Measuring mass. The change in mass of a reactant or product can be followed during a reaction. This method is useful when carbon dioxide is a product which leaves the reaction container.
It is not suitable for hydrogen and other gases with a small relative formula mass , M r. Measuring volume. The change in volume of a reactant or product can be followed during a reaction. This method is useful when a gas leaves the reaction container. The expression for K eq depends on how you write the direction of the reaction.
One other thing to note is that if a pure liquid or solid participates in the reaction, it is omitted from the equilibrium expression for K eq. This makes sense because the concentration of a pure solid or liquid is constant at constant temperature.
The equilibrium constant for any reaction at a particular temperature is a constant. This means that you can add reactants or products and the constant does not change.
The implications of this are quite profound. For example, if you add or take away products or reactants from a reaction, the amounts of reactants or products will change so that the reaction reaches equilibrium again—with the same value of K eq.
And because we know or can look up and calculate what the equilibrium constant is, we are able to figure out exactly what the system will do to reassert the equilibrium condition. The H2O term in the reactants can be omitted even though it participates in the reaction, because it is a pure liquid and its concentration does not change appreciably during the reaction. Can you calculate the concentration of pure water? We already know that a 0. A helpful way to think about this is to set up a table in which you note the concentrations of all species before and after equilibrium.
Note that we do not use a large number of significant figures to calculate K eq because they are not particularly useful, since we are making approximations that make a more accurate calculation not justifiable. In addition, note that K eq itself does not have units associated with it. Now we can calculate the equilibrium constant K eq , assuming that we can measure or calculate the concentrations of reactants and products at equilibrium.
All well and good, but is this simply an empirical measurement? It was certainly discovered empirically and has proven to be applicable to huge numbers of reactant systems. It just does not seem very satisfying to say this is the way things are without an explanation for why the equilibrium constant is constant.
How does it relate to molecular structure? What determines the equilibrium constant? What is the driving force that moves a reaction towards equilibrium and then inhibits any further progress towards products? You will remember we hope that it is the second law of thermodynamics that tells us about the probability of a process occurring. The criterion for a reaction proceeding is that the total entropy of the universe must increase.
So it should not be a surprise to you that there is a relationship between the drive towards equilibrium and the Gibbs free energy change in a reaction. We have already seen that a large, negative Gibbs free energy change from reactants to products indicates that a process will occur or be spontaneous, in thermodynamic terms [17] , whereas a large, positive equilibrium constant means that the reaction mixture will contain mostly products at equilibrium.
Think about it this way: the position of equilibrium is where the maximum entropy change of the universe is found. On either side of this position, the entropy change is negative and therefore the reaction is unlikely. If we plot the extent of the reaction versus the dispersion of energy in the universe or the free energy, as shown in the graph, we can better see what is meant by this. At equilibrium, the system sits at the bottom of an energy well or at least a local energy minimum where a move in either direction will lead to an increase in Gibbs energy and a corresponding decrease in entropy.
Remember that even though at the macroscopic level the system seems to be at rest, at the molecular level reactions are still occurring. This is also true for any phase change. Because any system will naturally tend to this equilibrium condition, a system away from equilibrium can be harnessed to do work to drive some other non-favorable reaction or system away from equilibrium.
On the other hand, a system at equilibrium cannot do work, as we will examine in greater detail. The relationship between the standard free energy change and the equilibrium constant is given by the equation:.
It allows us to calculate equilibrium constants from tables of free energy values see Chapter 9. In this equation, the variable Q is called the reaction quotient. Rather, they are the actual concentrations at the point in the reaction that we are interested in. It is easy to get mixed up and apply them incorrectly. Let us look at a chemical system macroscopically. If we consider a reaction system that begins to change when the reactants are mixed up that is, it occurs spontaneously , we will eventually see that the change slows down and then stops.
It would not be unreasonable to think that the system is static and assume that the molecules in the system are stable and no longer reacting. However, as we discussed earlier, at the molecular level we see that the system is still changing and the molecules of reactants and products are still reacting in both the forwards and reverse reactions. Some of these reactions will have enough energy to be productive; molecules of acetate will transfer protons to water molecules and the reverse reaction will also occur.
Although there is no net change at the macroscopic level, things are happening at the molecular level. Bonds are breaking and forming. This is the dynamic equilibrium we discussed earlier.
Now what happens when we disturb the system. We know that a 0. Now we add enough acetate [18] to make the acetate concentration 0. One way to think about this new situation is to consider the probabilities of the forward and backward reactions. If we add more product acetate , the rate of the backward reaction must increase because there are more acetate ions around to collide with.
But as we saw previously, as soon as more acetic acid is formed, the probability of the forward reaction increases and a new equilibrium position is established, where the rate of the forward reactions equal the rate of the backward reactions. Using this argument we might expect that at the new equilibrium position there will be more acetic acid, more acetate, and less hydronium ion than there was originally.
We predict that the position of equilibrium will shift backwards towards acetic acid. For that we have to look at Q and K eq. If we take the new initial reaction conditions 0.
This generates a value for Q as 1. Now, if we compare Q and K eq , we see that Q is larger than Keq 1. To do this, the numerator products must decrease, while the denominator reactants must increase. This approach leads us to the same conclusion as our earlier probability argument. This may not seem like much, but remember that each pH unit is a factor of 10, so this rise in pH actually indicates a drop in hydronium ion concentration of a bit less than a hundredfold.
In order to regain the most stable situation, the system shifts to the left, thereby reducing the amount of product:. There are a number of exercises that will allow you to better understand the calculations involved in defining the effects of perturbations changes in conditions, concentrations, and temperature on the equilibrium state of a system.
Many chemistry books are full of such buffer and pH problems. What is really important to note is that a system will return to equilibrium upon perturbation. This is where the system is most stable. And once the system is at equilibrium, further perturbations will lead to a new equilibrium state. However, it is important to understand why this principle works. What about temperature, volume, and partial pressure? How do they affect equilibrium?
We have also not specifically addressed equilibrium reactions that take place in the gas phase. As an example, important atmospheric reactions such as the formation and depletion of ozone take place in the gas phase. There is nothing particularly special or different about calculating the equilibrium constant for gas phase reactions.
The effect of increasing the volume is the same as decreasing the concentration, and increasing the pressure has the same effect as increasing the concentration. Note, however, that adding a gas that is not a participant in the reaction has no effect even though the total pressure is increased.
The effect of changing the temperature on the position of equilibrium is a little more complex. At first guess, you might predict that increasing the temperature will affect the rates of both the forward and backward reactions equally.
However, if we look more closely, we see that this is not true. Cast your mind back to the discussions of temperature and thermal energy. If the temperature of the system is raised, it means that thermal energy has been added to the system from the surroundings. In order to predict the effect of adding energy to the system, we need to have more information about the energy changes associated with that system.
We can measure or calculate enthalpy changes for many reactions and therefore use them to predict the effect of increasing the temperature adding thermal energy.
For example, take the reaction of nitrogen and hydrogen to form ammonia. The reaction is exothermic because for each mole of ammonia 17g , Now, if we heat this reaction up, what will happen to the position of equilibrium? Let us rewrite the equation to show that thermal energy is produced:. Sure enough, if we heat this reaction up, the position of equilibrium shifts towards ammonia and hydrogen—it starts to go backward! This is actually quite a problem, as this reaction requires a fairly high temperature to make it go in the first place.
The production of ammonia is difficult if heating up the reaction makes it go in the opposite direction to the one you want. To answer this question, let us consider the energy profile for an exothermic reaction. Stated in another way: more energy is required for molecules to react so that the reverse back reaction occurs than for the forward reaction.
Therefore, it makes sense that if you supply more energy, the reverse reaction is affected more than the forward reaction. There is an important difference between disturbing a reaction at equilibrium by changing concentrations of reactants or products, and changing the temperature.
When we change the concentrations, the concentrations of all the reactants and products change as the reaction moves towards equilibrium again, but the equilibrium constant stays constant and does not change. Now here is an interesting point: imagine a situation in which reactants and products are continually being added to and removed from a system. Such systems are described as open systems, meaning that matter and energy are able to enter or leave them. Open systems are never at equilibrium.
Assuming that the changes to the system occur on a time scale that is faster than the rate at which the system returns to equilibrium following a perturbation, the system could well be stable. Such stable, non-equilibrium systems are referred as steady state systems. Think about a cup with a hole in it being filled from a tap. If the rate at which water flows into the cup is equal to the rate at which it flows out, the level of water in the cup would stay the same, even though water would constantly be added to and leave the system the cup.
Living organisms are examples of steady state systems; they are open systems, with energy and matter entering and leaving. However, most equilibrium systems studied in chemistry at least those discussed in introductory texts are closed, which means that neither energy nor matter can enter or leave the system.
In addition, biological systems are characterized by the fact that there are multiple reactions occurring simultaneously and that a number of these reactions share components—the products of one reaction are the reactants in other reactions.
We call this a system of coupled reactions. An interesting coupled-reaction system aside from life itself is the Belousov—Zhabotinsky BZ reaction in which cesium catalyzes the oxidation and bromination of malonic acid. The typical BZ reaction involves a closed system, so it will eventually reach a boring macroscopically-static equilibrium state. The open nature of biological systems means that complex behaviors do not have to stop; they continue over very long periods of time.
The steady state systems found in organisms display two extremely important properties: they are adaptive and homeostatic. This means that they can change in response to various stimuli adaptation and that they tend to return to their original state following a perturbation homeostasis. Adaptation and homeostasis may seem contradictory, but in fact they work together to keep organisms alive and able to adapt to changing conditions.
Therefore, the greater the temperature, the higher the probability that molecules will be moving with the necessary activation energy for a reaction to occur upon collision. Even if two molecules collide with sufficient activation energy, there is no guarantee that the collision will be successful.
In fact, the collision theory says that not every collision is successful, even if molecules are moving with enough energy. The reason for this is because molecules also need to collide with the right orientation, so that the proper atoms line up with one another, and bonds can break and re-form in the necessary fashion.
For example, in the gas-phase reaction of dinitrogen oxide with nitric oxide, the oxygen end of N 2 O must hit the nitrogen end of NO; if either molecule is not lined up correctly, no reaction will occur upon their collision, regardless of how much energy they have. However, because molecules in the liquid and gas phase are in constant, random motion, there is always the probability that two molecules will collide in just the right way for them to react.
Of course, the more critical this orientational requirement is, like it is for larger or more complex molecules, the fewer collisions there will be that will be effective. An effective collision is defined as one in which molecules collide with sufficient energy and proper orientation, so that a reaction occurs.
According to the collision theory, the following criteria must be met in order for a chemical reaction to occur:. Some people place glow sticks in the freezer to make them last longer. Why do you think this works? The chemical reaction that happens in a light stick is slower when cold.
Do you think that starting with warmer reactants increases the rate of other chemical reactions? It is reasonable to think that temperature will affect the rate of other chemical reactions because temperature affected this reaction.
Explore Ask students how they could set up an experiment to find out if the temperature of the reactants affects the speed of the reaction. Ask students: How many sets of solutions should we use? Students should use two sets—one that is heated and one which is cooled. Tell students that they will use hot and cold water baths, like in the demonstration, to warm and cool the solutions.
Should the warmed samples of baking soda solution and calcium chloride solution be the same as the samples that are cooled? Samples of the same solution should be used and the same amount of cold solution as warm solution should be used.
In the glow stick demonstration, we could tell that the reaction was happening faster if the light was brighter. How can we tell if the reaction is happening faster in this chemical reaction? The chemical reaction is happening faster, if more products are produced. We should look for more bubbling carbon dioxide and more white precipitate calcium carbonate. Have students warm a pair of reactants and cool another and compare the amount of products in each reaction.
Question to Investigate Does the temperature of the reactants affect the rate of the chemical reaction? Use a graduated cylinder to add 20 mL of water to one of the baking soda solution cups. Swirl until as much of the baking soda dissolves as possible. There may be some undissolved baking soda in the bottom of the cup. Pour half of your baking soda solution into the other baking soda solution cup. Make the Calcium Chloride Solution Use a graduated cylinder to add 20 mL of water to one of the calcium chloride solution cups.
Swirl until the calcium chloride dissolves. Pour half of your calcium chloride solution into the other calcium chloride solution cup. The water should not be very deep.
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