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Chemistry & Chemical Reactivity: Chapter 19, Principles of Reactivity: Entropy & Free Energy Name______________________ Chapter Goals 1. Understand the concept of entropy and its relationship to spontaneity 2. Predict whether a process will be spontaneous 3. Understand and use a new thermodynamic function, free energy 4. Understand the relationship of a free energy change for a reaction, is equilibrium constant, and whether the reaction is product- or reactant-favored Chapters 6 and 19 together provide information on three laws of thermodynamics. Chapter 6 discusses heat and how heat is transferred which is basically solving for ∆H (law of conservation of energy), while chapter 19 discusses spontaneity (laws 2 and 3) which is solving for a new variable ∆S (whether a reaction occurs all on its own without outside interference), the driving forces that compel a reaction to achieve equilibrium, and teaches us how to determine whether a reaction favors more concentration of product or more concentration of reactant when it reaches equilibrium. A spontaneous change has no choice but to lead to a state of equilibrium. Heat will automatically transfer to a cooler substance, 10% of the gas put into a car’s gas tank actually makes the car move while the rest is wasted into the atmosphere (try ignoring that fact while you pay for your fuel), iron will rust if left unprotected, silver will tarnish, etc. We typically think of product-favored reactions as being spontaneous, but there are reactant-favored reactions like the insolubility of a substance that are also spontaneous. This chapter is to be treated as something of an historical timeline outlining the discoveries as they led first from enthalpy (Chapter 6); then to entropy of system, surroundings, and universe; and then finally to another new concept being introduced called Gibbs free energy which only refers to the reaction itself. This final equation really simplified the idea of spontaneity, but more on that later…. Heat and Spontaneity Usually exothermic reactions are spontaneous, but some endothermic or even non- energy changing reactions can be spontaneous as well: like dissolving NH4NO3 in water, changes in phase of substances at room temperature, etc. Entropy There is a new term called entropy, S, which means increasing disorder. All systems in our world are moving to greater disorder naturally or spontaneously. I notice this when I wear and wash my clothes…they always begin to look older, the colors fade, or the material stretches out. This section of the chapter, 19-3 (pg 906-910), is really not important to know as it goes into much more detail than anyone needs. I would avoid these pages at all cost!!! Geez Louise!!! All you really need to take away from this section is to know that spontaneous processes seek to disperse energy over more particles. Entropy allows a system to achieve a lower overall energy status, or state. 1 Chemistry & Chemical Reactivity: Chapter 19, Principles of Reactivity: Entropy & Free Energy Name______________________ A very simple example would be when you make lemonade. When you add the lemon “stuff” (whether it is dry or the frozen kind) to water, what happens to the lemon “stuff?” __________________________________________________________________ Ludwig Boltzmann (strange looking statue of the dude is on page 921) believed equilibrium of the lemon “stuff” would not be reached until the maximum disorder, or entropy, had been reached. He developed the following equation: S = k log W Where S = entropy, k is the Boltzmann constant, and W = the number of different ways energy can be distributed over the available energy levels. Okay…so what?! A summary of several painful pages of textbook is this: the final state of a system can be more probable (in other words “product-favored”) than the initial state by either 1. The atoms and molecules are more disordered in the final state 2. Energy is more dispersed in the final state a. If energy is more dispersed in the final state, the process is spontaneous b. If only matter is more dispersed in the final state, then quantitative data is necessary to determine the spontaneity of the system c. If energy is not dispersed in the final state…the process is NOT spontaneous and never will be Ludwig also decided that the perfect crystal at 0 K, or -273°C has no entropy, S = 0 (that’s the 3rd law of thermodynamics). The entropy of an element or compound under any set of conditions is the entropy gained when the substance undergoes a temperature change from 0 K to a warmer temperature. It should be obvious that as something warms up from absolute zero (where all molecular motion is supposed to stop), that the molecules will begin to move which signifies an increase in disorder, otherwise known as an increase in entropy. This means that all substances at temperatures above 0 K, have positive entropy values. Negative values mean the system is becoming neater (think about it!) ∆S = qrev T Where S = entropy, qrev = the heat absorbed, and T is temperature in Kelvin. Standard enthalpy, S°, is the entropy gained as the substance changes temperature from 0 K to standard conditions of 1 bar of pressure and 1 molal of concentration (more on molal in another chapter…you don’t need to know it here). Units for entropy are J/K x mol. You also need to know that entropy is another state function since it uses the ∆ sign. 2 Chemistry & Chemical Reactivity: Chapter 19, Principles of Reactivity: Entropy & Free Energy Name______________________ Two important concepts! 1. It should be obvious that entropy of a gas is higher than that of its liquid, and the entropy of a liquid is higher than that of its solid. If this is not obvious, just think about how much more the molecules can move around if the phase changes from solid to liquid and from liquid to gas. 2. Larger molecules have more entropy than smaller molecules. Think about a family of 9 children compared to a family with 2 children. The larger molecule has more atoms, more electron interaction, more anarchy! The larger molecule has more options in movement as they twist, rotate, zig and zag, and stretch, etc. The larger the molecule, the more space there is for the energy to be distributed. 3. For any substance, the higher the temperature, the more entropy the substance has. Think about this, the higher the temperature of a substance, the more chance that substance has to change phase, which leads back to the first idea above Problem 1: Which substance has the higher entropy under standard conditions? Explain yourself and then double-check your decision with Appendix L. a. NO2(g) or N2O4(g) b. I2(g) or I2(s) Entropy Changes in Physical and Chemical Processes Calculating the entropy with substances in their standard state requires the following equation and the standard enthalpy values found in Appendix L. Remember to keep track of your signs!! ∆S°f = Σ[n∆S°f(products)] – [n∆S°f(reactants)] Where n equals the number of moles of each substance on either side of the reaction arrow, ∆S°f stands for standard enthalpy (found in Appendix L). Problem 2: Predict whether each reaction below increases or decreases in entropy. Calculate the standard entropy changes for the following processes. Do the calculations match your predictions? a. Evaporation of 1.00 mole of liquid ethanol to ethanol vapor C2H5OH(l) C2H5OH(g) 3 Chemistry & Chemical Reactivity: Chapter 19, Principles of Reactivity: Entropy & Free Energy Name______________________ b. Oxidation of 1 mole of ethanol vapor C2H5OH(g) + 3 O2(g) 2 CO2(g) + 3 H2O(g) The second law of thermodynamics states that a spontaneous process increases the entropy of the universe. So we use the following equation and the values for a process under standard conditions in Appendix L, for this calculation: ∆S°universe = ∆S°system + ∆S°surrounding Using ∆S°universe as a guide 1. ∆S°universe > 0 indicates a possible spontaneous process 2. ∆S°universe = 0 indicates a system at equilibrium 3. ∆S°universe < 0 indicates a possible nonspontaneous process You will notice that the entropy values are in joules/K x mole rather than kilojoules per mole like enthalpy and Gibbs free energy (more on that in a moment). Make sure you convert all values to either Joules or kilojoules before you add the values together (you’ll understand after we get to the end). Also, where enthalpy and, later, Gibbs free energy values are equal to zero for atoms and diatomic molecules that stand alone, entropy is not equal to zero for these same species. Because entropy is greater than zero for any substance above zero Kelvin, there will always be values of entropy for all substances in Appendix L. Problem 3: Calculate the standard entropy of the system, ∆S°system (go above and find the formula ∆S°f = Σ[∆S°f(products)] – [∆S°f(reactants)]) for the formation of liquid methanol (CH3OH) from carbon monoxide gas and hydrogen gas. First write the balanced equation (because you need the coefficients, right?), and then obtain the entropy values from Table 19.1 on page 913 of your textbook. Now we learn to calculate ∆S°surrounding which is equal to: ∆S°surrounding = qsurroundings = - ∆Hsystem T T If you really look at the equation above, a ∆H less than zero (negative value) will result in a positive entropy (negative times a negative = positive) for the surroundings. Remember the entropy of a system should be opposite in sign from the entropy of the surroundings. 4 Chemistry & Chemical Reactivity: Chapter 19, Principles of Reactivity: Entropy & Free Energy Name______________________ To solve for the total entropy of the universe, we add the two entropies (system and surroundings) together: ∆S°universe = ∆S°system + ∆S°surrounding Problem 4: Show that ∆S°universe is positive (>0) for dissolving NaCl in water. Write the equation for dissolving solid NaCl into aqueous NaCl, then use the values in Appendix L to solve for ∆S°system, ∆S°surrounding, and ∆S°universe. Classify the following reactions as one of the four types of reactions summarized in the following table: (Note: do not use the numbers to answer your homework questions) Predicting Whether a Process Will Be Spontaneous Type ∆H°system ∆S°system Spontaneous? 1 Exothermic Less order Always ∆H°system < 0 ∆S°system > 0 2 Exothermic More order Depends on magnitudes of H/S ∆H°system < 0 ∆S°system < 0 More favorable at lower temperatures 3 Endothermic Less order Depends on magnitudes of H/S ∆H°system > 0 ∆S°system > 0 More favorable at higher temperatures 4 Endothermic More order Never ∆H°system > 0 ∆S°system < 0 Problem 5: Classify the following reactions as one of the 4 types of reactions summarized in the above table. a. CH4 (g) + 2 O2 (g) 2 H2O(l) + CO2 (g) H = -890.6 S = -242.8 5 Chemistry & Chemical Reactivity: Chapter 19, Principles of Reactivity: Entropy & Free Energy Name______________________ b. 2 Fe2O3(s) + 3 C(graphite) 4 Fe(s) + 3 CO2(g) H = +467.9 S = +560.7 c. C (graphite) + O2 (g) CO2 (g) H = -393.5 S = +3.1 d. N2 (g) + 3 F2 (g) 2 NF 3 (g) H = -264.2 S = -277.8 The problem with entropy and enthalpy is we can have reactions that decrease in entropy yet are spontaneous, and we can have endothermic reactions and have spontaneity as well. We still really cannot predict with certainty whether a process is spontaneous or not. Usually a negative enthalpy, ∆H, is spontaneous and usually an increase in entropy is spontaneous, but there are still a few reactions out there that are the opposite and still spontaneous. So, scientists went back to the drawing board to see what was missing in their understanding of thermodynamics. Gibbs Free Energy We use the same ideas already presented for enthalpy and entropy (even how they are calculated…summation of final values minus summation of initial values in Appendix L). ∆G°system = ∆H°system - T∆S°system To make a connection between ∆G°system (which equals ∆G°reaction) and predicting reaction spontaneity use the ideas below: 1. If ∆G°reaction < 0 (or negative), a reaction is spontaneous 2. If ∆G°reaction = 0, the reaction is at equilibrium 3. If If ∆G°reaction > 0 (or positive), a reaction is never spontaneous Finally!! We now have a prediction method that works every time. So what is Gibbs free energy? Free energy is the maximum amount of energy that is available to do useful work. Our bodies function on free energy. Our enzymes, proteins, systems, etc. would not work otherwise. Free energy is the total energy available from the dispersal of energy and matter as bonds break and reform among all the molecules/atoms in the system. Problem 6: Calculate the standard free energy change, If ∆G°reaction for the formation of methane at 298 K. Step 1: Calculate the summation of enthalpy (Appendix L) Step 2: Calculate the summation of entropy (Appendix L) Step 3: Put everything together into the formula: ∆G°system = ∆H°system - T∆S°system using the temperature given in the problem. 6 Chemistry & Chemical Reactivity: Chapter 19, Principles of Reactivity: Entropy & Free Energy Name______________________ Again, ∆G° of formation of any element in its standard state equals zero. You can also calculate ∆G° = ∆G°f products - ∆G°f reactants. Problem 7: Calculate the standard free energy change for the combustion of one mole of methane from the standard free energies of formation of the products and reactants. (Hint: go to Appendix L and use the third column and follow the same procedure as you have when solving for enthalpy of formation and entropy of formation) Really important information to understand: 1. Let’s watch temperature values change the value of G. If temperature is really large, what would happen to the second term of the formula: ∆G°system = ∆H°system - T∆S°system? ___________________________________ If the entropy value is larger than the enthalpy value, we say the reaction is entropy-favored. 2. If the temperature is really low, what would happen to the first term of the formula ∆G° = ∆G°f products - ∆G°f reactants? ________________________________________________ If the enthalpy value is larger than the entropy value, we say the reaction is enthalpy-favored. 3. What you need to understand about temperatures is their impact on the second term of the Gibbs equation. Low and high temperatures can predict the spontaneity of a reaction. Go back and look at the table of reaction types above. We can now re-visit the idea of whether a reaction is product-favored or reactant-favored as well as equilibrium. We don’t typically observe complete conversion of reactants to products in reality because some reactants will simply not encounter other reactants due to lack of close proximity, or proper speed or orientation when colliding. This creates a situation in which the free energy of the pure reactants is slightly different than the free energy of the pure products. Since the conditions are not standard in this case, we need to allow for a correction in the calculation: 7 Chemistry & Chemical Reactivity: Chapter 19, Principles of Reactivity: Entropy & Free Energy Name______________________ ∆G = ∆G° + RT ln Q Where R is the universal gas constant (8.3145 J/mol x K, T is in Kelvin, and Q is the reaction quotient. If ∆G° = 0, then the system is at equilibrium. Substituting zero for ∆G° in the above equation with zero allows us to solve for the equilibrium constant. ∆G = ∆G° + RT ln Q solve for ∆G° ∆G° = - RT ln Q replace Q with K ∆G° = - RT ln K Now we have the following equation where we solve for the Gibbs free energy for the reaction: ∆G°reaction = - RT ln K We can rearrange this equation further to solve for K ∆G°reaction = - RT ln K divide both sides by –RT ln K = - ∆G° now take the ex of the right side of your equation RT K = e - ∆G°/RT Problem 8: Determine the standard free energy change, ∆G°reaction, for the formation of 1.00 mol of ammonia from nitrogen and hydrogen, and use this value to calculate the equilibrium constant, Kp, for this reaction at 25°C. (Hint: the Gibbs free energy of formation of nitrogen and hydrogen to form ammonia equals zero, so you only need the ∆G°f of ammonia from Appendix L. The value for R = 8.3145 J/mol x K, and the value for T should be room temperature in Kelvin) 8 Chemistry & Chemical Reactivity: Chapter 19, Principles of Reactivity: Entropy & Free Energy Name______________________ When we know the value for ∆G°reaction and we solve for K we can predict whether a reaction if product- or reactant-favored. If ∆G°reaction is negative, or < 0, K is greater than 1 (products are favored) If ∆G°reaction is positive, or > 0, K is less than 1 (reactants are favored) If ∆G°reaction is = 0, K = 0 (happens rarely or not at all since that would mean the reaction doesn’t really occur??...what do you think??) Example problems for this concept can be found on pages 930-931. Check them out! Don’t forget to: 1. Read the chapter right away 2. Do your homework taking note of each due date 3. Check out Chapter Goals Revisited, page 933 4. Check out Key Equations, page 934 5. Compare Key Equations with AP Chemistry Equations and Constants handout 6. Review prior chapter goals, key equations, and chapter notes. 9