When is delta h favorable

Email Link. How do we find out if a system is favorable? Lavelle mentioned during Wednesday's class that we can use entropy in order to find out if a system is favorable. He said there where two ways in order to find find this out, but I was not able to catch it. Re: How do we find out if a system is favorable? When this happens, we need to use the fact that the free energy change is temperature-dependent in order to predict the outcome.

In other words, as the temperature rises, a process that involves an increase in entropy becomes more favorable. The idea that temperature affects the direction of some processes is perhaps a little disconcerting. It goes against common-sense that if you heat something up a reaction might actually stop and go backwards rest assured we will come back to this point later. But in fact there are a number of common processes where we can apply this kind of reasoning and find that they make perfect sense.

Up to this point, we have been considering physical changes to a system—populations of molecules going from solid to liquid or liquid to gaseous states and back. Not really what one commonly thinks of as chemistry, but the fact is that these transformations involve the making and breaking of interactions between molecules.

We can therefore consider phase transitions as analogous to chemical reactions, and because they are somewhat simpler, develop a logic that applies to both processes. So, let us assume for the moment that we do not already know that water boils changes from liquid to gas at o C. Click here to check your answer to Practice Problem 8.

Click here to see a solution to Practice Problem 8. G o for a reaction can be calculated from tabulated standard-state free energy data. Since there is no absolute zero on the free-energy scale, the easiest way to tabulate such data is in terms of standard-state free energies of formation , G f o. As might be expected, the standard-state free energy of formation of a substance is the difference between the free energy of the substance and the free energies of its elements in their thermodynamically most stable states at 1 atm, all measurements being made under standard-state conditions.

We are now ready to ask the obvious question: What does the value of G o tell us about the following reaction? By definition, the value of G o for a reaction measures the difference between the free energies of the reactants and products when all components of the reaction are present at standard-state conditions. G o therefore describes this reaction only when all three components are present at 1 atm pressure. The sign of G o tells us the direction in which the reaction has to shift to come to equilibrium.

The fact that G o is negative for this reaction at 25 o C means that a system under standard-state conditions at this temperature would have to shift to the right, converting some of the reactants into products, before it can reach equilibrium. The magnitude of G o for a reaction tells us how far the standard state is from equilibrium. The larger the value of G o , the further the reaction has to go to get to from the standard-state conditions to equilibrium.

Assume, for example, that we start with the following reaction under standard-state conditions, as shown in the figure below. The value of G at that moment in time will be equal to the standard-state free energy for this reaction, G o.

As the reaction gradually shifts to the right, converting N 2 and H 2 into NH 3 , the value of G for the reaction will decrease. If we could find some way to harness the tendency of this reaction to come to equilibrium, we could get the reaction to do work.

The free energy of a reaction at any moment in time is therefore said to be a measure of the energy available to do work. When a reaction leaves the standard state because of a change in the ratio of the concentrations of the products to the reactants, we have to describe the system in terms of non-standard-state free energies of reaction. The difference between G o and G for a reaction is important. There is only one value of G o for a reaction at a given temperature, but there are an infinite number of possible values of G.

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.