The equilibrium constant (Keq) of a reaction is related to the change in free energy (ΔG) through the equation ΔG = -RTlnKeq, where R is the gas constant and T is the temperature in Kelvin.

If Keq = 1, then lnKeq = ln(1) = 0. Therefore, ΔG = -RT*0 = 0.

So, if a reaction has Keq = 1, the change in free energy, ΔG, for that reaction is zero. This means the reaction is at equilibrium, with no net change in the concentrations of reactants and products over time.

Question

The equilibrium constant (Keq) of a reaction is related to the change in free energy (ΔG) through the equation ΔG = -RTlnKeq, where R is the gas constant and T is the temperature in Kelvin.

If Keq = 1, then lnKeq = ln(1) = 0. Therefore, ΔG = -RT*0 = 0.

So, if a reaction has Keq = 1, the change in free energy, ΔG, for that reaction is zero. This means the reaction is at equilibrium, with no net change in the concentrations of reactants and products over time.

Knowee AI · Accepted Answer

The equilibrium constant (Keq) of a reaction is related to the change in free energy (ΔG) through the equation ΔG = -RTlnKeq, where R is the gas constant and T is the temperature in Kelvin.

If Keq = 1, then lnKeq = ln(1) = 0. Therefore, ΔG = -RT*0 = 0.

So, if a reaction has Keq = 1, the change in free energy, ΔG, for that reaction is zero. This means the reaction is at equilibrium, with no net change in the concentrations of reactants and products over time.

If a reaction has Keq = 1, which best describes the change in free energy, ΔG, for that reaction?

Question

If a reaction has Keq = 1, which best describes the change in free energy, ΔG, for that reaction?

Solution

Similar Questions

Upgrade your grade with Knowee