Energy and Energy and Enthalpy Thermodynamics The internal energy (E) of a system consists of The energy change of a reaction the kinetic energy of all the particles (related to is measured at constant temperature) plus the potential energy of volume (in a bomb interaction between the particles and within the calorimeter). particles (eg bonding). We can only measure the change in energy of the system (units = J or Nm). More conveniently reactions are performed at constant Energy pressure which measures enthalpy change, ∆H. initial state final state ∆H ~ ∆E for most reactions we study. final state initial state ∆H < 0 exothermic reaction Energy "lost" to surroundings Energy "gained" from surroundings ∆H > 0 endothermic reaction < 0 > 0 2 o Enthalpy of formation, fH Hess’s Law o Hess's Law: The heat change in any reaction is the The standard enthalpy of formation, fH , is the change in enthalpy when one mole of a substance is formed from same whether the reaction takes place in one step or its elements under a standard pressure of 1 atm. several steps, i.e. the overall energy change of a reaction is independent of the route taken. The heat of formation of any element in its standard state is defined as zero. o The standard enthalpy of reaction, H , is the sum of the enthalpy of the products minus the sum of the enthalpy of the reactants. Start End o o o H = prod nfH - react nfH 3 4 Example Application – energy foods! Calculate Ho for CH (g) + 2O (g) CO (g) + 2H O(l) Do you get more energy from the metabolism of 1.0 g of sugar or 1.0 g of 4 2 2 2 the fatty acid derived from olive oil? How is energy stored in our bodies? o -1 Data: sucrose C12H22O11, molar mass = 342, H f = -2226 kJ mol o -1 o -1 given: fH (CO2) = -393 kJ mol oleic acid C18H34O2, molar mass = 283, H f = -783 kJ mol o -1 o -1 o -1 H f (CO2) = -393 kJ mol and H f (H2O) = -285 kJ mol fH (H2O) = -285 kJ mol Ho (CH ) = -75 kJ mol-1 f 4 Sugar: C H O (aq) + 12O (g) 12CO (g) + 11H O(l) o -1 12 22 11 2 2 2 fH (O2) = 0 kJ mol by definition Ho = {12(-393) + 11(-285)} – {-2226} = -7851 + 2226 = -5625 kJ mol-1 o o o o o H = fH (CO2) + 2 fH (H2O) - fH (CH4) - 2 fH (O2) Thus, each 1.0 g liberates -5625/342 = 16 kJ of energy. = -393 + 2(-285) - (-75) - 2(0) Fatty Acid: C18H34O2(aq) + 25½O2(g) 18CO2(g) + 17H2O(l) = - 888 kJ mol-1 Ho = {18(-393) + 17(-285)} – {-783} = -11,919 + 783 = -11,136 kJ mol-1 Thus, each 1.0 g liberates -11,136/283 = 39 kJ of energy. Hence fats contain more calories than sugar! 5 6 1 Exam Type Question Spontaneous Reactions In the absence of an adequate supply of oxygen, yeasts obtain Any reaction will have a preferred direction of metabolic energy by fermentation of glucose to produce change. ethanol. It is spontaneous in this direction. C6H12O6(s) 2C2H5OH(l) + 2CO2(g) Use the standard enthalpies of formation to calculate H for this reaction. –1 Thermochemical Data at 298 K Substance fH / kJ mol glucose(s) -1274 CO2(g) -393 C2H5OH(l) -278 2 marks 7 Entropy Entropy - S While most exothermic reactions (H < 0) are Entropy is a measure of the randomness or disorder. spontaneous, some endothermic ones (H > 0) are also The natural progression of things is from an ordered spontaneous. to a disordered state. Entropy (S ) is also important. Entropy is a thermodynamic quantity that describes the Entropy depends on how random a system is: this number of arrangements that are available to the system disorder can be positional or thermal in a given state. The main concept here is the more ways a particular state can be achieved the greater is the likelihood of finding it in that state. 9 10 Entropy Sgas >> Sliquid > Ssolid Temperature and Spontaneity A plot of S versus T Consider the process: H O (l) H O (g) shows the gradual 2 2 increase within a phase and the abrupt We could deduce that: H - endothermic increase with a phase and S iitiis positive change. The molecular views depict Experience shows us: the increase in T>100 C – vaporisation is spontaneous randomness of the T<100 C – condensation is spontaneous particles as the solid melts and in particular So conclude that: T is important, in the spontaneity of reactions as a liquid vaporises. 11 12 2 o Gibbs Free Energy - G Why is G important ? The effects of enthalpy (H) and entropy (S) on a reaction are Knowing the G o values for several reactions allows us to combined to give the Gibbs Free Energy compare the relative tendencies of reactions to occur. o The more negative the value of G the further the reaction will go G = H - TS towards completion. (We must use standard free energy for these comparisons since free energy changes with pressure or concentration). (G can not be measured directly but can be calculated from H and S.) If a reaction is not spontaneous (G o > 0) we can do two Note: G is dependent on temperature things to get it to work: o G > 0 Reaction is not spontaneous Change the temp until G < 0 (in the lab) o G = 0 System is at equilibrium Couple it with another reaction such that overall G < 0 G < 0 Reaction is spontaneous (living systems) 13 14 Example: G°= H°- TS° Question At what temperature will the following reaction become spontaneous? N2 (g) + O2 (g) 2 NO (g) Calculate G (298 K) for the reaction and determine whether it is spontaneous at this temperature. o -1 -1 o -1 Data: S / J K mol Hf / kJ mol N2 (g) 191.4 0 CO (g) + ½ O2 (g) CO2 (g) O2 (g) 204.9 0 NO (g ) 210. 5 90. 4 -1 Data: fG (CO) = -137.2 kJ mol , G (CO ) = -394.4 kJ mol-1 and, by definition, o -1 -1 f 2 S = (2 x 210.5) - (191.4 + 204.9) = + 24.7 J K mol fG (element) = 0. Ho = 2 x 90.4 = 180.8 kJ mol-1 Go = 180.8 - (T x 24.7 x 10-3) kJ mol-1 For spontaneity G < 0 (T x 24.7 x 10-3)> 180.8 T > 180.8 / (24.7 x 10-3) > 7320 K 15 16 The Role of ATP The Role of ATP 2- + - -1 The initial step in the metabolic breakdown of glucose is a Glucose + HPO4 + H [glucose phosphate] + H2O G = +13.8 kJ mol phosphoralation reaction. 4- 3- 2- + -1 ATP + H2O ADP + HPO4 + H G = - 30.5 kJ mol 2- + - Glucose + ATP4- [glucose phosphate]- + ADP3- G = - 16.7 kJ mol-1 Glucose + HPO4 + H [glucose phosphate] + H2O G = +13.8 kJ mol-1 But this is not spontaneous so it is coupled with a reaction that is spontaneous. All organisms use the hydrolysis of Reactions that release energy convert ADP to ATP adenosine triposphate (ATP) to ADP to drive this process. and reactions that require energy are coupled with hydrolysis of ATP to ADP. 4- 3- 2- + ATP + H2O ADP + HPO4 + H -1 G = - 30.5 kJ mol 17 18 3 The Role of ATP The Role of ATP Why is ATP a high energy ion and hence energy is released on hydrolysis of ATP? Coupling of two reactions can not occur if they physically separated from one another. An enzyme ensures they occur in the same vicinity. 4- 3- 2- + -1 ATP + H2O ADP + HPO4 + H G = - 30.5 kJ mol A: release of high charge repulsion B: formation of resonance stabilsed ion 19 20 Chemical Equilibrium Equilibrium The story so far… N2O4 is a colourless gas and decomposes into . Looked at bonding and what makes stable compounds. NO2, a brown gas. Energy change as the basis of chemical reactions. N2O4 (g) 2 NO2 (g) But NO can dimerise to form N O . But… 2 2 4 2 NO (g) N O (g) Many reactions don’t go to 2 2 4 completion. Eventually a balance is Equilibrium is established when reached in which the there is a unique balance of concentration of both reactants and products. 21 gases no longer changes. 22 Dynamic Equilibrium Why does a system reach equilibrium? The reaction has not stopped when equilibrium is reached; instead the rate of the forward reaction equals Equilibrium represents a minimum on a free energy curve. the rate of the reverse reaction. The spontaneous reaction is toward a free energy No net change in the concentrations of reactants and minimum not away from it. products occur. G N2O4 ()(g) 2 NO2 ()(g) A + B cf lowest point of an electric cable stretched between two pylons. C + D G = 0 A + B C + D 23 24 4 The Equilibrium Constant Kc and Kp Equilibrium constants calculated An equilibrium constant may be determined which is characteristic of a reaction at a particular temperature. from concentrations called Kc. [C]c[D]d Concentration is directly a A + b B c C + d D proportional to pressure. K a b [A] [B] Equ ilibri um reacti ons i nvol vi ng gases may b e expressed as Kp in which the partial pressure is [NO ]2 measured in atmospheres. N O (g) 2 NO (g) 2 2 4 2 K = 2 [N2O4] [NH3] Kc = 3 [N2][H2] K is always > 0 N2(g) + 3H2(g) 2NH3(g) p2 If the value of K > 1, products dominate.