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Unit 5 How do we predict chemical change? The central goal of this unit is to help you identify and apply the different factors that help predict the likelihood of chemical reactions. Chemistry XXI M1. Analyzing Structure M2. Comparing Free Energies Comparing the relative stability of different substances Determining the directionality and extent of a chemical reaction. M3. Measuring Rates Analyzing the factors that affect reaction rate. M4. Understanding Mechanism Identifying the steps that determine reaction rates. Unit 5 How do we predict chemical change? Chemistry XXI Module 2: Comparing Free Energies Central goal: To quantitatively determine the directionality and extent of chemical reactions. The Challenge Transformation How do I change it? Chemistry XXI Imagine that you were interested in QUANTITATIVELY determining which of two or more chemical processes could have occurred on the primitive Earth. What measurable properties of the system can be used to make the prediction? How could we actually quantify the directionality and extent of a chemical reaction? Relevant Factors Chemistry XXI The directionality and extent of a chemical reaction depend on three main factors: ENERGETIC FACTORS DHrxn ENTROPIC FACTORS DSrxn TEMPERATURE T But how? What Changes? Chemistry XXI Consider this chemical process: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) Is this a favored process? Why? The Facts It may be favored if the process: a) leads to more energetically stable compounds. energy is released (Exothermic: DHrxn < 0). q Chemistry XXI DHrxn < 0 (-) CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) The Facts It may be favored if the process: b) Leads to the formation of substances with more distinguishable configurations: DSrxn> 0 (+) Chemistry XXI CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) DSrxn > 0 (+) Certainly favored: DHrxn < 0 (-) (exothermic) DSrxn > 0 (+) Let’s Think Consider the possible decomposition of CO2 in the atmosphere: Chemistry XXI CO2(g) C(s) + O2(g) Analyze the sign of DHrxn and DSrxn and decide whether the process is favored or not. The Facts If, as in the previous case, DHrxn > 0 (+) (endothermic) DSrxn < 0 (-) the process is certainly unfavored. Chemistry XXI But what happens if: or DHrxn > 0 (+) (exothermic) DSrxn < 0 (-) ? DHrxn < 0 (-) (endothermic) DSrxn > 0 (+) ? The outcome depends on the TEMPERATURE. The Law In general, energetic effects become less relevant at high temperature, where entropic effects dominate. The sign of this quantity (Gibbs Free Energy, G): DGrxn = DHrxn – TDSrxn Chemistry XXI can be used to determine process directionality. 2nd Law of Thermodynamics At constant T and P: DGrxn < 0 for thermodynamically favored processes. Reaction Extent Chemistry XXI Experimental results indicate that processes at constant T and P occur in the direction in which the Gibbs free energy G of the system decreases. AB DGrxn = GB - GA < 0 Exergonic Favored CD DGrxn = GD – GC > 0 Endergonic Unfavored The more negative DGrxn, the larger the extent of the reaction. The more negative DGrxn, the more thermodynamically stable the products than the reactants. Evaluating Likelihood The quantitative analysis of the extent of reactions is crucial to determine, for example, what were the likely components of our primitive atmosphere. Chemistry XXI H2 is suspected to be an important component of the very early atmosphere. Was CO2 or CH4 more stable in this environment? How does temperature affect the answer? Let’s explore it! Hydrogen constitutes ~75% of the Universe’s elemental mass Let’s Think Consider this data: Reaction CO2(g) + 4 H2(g) CH4(g) + 2 H2O(l) DHorxn (kJ) DSorxn (J/K) -252.7 -410.3 Chemistry XXI Which compound of C (CO2 vs CH4) is most stable in the presence of H2 at 25 oC? Hint: Are reactants or products favored in the reaction? Let’s Think Reaction DHorxn (kJ) DSorxn (J/K) CO2(g) + 4 H2(g) CH4(g) + 2 H2O(l) -252.7 -410.3 DGrxn = DHrxn–TDSrxn = -252.7 – 298.15*0.4103 = -130.4 kJ Chemistry XXI CH4 is the stable forms at 25 oC in the presence of H2. Are there any temperatures at which CO2 is favored? Let’s Think Reaction DHorxn (kJ) DSorxn (J/K) CO2(g) + 4 H2(g) CH4(g) + 2 H2O(l) -252.7 -410.3 As T increases, the energetic contribution becomes less important. There is a temperature at which: Chemistry XXI DGrxn = DHrxn–TDSrxn = 0. Above this temperature, the reverse reaction is favored. DGrxn = -252.7 – T*0.4103 = 0 T = 615.9 K Quantitative Data Chemistry XXI Our discussion highlights the importance of measuring or calculating DHrxn, DSrxn, and DGrxn if we want to make predictions about the thermodynamic likelihood of a reaction. We have already A+BC+D discussed how to o o o DS = S – S rxn products reactants calculate DSrxn given available data for the Basic Assumption: standard molar entropies So (perfect crystal) = 0 at 0 K. of formation Sof. How do we get DHrxn and DGrxn for any process? Heat of Reaction We have seen that the heat of a reaction DHrxn can be measured by calorimetry. However, we can also calculate this quantity using measured standard enthalpies of reaction DHof for the substances involved in the process. Chemistry XXI Let’s consider a reaction involved in the production of CO2 in the primitive atmosphere: C(graphite) + 2 H2O(l) CO2(g) + 2 H2(g) How could we evaluate DHorxn given this information: C(graphite) + O2(g) CO2(g) H2(g) + ½ O2(g) H2O(l) DHof = -393.5 kJ/mol DHof = -285.8 kJ/mol Alternative Routes C(graphite) + 2 H2O(l) CO2(g) + 2 H2(g) We could think of the reaction as combination of a decomposition and a combination steps: 1st: 2 H2O(l) 2 H2(g) + O2(g) Chemistry XXI 2nd: C(graphite) + O2(g) CO2(g) DHorxn = 2 (+285.8) kJ DHorxn = -393.5 kJ + C(graphite) + 2 H2O(l) CO2(g) + 2 H2(g) DHorxn = 2 (285.8) - 393.5 = +178.1 kJ This approach is based on the central idea that, no matter what path we follow, the total energy transfer should be the same. Hess’s Law Generalizing All reactions can be conceived as a sequence of decompositions and combinations steps: Take the general reaction: aA+bBcC+dD Chemistry XXI Imagine it happens in the following steps: Reaction Decomposition of a moles of A Energy Transfer a x (-DHof,A) Decomposition of b moles of B b x (-DHof,B) Formation of c moles of C c x DHof,C Formation of d moles of D d x DHof,D DHorxn = c DHof,D + d DHof,D – a DHof,A – b DHof,B DHorxn = S DHof,products - S DHof,reactants Reference State DHorxn = S DHof,products - S DHof,reactants It is important to notice that in this relationship, we only need the DHof of the chemical COMPOUNDS involved, not of any elements. Chemistry XXI HOW SO? DH 0 Elements Reactants Products We are taking DHof = 0 for the elements in their standard state. Gibbs Free Energy An identical procedure can be used to calculate the DGorxn given the DGof for the different chemical COMPOUNDS involved in the reaction: DGorxn = S DGof,products - S DGof,reactants DGof = 0 for elements in their standard state. Chemistry XXI Alternatively, we can also calculate DGorxn as: DGorxn = DHorxn–TDSorxn This route is useful in estimating the effect of T on the value of DG. Let’s Think One central question in the theories about the origin of life is how complex organic compounds were synthesized from simpler molecules such as H2, N2, CH4, NH3, and H2O. Consider these possibilities in the synthesis of the simplest amino acid, glycine (C2H5NO2): Chemistry XXI 2 CO(g) + NH3(g) + H2(g) C2H5NO2(s) 2 CH4(g) + NH3(g) + 2 H2O(l) C2H5NO2(s) + 5 H2(g) 1. Calculate DHorxn, Sorxn, and DGorxn for these two processes. Let’s Think 2 CO(g) + NH3(g) + H2(g) C2H5NO2(s) Chemistry XXI 2 CH4(g) + NH3(g) + 2 H2O(l) C2H5NO2(s) + 5 H2(g) Substance DHof (kJ/mol) DGof (kJ/mol) Sof (J/(mol K) H2(g) 0 0 130.7 CH4(g) -74.6 -50.5 186.3 CO(g) -110.5 -137.2 197.7 NH3(g) -45.9 -16.4 192.8 H2O(l) -285.8 -237.1 70.0 C2H5NO2(s) -528.4 -368.8 103.5 Let’s Think DHorxn DSorxn DGorxn (kJ) (J/K) (kJ) 2 CO(g) + NH3(g) + H2(g) C2H5NO2(s) -261.5 -615.4 -78.0 2 CH4(g) + NH3(g) + 2 H2O(l) C2H5NO2(s) + 5 H2(g) 238.3 51.6 223. Chemistry XXI Reaction 2. Identify which reactions are product-favored and discuss the effect of temperature on the extent of each process. Let’s Think DHorxn DSorxn DGorxn (kJ) (J/K) (kJ) 2 CO(g) + NH3(g) + H2(g) C2H5NO2(s) -261.5 -615.4 -78.0 2 CH4(g) + NH3(g) + 2 H2O(l) C2H5NO2(s) + 5 H2(g) 238.3 51.6 223. Chemistry XXI Reaction The first reaction is product-favored at 25 oC; it becomes reactant-favored at high T. The second reaction is reactant-favored at 25 oC; it becomes more product-favored at high T. Reaction Extent Although the change in Gibbs free energy DGorxn allows us to make predictions about reaction directionality, the actual number tells us little about the final relative amounts of product to reactants. Chemistry XXI How can we better quantify “reaction extent”? Experimental results indicate that chemical reactions tend to reach an “equilibrium” state in which the concentration of products and reactants does not vary over time. C o n c e n t r a t i o n Time Equilibrium Constant At chemical equilibrium, the following ratio of the concentrations, or gas pressures, of reactants to products remains constant: Equilibrium Constant Chemistry XXI aA+bBcC+dD For reaction involving gases, the equilibrium constant may be expressed in terms of the partial pressures exerted by each gas: c d [C ] [ D] Kc a b [ A] [ B] c d C D a b A B P P Kp P P DGorxn vs. K The value of the equilibrium constant K is directly related to the DGorxn for the process: o DGrxn ( ) o RT DGrxn RT ln K K e Chemistry XXI Then: K e o o DH rxn DS rxn ( ) RT R In general: R = 8.314 J/(K mol) Product-Favored Process: DGorxn < 0 K>1 Reactant-Favored Process: DGorxn > 0 K<1 Chemistry XXI Let’s Think Predict the effect of changing DHorxn, DSorxn, and T on the value of K and the reaction extent. K e o o DH rxn DS rxn ( ) RT R Reaction Extent As we have seen, different reaction reach equilibrium states with different proportion of products and reactants in the system. For example, combustion reactions tend to have very large K values: CH4(g) + O2(g) CO2(g) + 2 H2O(g) DGorxn = -800.9 kJ Chemistry XXI K = 2.07 x 10140 at 25 oC For all practical purposes, this reaction fully goes to completion at all temperatures (DHorxn < 0, DSorxn> 0): CH4(g) + O2(g) CO2(g) + 2 H2O(l) Reaction Extent But what about a reaction such as this: CO(g) + H2O(l) CH2O2(l) DGorxn = -12.9 kJ K = 182.0 at 25 oC (DHorxn < 0, DSorxn < 0) Chemistry XXI which may have played a central role in the formation of complex molecules in our planet? Whether this reaction is product- or reactant-favored depends on the actual conditions of the process. Thus, it is better to represent it as: CO(g) + H2O(l) CH2O2(l) to highlight the importance of chemical equilibrium. Chemical Evolution In 1956, Urey and Miller conducted a classic experiment on the origin of life. Chemistry XXI The experiment showed that biological molecules, such as amino acids, can form from simple reactants. Since then, many experiments have been performed with different reactant mixtures and sources of energy (heat, UV, X-rays, etc.) Primitive Mixtures These are two of the reactant mixtures that have been tested: H2O, CH4, NH3, H2 CO, N2, H2 Results indicate that the elemental composition (presence of C, H, N, O) of the mixture is more relevant than the kinds of molecules used. Why? Chemistry XXI No matter the mixture, the concentration of species seems to be controlled by these equilibriums: CH4(g) + 2 H2O(g) CO2(g) + 4 H2(g) CO2(g) + H2(g CO(g) + 3 H2O(g) N2(g) + 3 H2(g) 2 NH3(g) Likely occurring between 500-1000 oC. Let’s Think CH4(g) + 2 H2O(g) CO2(g) + H2(g) N2(g) + 3 H2(g) CO2(g) + 4 H2(g) CO(g) + H2O(g) 2 NH3(g) Chemistry XXI Substance DHof (kJ/mol) Sof (J/(mol K) Estimate K for these processes at 500 oC and 800 oC. H2(g) 0 130.7 CH4(g) -74.6 186.3 CO2(g) -393.5 213.8 CO(g) -110.5 197.7 NH3(g) -45.9 192.8 H2O(g) -241.8 188.8 N2(g) -108.6 191.6 Let’s Think o o DH rxn DS rxn K exp( ) RT R DHor (kJ) DSor K500 (J/K) K800 CO2(g) + 4 H2(g) 164.7 172.7 0.0078 10.1 42.0 1.54 Reactions CH4(g) + 2 H2O(g) Chemistry XXI CO2(g) + H2(g) N2(g) + 3 H2(g) R = 8.314 J/(K mol) CO(g) + H2O(g) 2 NH3(g) 41.2 -91.8 0.257 -198.1 7.16 x 1.32x 10-5 10-6 Discuss which compounds 500 oC N2, H2, H2O, CH4 of C, H, N, and O are more oC N , H , CO , CO 800 2 2 2 likely to exist at each T. Chemistry XXI Let′s apply! Assess what you know Let′s apply! Going Up-Hill The synthesis reactions for the formation of many amino acids, proteins, carbohydrates and fats starting from simple molecules have DGorxn > 0. Chemistry XXI One central question in the origin of life is how it was possible to induce and sustain the synthesis of these types of substances. 6 CO2(g) + 6 H2O(l) C6H12O6(s) + 6 O2(g) DGorxn = 2880 kJ Glucose Let′s apply! Forced Reactions A chemical reaction with DGorxn > 0 is not favored thermodynamically, but can be forced to happen by coupling it with other favored reactions. This commonly happens in biological systems. For example, consider an hypothetical reaction: Chemistry XXI A + B AB DGo1 > 0. Reactions like this in living organisms are often coupled with the favored hydrolysis of ATP: ATP(aq) + H2O(l) ADP(aq) + Pi(aq) DGo2 = -30.5 kJ Let′s apply! Forced Reactions Chemistry XXI The coupling is often accomplished by the following mechanism: The formation of A-Pi facilitates the addition of B. A + ATP + H2O A-Pi + ADP A-Pi + B AB + Pi A + B + ATP + H2O AB + ADP + Pi DGorxn = DGo1 + DGo2 < 0 Predict Let′s apply! For example, glutamine is an amino acid synthesized many organisms from another amino acid, called glutamic acid. B C Chemistry XXI AB DGorxn = 14.2 kJ Glutamic Acid AB + C AC + B AC Glutamine Express and calculate K for this reaction at 25 oC. K = [AB][C]/[AC][B] = 3.25 x 10-3 Let′s apply! Predict Chemistry XXI Write the overall reaction when the synthesis of glutamine is coupled to the hydrolysis of ATP and calculate its K at 25 oC. AB + C AC + B DGo1 = 14.2 kJ ATP(aq) + H2O(l) ADP(aq) + Pi(aq) DGo2 = -30.5 kJ AB + C + ATP + H2O AC + B + ADP + Pi DGorxn = DGo1 + DGo2 = -16.3 kJ K = 717 Propose a mechanism for the coupled process. Predict Let′s apply! Glutamic Acid (AB) Glutamine (AC) AB + C AC + B DGo1 = 14.2 kJ ATP(aq) + H2O(l) ADP(aq) + Pi(aq) DGo2 = -30.5 kJ Chemistry XXI AB + ATP + H2O AB-Pi + ADP AB-Pi + C AC + Pi + B AB + C + ATP + H2O AC + B + ADP + Pi Chemistry XXI Discuss with a partner how the values of DHorxn, DSorxn, and DGorxn affect the directionality and extent of a chemical process. Comparing Free Energies Summary At constant T and P: DGrxn = DHrxn – TDSrxn < 0 for thermodynamically favored processes. Chemistry XXI a A + b B c C + d D DHorxn DSorxn + + + + DGorxn - DIRECTIONALITY? Product-favored + Reactant favored Depends on T P fav at low T Depends on T P fav at high T Equilibrium Constant a A + b B c C + d D c EXTENT? d [C ] [ D] Kc a b [ A] [ B] DG o rxn Chemistry XXI K e RT ln K o o DH rxn DS rxn ( ) RT R In general: Product-Favored Process: DGorxn < 0 K>1 Reactant-Favored Process: DGorxn > 0 K<1 Chemistry XXI For next class, Investigate how the rate of a chemical reaction can be measured or calculated. How can we use experimental data to track the rate of a chemical reaction as a function of time?