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4 The cell is a very complex system and is not simply evolved from simple components. In the last lecture, we talked about the types of molecules in a cell and also the forces that hold these molecules together and we said that the molecules and forces that we see in a cell are in fact no different from those that are in a flask. So although it is necessary to understand molecules and forces to understand what makes a cell alive, it is not sufficient. You all know that living systems require energy. If you stop eating, you will eventually stop living. It may take awhile because your body can live on stored fat and on excess muscle for weeks, but sooner or later you will die. You might wonder why a cell needs energy. Where does the energy come from and where does it go? Before we can answer that question, we need to introduce thermodynamics. Thermodynamics deals with the energy flow in any system. Thermodynamics helps us to predict how systems will behave under different sets of circumstances and whether a reaction will occur spontaneously -- that is, without an input of energy. page 4

5 You have probably all heard the first law of thermodynamics, which states that energy cannot be created or destroyed. It can only be converted from one form to another, or used to do work. When we add energy to solid water (ice) by heating (thermal energy), the first thing that happens is that the energy of the molecules within the solid ice increases and the molecules start to wiggle around more. The thermal energy is converted to kinetic energy (motion). As the heating continues, some water molecules gain enough energy to break away from the other water molecules. Since water molecules interact through hydrogen bonding, and we need energy to break those interactions, we can say that the water molecules are using the kinetic energy that they have gained to do work. One of the hallmarks of a living system is that it can metabolize, i.e. make use of, food (carbohydrates, proteins, fats) as an energy source. One of the main things that occurs in metabolism is that glucose (C 6 H 12 O 6 ) combines with oxygen through multiple processes (including the glycolytic pathway, which is shown on this slide) to give carbon dioxide and water. As you will see in the next slide, energy is released when glucose and oxygen are converted to carbon dioxide and water. Some of the energy that is released by the transformation can be converted to other forms of energy. A key function of the glycolytic pathway (and later metabolic processes) is to convert the released energy to another energy store known as adenosine triphosphate (ATP colored red in diagram) in the cell, which can later be used by cellular machinery to help carry out important reactions of metabolism. You will learn about ATP later in the page 5

6 course. (You do not need to know the details of the glycolytic pathway for this course). page 5

7 Cells need energy to make biomolecules to carry out cellular transformations. To get energy, cells burn fuel. One of the universal fuels used by all cells from bacteria to man is glucose. To burn any fuel requires oxygen. Oxygen reacts with glucose (or any fuel) to form carbon dioxide and water. The process by which this occurs in cells involves multiple steps. Those steps are not important. What is important is that the overall transformation involves the release of energy. On this slide, glucose and oxygen are depicted as being at a higher energy level than carbon dioxide and water to capture the idea that there is a release of energy when this reaction occurs. We say that this reaction is energetically favorable or that it occurs spontaneously. (However, as you will learn, some spontaneous reactions are very fast while others occur at a glacially slow pace. We will introduce kinetics-- the rates at which reactions occur -- after thermodynamics. It is another concept that will be used throughout the course). page 6

8 Some reactions release energy. Others require energy. The reaction of carbon dioxide and water to form glucose is not one that can occur spontaneously because it requires going from a lower energy state to a higher energy state. That energy ultimately comes from the sun. We aren t going to focus now on how the energy of the sun is harnessed to drive reactions towards disfavored (higher energy) products. Instead, we are going to ask how one can tell whether a reaction is favored or disfavored. Once we know that, we can begin to explain better how Nature accomplishes the impressive feat of driving reactions in either the favored or the disfavored directions depending on the immediate needs of the cell. page 7

9 How do we know which way is uphill in energy and which way is downhill in energy for a chemical reaction? Do we simply measure the amount of energy in the individual molecules by some means? The answer is NO. Energy itself is not a measurable entity, but other quantities related to energy are measurable. For example, in order to know how much kinetic energy (K.E.) a person has while running, we can measure the speed of the person and his or her mass. Then we can crudely (neglecting various complicating factors) calculate the K.E. that this person possesses using the relationship K.E. = 0.5mv 2. Similarly, to know the chemical energy in a system, we must use measurable observables that we can relate to energy. These observables include concentrations and temperature. In order to relate the measured observables to the energy in a system, we need to introduce the concept of chemical equilibrium. To do that, we will consider a simple chemical reaction, the hydration of carbon dioxide. Carbon dioxide is taken in by plants and converted into glucose. An early step in the process of converting CO 2 to glucose, which is shown on the next slide, involves the reaction of carbon dioxide with water. We will use this reaction to understand the concept of chemical equilibrium. page 8

10 This slide depicts two reactions. The first is one we talked about in earlier lectures. The second is the one I just introduced. Note that the reactions are drawn with two stacked arrows, one pointing to the left and the other pointing to the right. In the top reaction, the arrow pointing to the right represents the conversion of liquid water to gaseous water while the arrow pointing to the left represents the conversion of gaseous water to liquid water. In the bottom reaction, the arrow pointing to the right represents the addition of water to carbon dioxide to form carbonic acid while the arrow pointing to the left represents the decomposition of carbonic acid to form carbon dioxide and water. If these molecules are allowed to interconvert under a given set of conditions, they will reach a point where the concentrations of the interconverting species are stable. We say they have reached equilibrium; that is, the rate at which A is converted to B (i.e., the number of molecules in state A that are converted to state B per minute) is identical to the rate at which B is converted back to A. Although reactions occur in both directions, no net interconversion takes place. We will use the example of hydration of carbon dioxide to further discuss chemical equilibrium and thermodynamics. This reaction is simple, but it also happens to be important. CO 2 is one of the greenhouse gases that contribute to global warming. That is, CO 2 produced by industrial activity (burning things to get energy) absorbs radiation reflected from the earth s surface, trapping it in the atmosphere and warming the planet. A significant portion of the CO 2 that is formed is absorbed into the oceans where some of it combines with water to form carbonic acid. For a long time, people thought this was a good thing because it reduced the amount of CO 2 in the atmosphere. The problem is that the carbonic acid that is formed makes the water more acidic. That, in turn, causes all kinds of sea creatures - page 9

11 corals and lots of other things -- that have calcium carbonate exoskeletons to dissolve. This is an example of something we call Le Chatelier s Principle, which we will also talk about. page 9

12 In the reaction shown on this slide, the hydration of carbon dioxide, carbon dioxide (CO 2 ) and water (H 2 O) exist in equilibrium with carbonic acid (H 2 CO 3 ). CO 2 (aq.) refers to carbon dioxide in an aqueous environment. As depicted in the slide, carbon dioxide interacts with water through hydrogen bonding (a type of permanent dipolepermanent dipole interaction). The same is true for H 2 CO 3 (aq.). The symbol (l) in the equation indicates that the water referred to here is in the liquid state. (aq.) and (l) are called state symbols and they are very important and should be included in equations when we are talking about equilibrium and thermodynamics. Other state symbols include (s) for solid, (g) for gas, etc. You can see that when carbon dioxide is converted to carbonic acid, a C-O bond is formed between the C atom of carbon dioxide and the O atom of a water molecule. In the process, one of the C=O bonds (a double bond) in carbon dioxide becomes a C-O (a single bond). The C atom in the molecule still forms four bonds and maintains the octet configuration. You might wonder: which side of the reaction shown above is favored? page 10

13 To tell which side is favored, you measure the concentrations of the chemical species at equilibrium -- that is, after the reaction has reached the point where the concentrations of the products and the reactants have stopped changing. All reactions have a characteristic equilibrium constant,k eq, under a given set of conditions. K eq is defined as the ratio of the product of the concentrations of the species on the right (by convention, we refer to these as the products) over the product of the concentrations of the species on the left (by convention, we call these the reactants). That is, for a reaction A + B <==> C + D, K eq = [products]/[reactants] = [C][D]/[A][B]. If K eq is less than 1, then the concentrations of the products are lower than the concentrations of the reactants at equilibrium. Conversely, if K eq is greater than 1, then the concentrations of products are greater than the concentrations of the reactants. If K eq is exactly 1, then there is a fifty:fifty mixture of reactants and products at equilibrium. In addition to the conditions given on the slide, there are two extremes of K eq that we can also consider. If K eq ~ 0 (close to zero), it means that at equilibrium, there is almost no product. That is, the reaction does not proceed. If K eq ~ (close to infinity), it means that at equilibrium, almost all the reactants will have become the products. In other words, the reaction has proceeded to completion. For the hydration of carbon dioxide, the concentrations of the products and reactants have been measured. It is not important how, but it is straightforward. The concentration of water is so high compared to the other species that it is essentially constant and has the value of 55 M at atmospheric pressure. From the concentrations of the products and reactants, an equilibrium constant, K eq, of 3.09 x 10-5 M -1 at 25 o C has been measured. This means that the page 11

14 left side of the reaction is (strongly) favored, and when carbon dioxide is dissolved in water, only a small fraction of it is converted to carbonic acid at equilibrium. page 11

15 This slide depicts what happens if you start out with very different concentrations or ratios of reactants and products. If you start out with a 100% solution of aqueous carbonic acid (H 2 CO 3 ), you will observe that the concentration of H 2 CO 3 decreases with time while the concentration of CO 2 increases. At some point in time, the concentrations of H 2 CO 3 and CO 2 will be equal and [H 2 CO 3 ]/[CO 2 ] = 1 (state 1). Since I told you what K eq for the hydration of carbon dioxide is, you know that state 1 is not at equilibrium since K eq [H 2 O] 1 (i.e., K eq [H 2 O] [H 2 CO 3 ]/[CO 2 ] at state 1). The concentrations of H 2 CO 3 and CO 2 continue to change with time until the solution reaches the state of chemical equilibrium (state 2), at which point the ratio of products to reactants remains constant and [H 2 CO 3 ]/[CO 2 ] = K eq [H 2 O]. If you start with 100% CO 2, you will observe that the concentration of CO 2 begins to decrease with time while the concentration of H 2 CO 3 starts to build up. Although the concentrations of reactants and products started out much closer to their equilibrium values for this scenario, the system reaches the same equilibrium as in the first scenario. This thought experiment illustrates a key point: the final state -- i.e., the chemical equilibrium -- does not depend on the initial state. page 12

16 From the K eq for the hydration of carbon dioxide, you know that the reaction strongly favors CO 2 and H 2 O. Therefore, you know that CO 2 (aq.) is at a lower energy than H 2 CO 3 (aq.). We define a quantity called the Gibbs free energy (G), which is the energy in a particular state. This example shows two states: 100% CO 2 (aq.) and 100% H 2 CO 3 (aq.). For any reaction, we can define the change in G of a reaction that proceeds from a state with 100% reactant to a state with 100% product as ΔG rxn = G (product) G (reactant). Because the value of G changes with temperature and concentration, we define a set of conditions under which ΔG rxn s can be compared to one another. Under these conditions, called the standard state, concentrations are 1 M, the pressure is 1 atmosphere (atm), and the temperature is constant. The difference in the Gibbs energies of two states under standard state conditions is denoted as ΔG o rxn. In the case of the hydration of carbon dioxide, there is a ΔG o rxn associated with the hypothetical reaction in which all the carbon dioxide (1 M, aqueous) is converted to carbonic acid (1 M, aqueous) at 298 K (room temperature). We know from the position of the equilibrium (it favors reactants) that the state of 100% CO 2 (aq.) is more stable (has less G, as depicted in the slide) than the state of 100% carbonic acid (aq.). You might wonder why there is any conversion of CO 2 to carbonic acid if the state of 100% CO 2 is more stable. page 13

17 The answer is that the free energy of the system at equilibrium, which contains a mixture of the different chemical species, has an even lower G associated with it. This graph depicts the relative free energy levels of the system for different ratios of reactants and products. At equilibrium (where [H 2 CO 3 ]/[CO 2 ] = K eq [H 2 O]), the amount of G in the system is lower than both the state of 100% CO 2 and the state of 100% H 2 CO 3. The take home point is that all systems approach the state that possesses the least Gibbs free energy. This state is the equilibrium state, where the concentrations of products and reactants stop changing. No matter what ratio you start with, the species react to give the same equilibrium ratio of concentrations because that is the lowest energy state. page 14

18 Now that you know that the position of the chemical equilibrium is the state of lowest Gibbs energy, let s take a closer look at the graph that we constructed to describe the Gibbs free energy diagram for the interconversion of CO 2 and carbonic acid in aqueous solution. The line represents the Gibbs free energies at any state between the two pure states. For each state (or point) along the line, we can define a ΔG as the slope (gradient) of the curve at that state. We shall not be concerned with how we can accurately determine the slope of each point on the curve as that would require more advanced mathematical treatments. Instead we can quite accurately estimate the slopes by redefining ΔG as the change in G of the system as it proceeds from one state to another state via an infinitesimally small step. Therefore, for any two states along the line that vary by very small increments, ΔG = G final - G initial. Based on that we can predict the direction in which the concentrations will change. The concentrations will always change towards the state of lower free energy. For example, the ΔG between A final and A iintial in the plot above is negative; therefore, the step will occur spontaneously towards A final. The ΔG between C final and C iintial is positive; therefore, the step is disfavored in that direction (but is favored in the reverse direction). For each point along the curve (i.e., each set of concentrations of products over reactants) you can do the same type of analysis and figure out whether reaction will occur in the defined direction. Steps occur spontaneously in the direction for which ΔG is negative until equilibrium is reached, at which point ΔG = 0. At equilibrium, a change in the net concentrations in any direction will result in a positive ΔG. You have reached a dynamic state in which molecules are interconverting between species, but there is no net change in concentration because there is no driving force. page 15

19 We started this lecture by asking how one can tell which direction is favored for a chemical reaction. We said that it is necessary to understand the concept of chemical equilibrium in order to answer that question. Over the last few slides, we have intuitively related K eq to the value of ΔG o rxn. We said that when K eq is less than 1 (in the case of the hydration of carbon dioxide), it means that the state of 100% CO 2 is more stable and hence has lower G than the state of 100% H 2 CO 3. We now want to understand the relationship between K eq and ΔG o rxn in a more quantitative manner. On the last slide, we introduced another term called ΔG, the slope (gradient) of the curve at any state (or the change in Gibbs free energy associated with an infinitesimally small step from one state to another). We learned that reactions proceed towards a state of chemical equilibrium in which ΔG = 0. We can employ a very useful equation that relates K eq to ΔG o rxn. For the representative reaction of A + B <==> C + D, G = G o rxn + RT ln ([C][D]/[A][B]), where T is temperature in degrees Kelvin and R is the ideal gas constant. We know that when the system is at equilibrium, G = 0 and [C][D]/[A][B] = K eq ; therefore, at equilibrium this equation becomes G o rxn = RTln(K eq ). page 16

20 Using the expression derived in the last slide (ΔG o rxn = RT ln K eq ), we can now work out conditions involving ΔG o rxn that will tell us which side of a reversible reaction is favored. If the equilibrium constant for a reaction is greater than 1, then ΔG o rxn would be negative (< 0). This tells us that the final state with 100% B has less G compared to the initial state with 100% A and the reaction will proceed spontaneously with a release of energy. For a reaction with K eq < 1, ΔG o rxn is positive (> 0). Therefore, the final state with 100% B now has more G compared to the initial state with 100% A. In order to convert a substantial amount of reactants A to products B, energy is required. The reaction is disfavored relative to the products. Finally, when K eq = 1, ΔG o rxn = 0, and the amount of G in the initial and final states are equal. Neither the initial state and the final state is favored, and the reaction proceeds to give equal amounts of products and reactants. page 17