Thermochemistry Chapter 4 Thermochemistry is the study of energy changes that occur during chemical reactions Focus is on heat and matter transfer between the system and the surroundings
Energy The ability to do work or transfer heat. Work is done when the energy applied causes an object that has mass to move. kinetic energy - energy of motion potential energy energy due to position energy associated with forces of attraction and repulsion between objects
Units of Energy The SI unit of energy is the joule (J). kg m 2 1 J = 1 s 2 An older, non-si unit is still in widespread use called the calorie (cal). 1 cal = 4.184 J
Conservation of Energy Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. E = E final - E initial
The energy of a system depends only on its present state, not on the path by which the system arrived at that state. So, E depends only on E initial and E final. In the system below, the water could have reached room temperature from either direction.
Exchange of Heat - System and When heat is released by the system to the surroundings, the process is exothermic Surroundings When heat is absorbed by the system from the surroundings, the process is endothermic.
Endothermicity and Exothermicity Process is exothermic when H is negative. Process is endothermic when H is positive.
Atoms and molecules possess energy kinetic energy proportional to the absolute temperature potential energy chemical bonds hold atoms together, forces of attraction give rise to a compound s potential energy (chemical energy) bonds break - increase PE of system bonds made - lower PE of system Changes in chemical energy occur during chemical reactions
Heat Capacity and Specific Heat heat capacity - the amount of energy required to raise the temperature of a substance by 1 K (1 C). specific heat capacity - the amount of energy required to raise the temperature of 1 g of a substance by 1 K.
Heat Capacity and Specific Heat To calculate a quantity of heat (q) heat = specific heat mass temperature change q = s m T J = J g -1 K -1 g K
Specific Heats q = s m T If T > 0, then q > 0 and heat is gained by the system If T < 0, then q < 0 and heat is lost by the system Molar heat capacity is the product of specific heat times the molar mass of a substance (units are J mol 1 K 1 )
Calorimetry Calorimetry is a technique used to measure heat exchange in chemical reactions A calorimeter is the device used to make heat measurements. Calorimetry is based on the law of conservation of energy. qwater = - qsurroundings
Coffee Cup Calorimeter We indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter The specific heat for water is well known (4.184 J/g/K), We can measure H for the reaction with this equation: q = m s T
Calorimetry example How much heat is given off when an 869 g iron bar cools from 94 C to 5 C? (specific heat of Fe = 0.444 J g -1 K -1 )? q= m x s x t t = t final t initial = 5 C 94 C = -89 C q = 869 g x 0.444 J/g/K x -89 K = -34,000 J (q < 0, heat lost!)
Example 7-4 In a neutralisation reaction between a strong acid and a strong base, the following reaction forms water. H + (aq) + OH - (aq) H 2 O(l) If 25.00 ml of 2.50 M HCl and 25.00 ml of 2.50 M NaOH, both at 21.1 0 C react and the final temperature is 37.8 0 C, determine the heat of the neutralisation reaction expressed per mole of H 2 O formed.
Solution: Assume that 50.00 ml of water forms which absorbs all the heat. Heat of reaction = q neutr = -q calorim q calorim = m x s x T = = q neutr = -q calorim =
n(h + ) = C x V = = H + (aq) + OH - (aq) H 2 O(l) n(h + ) = n(oh - ) = n(h 2 O) = 0.0625 mol Amount of heat produced per mole of H 2 O is q neutr = = (Since q < 0, heat is lost, that is neutralisation reaction is exothermic!)
Enthalpy Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure standard enthalpy of reaction ( H rxn): the enthalpy change for the transformation of reactants in their standard states to products in their standard states. There are many types of heats of reaction, e.g. heat of formation, heat of combustion, heat of hydrogenation.
Standard State Standard state is the reference state for the material's thermodynamic state properties. Required for comparison purposes. Standard state of any substance the physical state at which it is most stable at 1 bar and 298 K. When a reaction occurs at constant pressure, the heat of reaction is equal to the enthalpy change.
Enthalpies of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = H products H reactants
Hess s law states that If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps. Hess s Law
Hess s Law The total enthalpy change depends only on the initial state of the reactants and the final state of the products.
Calculation of H We can use Hess s law in this way: H o rxn = Σ n Ho f(products) - Σ m Ho f(reactants) where n and m are the stoichiometric coefficients.
Heat of Combustion The heat of combustion is the heat released in the reaction of one mole of a substance in its standard state with oxygen. The equation for the combustion of a compound must show the reaction of one mole of the compound with sufficient oxygen to convert all of the carbon and hydrogen present to gaseous CO 2 and liquid H 2 O. C 6 H 6 (l) + 7½O 2 (g) 6CO 2 (g) + 3H 2 O(l) H = 3274 kj CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H = 890.2 kj
Example The heat of combustion of ethane gas, C 2 H 6 (g), is 1560.1 kj mol -1. If H for CO 2 (g) and H 2 O(l) are 393.5 kj mol -1 and 285.8 kj mol -1 respectively, calculate the heat of formation of ethane. combustion equation: C 2 H 6 (g) + 3½O 2 (g) 2CO 2 (g) + 3H 2 O(l) H o rxn = Σ H o f (products) Σ Ho f (reactants) = [(2 mol) H o f (CO 2 ) + (3 mol) H o f (H 2 O)] [(1 mol) H o f (C 2 H 6 ) + (3½ mol) H o f (O 2 )] 1560.1 kj = (2 x 393.5) + (3 x 285.8) H o f (C 2 H 6 ) 0 kj H o f (C 2 H 6 ) = 84.3 kj mol 1
Enthalpy of Formation The standard heat of formation of a compound ( H): the heat change when one mole of the compound in its standard state is formed from its elements in their standard states at a specified temperature. The standard enthalpy of formation of an element in its standard state is taken to be zero.
Enthalpies of Formation
Standard Enthalpies of Formation Standard enthalpies of formation, H 0 f, are measured under standard conditions (25 C and 1.00 atm pressure).
Calculate the standard enthalpy of formation of CS 2 (l) given that: C(graphite) + O 2 (g) CO 2 (g) H 0 = -393.5 kj rxn S(rhombic) + O 2 (g) SO 2 (g) H 0 rxn= -296.1 kj CS 2 (l) + 3O 2 (g) CO 2 (g) + 2SO 2 (g) H 0 rxn= -1072 kj 1. Write the enthalpy of formation reaction for CS 2 C(graphite) + 2S(rhombic) CS 2 (l)
2. Add the given rxns so that the result is the desired rxn. (molar ratios!) + C(graphite) + O 2 (g) CO 2 (g) H 0 rxn = -393.5 kj 2S(rhombic) + 2O 2 (g) 2SO 2 (g) H 0 CO 2 (g) + 2SO 2 (g) CS 2 (l) + 3O 2 (g) H 0 C(graphite) + 2S(rhombic) CS 2 (l) rxn rxn = -296.1x2kJ = +1072kJ H 0 rxn = -393.5 + (2x-296.1) + 1072 = 86.3 kj
Benzene (C 6 H 6 ) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kj/mol. 2 C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) H rxn 0 = Σn H 0 f (products)- Σm H 0 f (reactants) H rxn 0 = [ 12 H 0 f(co 6 H 0 2 ) + f(h 2 O) ] - [ 2 H 0 f(c 6 H 6 )] H 0 rxn = [ 12x 393.5 + 6x 285.8 ] [ 2x49.04 ] = -6535 kj -6535 kj 2 mol = -3267 kj/mol C 6 H 6
Application of Hess s Law Find H of NO 2 (g) given that ½N 2 (g) + ½O 2 (g) NO(g) 2NO 2 (g) 2NO(g) + O 2 (g) H f = 90.25 kj..(1) H f = 114.14 kj (2) Desired equation: ½N 2 (g) + O 2 (g) NO 2 (g) H =? 2NO(g) + O 2 (g) 2NO 2 (g) NO(g) + ½O 2 (g) NO 2 (g) ½N 2 (g) + ½O 2 (g) NO(g) NO(g) + ½O 2 (g) NO 2 (g) ½N 2 (g) + O 2 (g) NO 2 (g) H = 114.14 kj H = 57.07 kj..(-2)/2 H = 90.25 kj H = 57.07 kj H = 33.18 kj
Bond Energies bond breaking - endothermic bond formation exothermic Bond energies indicate the amount of energy required to break a particular bond and are therefore positive. H = [sum of bond energies of all bonds in reactants (bonds broken)] [sum of bond energies of all bonds in products (bonds formed)]
Example - hydrogenation reaction C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) Bond energies of reactants = E C=C + 4E C H + E H H = 606.1 + 4(409.64) + 430.5 kj = 2675.16 kj Bond energies of products = E C C + 6E C H = 334.4 + 6(409.64) kj = 2792.24 kj H = 2675.16 2792.24 kj = 117.08 kj mol 1