = -RT ln(K)
ΔG = -(8.314 J/mol*K)(298.15 K) ln(50)
Now, you can calculate the free energy change (ΔG) using the above equation. Please note that the answer will depend on the specific reaction and its stoichiometry.
2. a) When the ΔG = 0, it implies that the reaction is at equilibrium. Consequently, the position of the equilibrium is balanced, and the forward and backward rates of the reaction are equal. The product and reactant concentrations remain constant. The temperature multiplied by the change in entropy (ΔS) would also be zero, as ΔG = ΔH - TΔS, and when ΔG = 0, ΔH = TΔS.
b) When the ΔG = 1, it suggests that the reaction is not at equilibrium. If ΔG is positive (greater than zero), it means the reaction is not spontaneous and will proceed in the reverse direction. The product concentration is lower than the reactant concentration. The temperature multiplied by the change in entropy (ΔS) would depend on the specific reaction and the values of ΔH and ΔG.
c) When ΔG > 1, the reaction is still not at equilibrium, and it is not spontaneous in the forward direction. The position of the equilibrium will depend on the specific reaction and the actual values of ΔG, ΔH, and ΔS. The temperature multiplied by the change in entropy (ΔS) would also depend on the specific reaction and the values of ΔH and ΔG.
In summary, the position of the equilibrium, the temperature multiplied by the change in entropy, and the free energy change at standard states would vary depending on the value of ΔG and the specific reaction under consideration.