Introductory Physical Chemistry Course Notes PDF

Introductory Physical Chemistry CHM 101 / Chemistry Department Dr. F. Akinwunmi 1 Thermochemistry Description 2  Thermochemistry Thermodynamics is the science of the relationships between heat and other forms of energ...

Introductory Physical Chemistry CHM 101 / Chemistry Department Dr. F. Akinwunmi 1 Thermochemistry Description 2  Thermochemistry Thermodynamics is the science of the relationships between heat and other forms of energy. Thermochemistry deals with the study of the quantities of heat absorbed or evolved (released) by chemical reactions. Chemical reactions occurs generally either by gaining heat from the surrounding or losing heat (a form of energy) to the surroundings. In chemical reactions, energy changes occur when the reactant transform to product as a result of the difference in chemical energy (internal energy) between the reactant(s) of a given reaction. 3  Energy changes Energy is defined as the potential or the capacity to move matter. It is the ability to do work. [Unit: Joule (J)]. Forms of Energy Heat Light Electrical Chemical Mechanical These forms are interconvertible The transformation of Chemical energy (energy of substances) into heat during chemical reactions is vital The internal energy of substances is usually defined in terms of kinetic and potential energies of the particles making up the substance. 4 Kinetic energy [Ek] ▪ [Ek] is the energy associated with an object by virtue of its motion. ▪ Mathematically: E = 1/2mv2 k where m = mass of object (kg) and v is its velocity or speed (m/s). Example Determine the K.E of an object with mass of 60 kg whose speed is 26.8m/s. Ek = 1/2mv2 = 0.5 x 60kg x (26.8m/s)2 = 21547.2 kg m2/s2 = 21547.2 J = 2.15472 x 104 J = 21.5472 kJ. Note: 1 cal = 4.184J [Also 1 kcal = 4.184 kJ] Thus, 21547.2 J = 21547.2/4.184 cal = 5150 cal = 5.15 kcal. Potential energy [Ep] ▪ [Ep] is the energy an object has by virtue of its position in a field of force. ▪ Mathematically: Ep = mgh where m = mass of object (kg), g = gravitational constant (m/s2), h = height (m) 5  Heat content The heat content of a substance is the internal energy possessed by the substance due to its structure and physical state. It is also referred to as ENTHALPY (H). The change in enthalpy for a reaction at a given temperature and pressure (called the enthalpy of reaction) is obtained by subtracting the enthalpy of the reactants from the enthalpy of the products. Enthalpy change = Heat content of product – Heat content of reactant ΔH = Hp – Hr ------------------------------------------- (1) Where ΔH is the Enthalpy change, Hp is the heat content of product and Hr is the heat content of reactant ΔH = mcΔϴ ---------------------------------------------- (2) Where m = mass, c = specific heat capacity and Δϴ is change in temperature 6  Heat of reaction (qp) ▪ This is the amount of heat evolved or absorbed when a chemical reaction occurs between molar quantities of the substances as represented in the equation of reaction under standard conditions. ❖ The enthalpy of reaction equals the heat of reaction at constant pressure. ie ΔH = qp ▪ Chemical reaction can be divided into 2 types [in terms of heat of reaction and the surrounding]: ✓ Endothermic reaction ✓ Exothermic reaction Endothermic reaction ❑ A chemical reaction in which energy/heat is absorbed from the surrounding for product to be formed. ❑ The heat absorbed during the reaction results in a decrease in temperature of its surroundings (q is positive). o E,g thermal decomposition of potassium trioxochlorate (V). 2KClO3 → 2KCl + 3O2 ΔH = +ve 7  Ea Fig 1: Energy Profile diagram for Endothermic reaction [ NO + Cl2 → NOCl + Cl ] 8  Exothermic reaction ▪ A chemical reaction in which energy/heat is evolved to the surrounding for product to be formed (q is negative). ❑ Examples ✓ Reaction between CaO and water CaO(s) + H2O(l) → Ca(OH)2(aq) Ea ✓ neutralization reaction between an acid and base. NaOH(aq) + HCl(aq) → NaCl(s)+ H2O(l) ---------------- ✓ Suppose that in an experiment, 1 mol of methane burns in oxygen and evolves 890 kJ of heat: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). The reaction is exothermic. Fig 2: Energy Profile diagram for Therefore the heat of reaction, q, is -890 kJ. Exothermic reaction 9 In Summary: Practice Question Ammonia burns in the presence of a platinum catalyst to give nitric oxide, NO. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) In an experiment, 4 mol NH3 is burned and evolves 1170 kJ of heat. Is the reaction endothermic or exothermic? What is the value of q? 10  Thermochemical Equation ▪ A thermochemical equation is the chemical equation for a reaction (including phase labels) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation. ❑ The enthalpy change, ΔH, depends on the phase of the substances. Examples 2Na(s) + 2H2O(l ) → 2NaOH(aq) + H2(g); ΔH = -368.6 kJ 2H2(g) + O2(g) → 2H2O(g); ΔH = -483.7 kJ 2H2(g) + O2(g) → 2H2O(l ); ΔH = -571.7 kJ Solved Examples 1. Aqueous sodium hydrogen carbonate solution (baking soda solution) reacts with hydrochloric acid to produce aqueous sodium chloride, water, and carbon dioxide gas. The reaction absorbs 12.7 kJ of heat at constant pressure for each mole of sodium hydrogen carbonate. Write the thermochemical equation for the reaction. Solution Na2CO3(aq) + HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g), ΔH = +12.7 kJ 11 2. A propellant for rockets is obtained by mixing the liquids hydrazine, N2H4, and dinitrogen tetroxide, N2O4. These compounds react to give gaseous nitrogen, N2 and water vapor, evolving 1049 kJ of heat at constant pressure when 1 mol N2O4 reacts. Write the thermochemical equation for this reaction. Solution 2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g) , ΔH = -1049 kJ Practice Question When 1 mol of iron metal reacts with hydrochloric acid at constant temperature and pressure to produce hydrogen gas and aqueous iron(II) chloride, 89.1 kJ of heat evolves. Write a thermochemical equation for this reaction. MANIPULATING THERMOCHEMICAL EQUATIONS ❑ When a thermochemical equation is multiplied by any factor, the value of ΔH for the new equation is obtained by multiplying the value of ΔH in the original equation by that same factor. ❑ When a chemical equation is reversed, the value of ΔH is reversed in sign. 12 Example Consider the thermochemical equation for the synthesis of ammonia N2(g) +3H2(g) → 2NH3(g); ΔH = -91.8 kJ Doubling the previous equation, you obtain 2N2(g) + 6H2(g) → 4NH3(g); ΔH = -184 kJ Suppose you reverse the first equation we wrote for the synthesis of ammonia. Then the reaction is the dissociation of 2 mol of ammonia into its elements. The thermochemical equation is 2NH3(g) → N2(g) + 3H2(g); ΔH = +91.8 kJ Practice Question When white phosphorus burns in air, it produces phosphorus(V) oxide. P4(s) + 5O2(g) → P4O10(s); ΔH = -3010 kJ What is ΔH for the following equation? P4O10(s) → P4(s) + 5O2(g) 13  Calculating the Heat of Reaction from Stoichiometry Worked Examples ❑ How much heat is evolved when 9.07 x 105 g of ammonia is produced according to the following equation? (Assume that the reaction occurs at constant pressure.) N2(g) + 3H2(g) → 2NH3(g); ΔH = - 91.8 kJ Solution Production of 2 moles of ammonia involved the release of 91.8 kJ of heat Therefore, 1 mole = -91.8/2 = -45.9 kJ Mole = m/M = 9.07 x 105 g/17gmol-1 Heat produced when 9.07 x 105 g of ammonia is produced: 9.07 x 105 g/17gmol-1 X -45.9 kJ = -24.489 x 105 kJ = -2.45 x 106 kJ 14 ❑ How much heat evolves when 10.0 g of hydrazine reacts according to the reaction: 2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g) , ΔH = -1049 kJ 2 moles of hydrazine reacts to liberate -1049 KJ of heat Therefore, 1 mole will produce 1049/2 = -528.5 kJ Molar mass of hydrazine = (2 x 14) + (4 x 1) = 28 + 4 = 32g/mol No of mole of hydrazine when 10 g reacts: 10/32 mol Amount of heat liberated: 10/32 x [-528.5] kJ = 165.2 kJ Practice Question Hydrogen sulfide, H2S, is a foul-smelling gas. It burns to form sulfur dioxide. 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g); ΔH = -1036 kJ Calculate the enthalpy change to burn 28.5 g of hydrogen sulfide. 15  Heat capacity ▪ The heat capacity (C) of a sample of substance is the quantity of heat needed to raise the temperature of the sample of substance by one degree Celsius (or one kelvin). ▪ Mathematically, q = CΔt where Δt = final temp – initial temp (tf - ti) Example: A piece of iron requires 6.70 J of heat to raise the temperature by one degree Celsius. Its heat capacity is therefore 6.70 J/⁰C. The quantity of heat required to raise the temperature of the piece of iron from 25.0⁰C to 35.0⁰C is q = CΔt = (6.70 J/⁰C) = (35.0⁰C - 25.0⁰C) = 67.0 J ❑ The molar heat capacity of a substance is its heat capacity for one mole of substance. Specific heat capacity ▪ The specific heat capacity (s) (or simply specific heat) is the quantity of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one kelvin) at constant pressure. ▪ Mathematically, q = s x m x Δt where s = specific heat capacity, m = mass, 16 Δt = final temp – initial temp (t - t ) Example Calculate the heat absorbed by 15.0 g of water to raise its temperature from 20.0⁰C to 50.0⁰C (at constant pressure). The specific heat of water is 4.18 J/(g.⁰C). q = s x m x Δt = 4.18 J/g.⁰C x 15 g x (50 - 20) ⁰C = 4.18 x 15 x 30 J = 1881J = 1.88 x 103 J = 1.88 kJ Practice question Iron metal has a specific heat of 0.449 J/(g. ⁰C). How much heat is transferred to a 5.00-g piece of iron, initially at 20.0⁰C, when it is placed in a pot of boiling water? Assume that the temperature of the water is 100.0⁰C, and that the water remains at this temperature, which is the final temperature of the iron. 17  Measurement of Heat of Reaction ▪ Heat of reaction is measured by using a calorimeter, a device used to measure the heat absorbed or evolved during a physical or chemical change. ▪ For reactions involving gases, a bomb calorimeter is generally used. Bomb calorimeter 18  Calculating ΔH from Calorimetric Data Example 0.562 g of graphite is placed in a calorimeter with an excess of oxygen at 25.00⁰C and 1 atm pressure. The graphite is ignited, and it burns according to the equation: C(graphite) + O2(g) → CO2(g) On reaction, the calorimeter temperature rises from 25.00⁰C to 25.89⁰C. The heat capacity of the calorimeter and its contents was determined in a separate experiment to be 20.7 kJ/⁰C. What is the heat of reaction at 25.00⁰C and 1 atm pressure? Express the answer as a thermochemical equation. Solution The heat released by the reaction is absorbed by the calorimeter and its contents. Let qrxn be the quantity of heat from the reaction mixture, and let Ccal be the heat capacity of the calorimeter and contents. The quantity of heat absorbed by the calorimeter is CcalΔt. This will have the same magnitude as qrxn, but the opposite sign: qrxn = - CcalΔt. The heat from the graphite sample is: qrxn = -CcalΔt = -20.7 kJ/°C x (25.89°C - 25.00°C) = -20.7 kJ/°C x 0.89°C = -18.4 kJ The factor to convert grams C to kJ heat is -18.4 kJ/0.562 g C. The conversion of 1 mol C to kJ heat for 1 mol (ΔH) is: -18.4 kJ x 12/0.562 = -3.9 x 102 kJ When 1 mol of carbon burns, 3.9 x 102 kJ of heat is released 19 The thermochemical equation: C(graphite) + O2(g) → CO2(g); ΔH = -3.9 x 102 kJ Practice question Suppose 33 mL of 1.20 M HCl is added to 42 mL of a solution containing excess sodium hydroxide, NaOH, in a coffee-cup calorimeter. The solution temperature, originally 25.0⁰C, rises to 31.8⁰C. Give the enthalpy change, ΔH, for the reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 20  HESS LAW OF CONSTANT HEAT SUMMATION ❑ states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps. Worked Examples 1. Find the enthalpy change for the combustion of graphite (carbon) to carbon monoxide. 2C(graphite) + O2(g) → 2CO(g) ▪ The combustion of graphite to carbon monoxide takes place in two separate steps: 2C(graphite) + 2O2(g) → 2CO2(g) (first step) 2CO2(g) → 2CO(g) + O2(g) (second step) Given: C + O2 → CO2, ΔH =-393.5 kJ per mole of CO2 and 2CO(g) + O2(g) → 2CO2(g); ΔH = -566.0 kJ ▪ By Hess’s law, the enthalpy change for the overall equation (which is the equation you want) equals the sum of the enthalpy changes for the two steps. 2C(graphite) + 2O2 → 2CO2 2CO2 → 2CO + O2 21 2C(graphite) + O2 → 2CO (Overall step) first step; ΔH = -393.5 x 2 = -787 kJ Second step; ΔH = -566.0 x (-1) = +566.0 kJ [reversing the equation] Adding these steps algebraically: ΔH = -787 + 566 kJ = - 221 kJ Thus, the combustion of 2 mol of graphite to produce Carbon monoxide has an enthalpy change of -221kJ i.e evolved 221kJ of heat. 22 2. Suppose you are given the following data: S(s) + O2(g) → SO2(g); ΔH = -297 kJ (1) 2SO3(g) → 2SO2(g) + O2(g); ΔH =+198 kJ (2) How could you use these data to obtain the enthalpy change for the following equation? 2S(s) + 3O2(g) → 2SO3(g) (3) Solution By Hess Law: Multiply (1) by 2: 2S(s) + 2O2(g) → 2SO2(g); ΔH = -297 kJ x 2 =-594 kJ Reverse equation (2): 2SO2(g)+ O2(g) → 2SO3(g); ΔH = -198 kJ Combining the two equations, we have: 2S(s) + 2O2(g) → 2SO2(g); ΔH = -297 kJ x 2 =-594 kJ 2SO2(g)+ O2(g) → 2SO3(g); ΔH = -198 kJ 2S(s) + 3O2(g) → 2SO3(g), ΔH = - 594 + (- 198)kJ = - 594 – 198 kJ = - 792 kJ 23 Thus, the enthalpy change for the reaction is -792 kJ 3. What is the enthalpy of reaction, ΔH, for the formation of tungsten carbide, WC, from the elements? W(s) + C(graphite) → WC(s) The heats of combustion of the elements and of tungsten carbide can be measured easily: 2W(s) + 3O2(g) → 2WO3(s); ΔH = -1685.8 kJ (1) C(graphite) + O2(g) → CO2(g); ΔH = -393.5 kJ (2) 2WC(s) + 5O2(g) → 2WO3(s) + 2CO2(g); ΔH = - 2391.8 kJ (3) Solution Recall, the overall reaction is: W(s) + C(graphite) → WC(s) Multiply equation (1) by 1: 2W(s) + 3O2(g) → 2WO3(s); ΔH = -1685.8 kJ Multiply equation (2) by 2: To obtain the overall reaction: 2C(graphite) + 2O2(g) → 2CO2(g); ΔH = -393.5 kJ x 2 = - 787 kJ Divide equation (4) by 2: Reverse equation (3): W(s) + C(graphite) → WC(s) ΔH = - 81kJ/2 2WO3(s) + 2CO2(g) → 2WC(s) + 5O2(g), ΔH = + 2391.8 kJ = - 40.5kJ Combining these equations yields: Thus, the enthalpy of reaction (ΔH) for the formation of 2W(s) + 3O2(g) → 2WO3(s); ΔH = -1685.8 kJ tungsten carbide is – 40.5kJ. 2C(graphite) + 2O2(g) → 2CO2(g); ΔH = -393.5 kJ x 2 = - 787 kJ 2WO3(s) + 2CO2(g) → 2WC(s) + 5O2(g), ΔH = + 2391.8 kJ 2W(s) + 2C(graphite) → 2WC(s) ΔH = -1685.8 kJ + (- 787 kJ) + 2391.8 kJ ΔH = - 81kJ 2W(s) + 2C(graphite) → 2WC(s) ΔH = - 81kJ (4) 24  Standard enthalpy of formation ❑ The standard enthalpy of formation (also called the standard heat of formation) of a substance, denoted ΔH⁰f, is the enthalpy change for the formation of one mole of the substance in its standard state from its elements in their reference (stablest) form and in their standard states. ✓ The formation reaction for 1 mol of liquid water as follows: H2(g) + 1/2O2(g) → H2O(l) The standard enthalpy change for this reaction is - 285.8 kJ per mole of H2O. Therefore, the thermochemical equation is H2(g) + 1/2O2(g) → H2O(l); ΔH⁰f, = -285.8 kJ ✓ In general, you can calculate the ΔH⁰ for a reaction by the equation ✓ Here E is the mathematical symbol meaning “the sum of,” and m and n are the coefficients of the substances in the chemical equation. 25 Example Large quantities of ammonia are used to prepare nitric acid. The first step consists of the catalytic oxidation of ammonia to nitric oxide, NO. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) Given ΔH⁰f of NH3, O2, NO and H2O(g) as -45.9, 0, +90.3 and – 241.8 kJ respectively Solution = [4(90.3) + 6(-241.8)] – [ 4(-45.9)+5(0)] = - 906kJ 26  Practice Questions 1. Consider the equation 3. Hydrogen peroxide, H2O2, is a colorless liquid whose CH4(g) + 4Cl2(g) → CCl4(l ) + 4HCl(g); ΔH⁰ ? solutions are used as a bleach and an antiseptic. H2O2 can Given be prepared in a process whose overall change is C(graphite) + 2H2(g) → CH4(g); ΔH⁰f = -74.9 kJ (1) H2(g) + O2(g) → H2O2(l ) C(graphite) + 2Cl2(g) → CCl4(l); ΔH⁰f = -135.4 kJ (2) Calculate the enthalpy change using the following data: 1/2H2(g) + 1/2Cl2(g) → HCl(g); ΔH⁰f = -92.3 kJ (3) 2H2O2(l ) → 2H2O(l ) + O2(g); ΔH = -196.0 kJ By Hess Law: H2(g) + 1/2 O2(g) → H2O(l ); ΔH = -285.8 kJ Answer = -429kJ 2. Manganese metal can be obtained by reaction of 4. Hydrogen sulfide gas is a poisonous gas with the manganese dioxide with aluminum. odor of rotten eggs. It occurs in natural gas and is 4Al(s) + 3MnO2(s) → 2Al2O3(s) + 3Mn(s) produced during the decay of organic matter, which contains sulfur. The gas burns in oxygen as follows: What is ΔH for this reaction? Use the following data: 2H2S(g) + 3O2(g) → 2H2O(l ) +2SO2(g) 2Al(s) + 32 O2(g) → Al2O3(s); ΔH = -1676 kJ Mn(s) + O2(g) → MnO2(s); ΔH = -520 kJ Calculate the standard enthalpy change for this reaction using standard enthalpies of formation. ΔH⁰f H2S = -20.5, O2 = 0, H2O(l) = -285.8, SO2 = -296.8 27 5. The molar enthalpies of combustion of hydrogen, carbon and methane are -285.8, - 393.5 and -890.4 kJ respectively. Calculate the molar enthalpy of methane. 6. The molar enthalpy of combustion of ethanol is – 1430 kJ/mol and the molar enthalpies of formation of CO2 and water are -393.4 and -285.8 kJ/mol respectively. What is the molar enthalpy of formation of ethanol under the same conditions. 7. Calculate the enthalpy change for the following reaction: 3NO2(g) + H2O(l ) → 2HNO3(aq) + NO(g) Given ΔH⁰f of NO2, H2O(l), HNO3 and NO as 33.10, -285.8, -207.4 and 90.3 kJ respectively 8. When 1 mol of iron metal reacts with hydrochloric acid at constant temperature and pressure to produce hydrogen gas and aqueous iron (II) chloride, 89.1 kJ of heat evolves. Write a thermochemical equation for this reaction. 9. When 2 mol of potassium chlorate crystals decompose to potassium chloride crystals and oxygen gas at constant temperature and pressure, 78.0 kJ of heat is given off. Write a thermochemical equation for this reaction. 28  10. Hydrazine, N2H4, is a colorless liquid used as a rocket fuel. What is the enthalpy change for the process in which hydrazine is formed from its elements? N2(g) + 2H2(g) → N2H4(l ) Use the following reactions and enthalpy changes: N2H4(l ) + O2(g) → N2(g) + 2H2O(l ); ΔH = -622.2 kJ H2(g) + 1/2O2(g) → H2O(l ); ΔH = -285.8 kJ 11] Ammonia will burn in the presence of a platinum catalyst to produce nitric oxide, NO. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) What is the heat of reaction at constant pressure? Use the following thermochemical equations: N2(g) + O2(g) → 2NO(g); ΔH = + 180.6 kJ N2(g) + 3H2(g) → 2NH3(g); ΔH = -91.8 kJ 2H2(g) + O2(g) → 2H2O(g); ΔH = -483.7 kJ CS2(l ) + 3O2(g) → CO2(g) + 2SO2(g) Calculate the standard enthalpy change for this reaction using standard enthalpies of formation. ΔH⁰f CS2(l) = 89.7, O2 = 0, CO2 = -393.5, SO2 = -296.8 29 12] The cooling effect of alcohol on the skin is due to its evaporation. Calculate the heat of vaporization of ethanol (ethyl alcohol), C2H5OH. C2H5OH(l) → C2H5OH(g); ΔH⁰ ? The standard enthalpy of formation of C2H5OH(l) is - 277.7 kJ/mol and that of C2H5OH(g) is - 235.4 kJ/mol. 13] Carbon tetrachloride, CCl4, is a liquid used as an industrial solvent and in the preparation of fluorocarbons. What is the heat of vaporization of carbon tetrachloride? CCl4(l ) → CCl4(g); ΔH⁰ ? Use standard enthalpies of formation ΔH⁰f. CCl4 (l)= - 135.4 kJ/mol, CCl4(g) = -95.98 kJ/mol 14] Carbon disulfide is a colorless liquid. When pure, it is nearly odorless, but the commercial product smells vile. Carbon disulfide is used in the manufacture of rayon and cellophane. The liquid burns as follows: CS2(l ) + 3O2(g) → CO2(g) + 2SO2(g) Calculate the standard enthalpy change for this reaction using standard enthalpies of formation. ΔH⁰f CS2(l) = 89.7, O2 = 0, CO2 = -393.5, SO2 = -296.8 kJ/mol 30 Thank you