The figure below shows the relationship between G for the following reaction and the logarithm to the base e of the reaction quotient for the reaction between N 2 and H 2 to form NH 3. Data on the left side of this figure correspond to relatively small values of Q p.
They therefore describe systems in which there is far more reactant than product. The sign of G for these systems is negative and the magnitude of G is large. The system is therefore relatively far from equilibrium and the reaction must shift to the right to reach equilibrium. Data on the far right side of this figure describe systems in which there is more product than reactant.
The sign of G is now positive and the magnitude of G is moderately large. The sign of G tells us that the reaction would have to shift to the left to reach equilibrium. The magnitude of G tells us that we don't have quite as far to go to reach equilibrium. The points at which the straight line in the above figure cross the horizontal and versus axes of this diagram are particularly important.
The straight line crosses the vertical axis when the reaction quotient for the system is equal to 1. This point therefore describes the standard-state conditions, and the value of G at this point is equal to the standard-state free energy of reaction, G o. The point at which the straight line crosses the horizontal axis describes a system for which G is equal to zero. Because there is no driving force behind the reaction, the system must be at equilibrium.
The relationship between the free energy of reaction at any moment in time G and the standard-state free energy of reaction G o is described by the following equation. We can therefore solve this equation for the relationship between G o and K. This equation allows us to calculate the equilibrium constant for any reaction from the standard-state free energy of reaction, or vice versa. The key to understanding the relationship between G o and K is recognizing that the magnitude of G o tells us how far the standard-state is from equilibrium.
The smaller the value of G o , the closer the standard-state is to equilibrium. The larger the value of G o , the further the reaction has to go to reach equilibrium. The relationship between G o and the equilibrium constant for a chemical reaction is illustrated by the data in the table below.
Use the value of G o obtained in Practice Problem 7 to calculate the equilibrium constant for the following reaction at 25C:. Click here to check your answer to Practice Problem 9. Click here to see a solution to Practice Problem 9. The equilibrium constant for a reaction can be expressed in two ways: K c and K p. We can write equilibrium constant expressions in terms of the partial pressures of the reactants and products, or in terms of their concentrations in units of moles per liter.
For gas-phase reactions the equilibrium constant obtained from G o is based on the partial pressures of the gases K p. For reactions in solution, the equilibrium constant that comes from the calculation is based on concentrations K c. Use the following standard-state free energy of formation data to calculate the acid-dissociation equilibrium constant K a at for formic acid:. HCO 2 aq HCO 2 - aq Click here to check your answer to Practice Problem Click here to see a solution to Practice Problem The Temperature Dependence of Equilibrium Constants.
Equilibrium constants are not strictly constant because they change with temperature. We are now ready to understand why. The standard-state free energy of reaction is a measure of how far the standard-state is from equilibrium. But the magnitude of G o depends on the temperature of the reaction. As a result, the equilibrium constant must depend on the temperature of the reaction. May 9, Below is a table to summarize it up! Related questions What conditions are spontaneous processes trying to satisfy?
Why are spontaneous processes irreversible? Why are all spontaneous processes not exothermic? Why do spontaneous processes increase entropy? Why is diffusion a spontaneous process? You could then say that the reaction shifts to the left. Re: sign of delta G Post by Joowon Seo 3A » Tue Feb 18, am When delta G is negative it becomes spontaneous and the reaction will favor the products if delta G is positive the reaction will favor the reactants. Delta G will be negative, thus the reaction will be more spontaneous.
If you decrease the amount of product, the lnK part of the expression will become smaller, making the G value more positive. Delta G will be positive, thus the reaction will be less spontaneous.
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