Thermochemistry Chapter 5

Thermochemistry Thermochemistry: Relationships between chemical reactions and energy changes Thermodynamics: Therm: heat Dynamics: power

Energy Radiant energy Thermal energy Chemical energy Nuclear energy Energy: the capacity to do work Radiant energy Thermal energy Chemical energy Nuclear energy Potential energy- relative to position- stored in bonds electrostatic potential energy (C) = kQ1Q2 d Kinetic energy Ek = ½ mv2 Units of Energy: J = 1kg-m2/s2 1 cal= 4.184 J

Energy Changes in Chemical Reactions Which one has more thermal energy? 900C 900C Coffee cup? Bathtub?

Energy Changes in Chemical Reactions Which one has more thermal energy? 400C 900C What additional information is needed? How does it relate to heat?

Energy Changes in Chemical Reactions Heat: the transfer of _______ between two bodies that are at ________temperatures. Temperature is a measure of the ________. Measure of the ______________ Temperature = Thermal Energy

Energy Changes SURROUNDINGS SYSTEM The system is the specific part of the universe that is of interest in the study. SURROUNDINGS SYSTEM open closed isolated Exchange: mass & energy energy nothing

Energy Changes

Enthalpy Exothermic process: is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat is supplied to the system from the surroundings. energy + H2O (s) H2O (l) energy + 2HgO (s) 2Hg (l) + O2 (g)

ΔHreaction = ΔHproducts - ΔHreactants Enthalpy Enthalpy: a thermodynamic quantity used to describe heat changes(absorbed/released) at constant pressure (most reactions occur at constant pressure). H = E + PV Equation: ΔHreaction = ΔHproducts - ΔHreactants

Enthalpy: Enthalpy Diagrams DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0

Enthalpy: Enthalpy Diagrams ΔHrxn= ΔHproducts – ΔHreactants

Enthalpy: Examples Indicate the sign of enthalpy change in the following processes carried under atmospheric pressure, and indicate whether the process is exothermic or endothermic: An ice cube melts 1 g of butane (C4H10) is combusted in sufficient oxygen to give complete combustion to CO2 and H2O A bowling ball is dropped from a height of 8 ft into a bucket of sand.

Thermochemical Equations Is DH negative or positive? System absorbs heat Endothermic DH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. H2O (s) → H2O (l) DH = 6.01 kJ 6.3

Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) DH = -890.4 kJ 6.3

Thermochemical Equations: Rules Stoichiometric coefficients: equal the number of moles of a substance H2O (s) H2O (l) DH = 6.01 kJ If you reverse a reaction, the sign of DH changes H2O (l) H2O (s) DH = - 6.01 kJ If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. 2H2O (s) 2H2O (l) DH = 12.0 kJ 2 x 6.01kJ/mol =12.0 KJ/2 mol

Thermochemical Equations: Rules When a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes.

Thermochemical Equations: Rules The physical states of all reactants and products must be specified in thermochemical equations. H2O(s) → H2O(l) ΔH = 6.00 kJ H2O(l) → H2O(g) ΔH = 44.0 kJ What will be the ΔH when 1 mole of ice at 0ºC is changed into one mole of steam at the boiling point?

Thermochemical Equations: Rules How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) DH = -3013 kJ 1 mol P4 123.9 g P4 x -3013 kJ 1 mol P4 x 266 g P4 [ -6470 kJ]

Enthalpy: Examples 4. The complete combustion of liquid octane, C8H18, to produce carbon dioxide and liquid water at 25 ºC and at constant pressure gives off 47.9 kJ per gram of octane. Write a chemical equation to represent this information. [2C8H18(l) + 25 O2(g) → 16 CO2(g) + 9H2O(l) ΔH = -1.09 x 104 kJ]

Enthalpy: Examples SO2(g) + ½ O2(g) →SO3(g) ΔH = -99.1 kCal 5. Given the thermochemical equation: SO2(g) + ½ O2(g) →SO3(g) ΔH = -99.1 kCal Calculate the heat evolved when 74.6 g of SO2 (MM = 64.07 g/mol) is converted to SO3. [ -115 kJ] 6. Hydrogen peroxide can decompose to water and oxygen by the following reaction: 2H2O2(l) → 2H2O(l) + O2(g) ΔH = -196 kJ Calculate the value of q (heat evolved or absorbed) when 5.00 g of H2O(l) decomposes at constant pressure. [ -14.4 kJ]

Enthalpy: Examples 7. Given the equation H2(g) + I2(s) → 2HI(g) ΔH = +52.96 KJ Calculate ΔH for the reaction HI(g) → ½ H2(g) + ½ I2(s) [ -26.48 kJ] 8. Given the equation: 3O2(g) → 2O3(g) ΔH = +285.4 kJ Calculate ΔH for the following reaction: 3/2 O2(g) → O3(g) [ 142.7 kJ]

Calorimetry and Heat Exchanges in a Chemical Reaction Read Section 6.4 (Pages 213-2200) Examples: pp. 216 – 220: 6.1 – 6.3 Exercises: pp. 240: 6.19, 6.23, 6.26, 6.28

Calorimetry Calorimetry is performed in devices called calorimeters Calorimetry is based on the Law of Conservation of Energy

Calorimetry 9. The specific heat of aluminum is 0.895 J/g ºC. Calculate the heat necessary to raise the temperature of 40.0 g of aluminum from -20.0 ºC to 32.3 ºC. Specific heat of aluminum is 0.215-cal/g ºC. [1.87 x 103 Joules]

Calorimetry: Heat Capacity Calorimetry: Experimental measurement of heat produced in chemical and physical processes Heat Capacity, C, of a system: the quantity of heat needed to raise the temperature of matter 1 K. Units: J/ºC Molar Heat Capacity: The heat required to raise 1 mole of a substance, 1 K.

Calorimetry: Specific Heat Specific heat of a substance (also represented by cp): the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius ( or 1 K).

Calorimetry: Definitions Heat (q) absorbed or released: q = mcpDt q = CDt Dt = tfinal - tinitial When q>0 endothermic reaction When q<0 exothermic reaction

Calorimetry: Specific Heat 10. How much heat, in joules and kilojoules, does it take to raise the temperature of 225 g of water from 25.0 ºC to 100.0 ºC [ 7.05 x 104 J; 70.5 kJ] 11. Calculate the heat capacity, specific heat and the molar heat capacity of an aluminum block that has mass of 5.7 g that must absorb 629 J of heat from its surroundings for its temperature to rise from 22 ºC to 145 ºC. [C = 5.11 J/ºC; cp = 0.90 J g-1 ºC-1, 24 J mol-1 ºC-1 ]

Calorimetry: Problems 12. What will be the final temperature if a 5.00-g silver ring at 37.0 ºC gives off 25.0 J of heat to the surroundings? Specific heat of silver = 0.235 J g-1 ºC-1 [15.7 ºC] 13. A 15.5 g sample of a metallic alloy is heated to 98.9 ºC and then dropped into 25.0 g of water in a calorimeter. The temperature of the water rises from 22.5 ºC to 25.7 ºC. Calculate the specific heat of the alloy. [0.29 J g-1 ºC-1]

Calorimetry: Problems 14. Suppose a piece of gold with a mass of 21.5 g at temperature of 95.00 ºC is dropped into an insulated calorimeter containing 125.0 g of water at 22.00 ºC. What will be the final temperature of the water? [22.4 ºC]

Measuring Enthalpy Changes for Chemical Reactions: Constant-Volume Calorimetry Bomb Calorimeter

Constant Volume Calorimeter: Isolated System No exchange of matter and energy Uses Cp of water

Constant-Volume Calorimetry: Isolated system qsys = qwater + qbomb + qrxn = 0 There is no heat exchange with the surroundings qrxn = - (qwater + qbomb) Cbomb= mbombx cbomb qbomb = CbombDt qwater = mwcpΔt qrxn = -(qwater + CbombΔt) = -(mwcpΔt + CbombΔt) qrxn = - (mwcp + Cbomb)Δt Measured heat is not enthalpy: constant volume

Constant Volume Calorimetry: Problems 15. In a preliminary experiment, the heat capacity of a bomb calorimeter assembly is found to be 5.15 kJ/°C. In a second experiment. A 0.480 g sample of graphite (carbon) is placed in the bomb with an excess of oxygen. The water, bomb, and other contents of the calorimeter are in thermal equilibrium at 25.00 °C. The graphite is ignited and burned, and the water temperature rises to 28.05 °C. Calculate ΔH for the reaction. C(graphite) + O2(g) → CO2(g) [ΔH = -393 kJ]

Measuring Enthalpy Changes for Chemical Reactions Coffee-Cup Calorimeter: Constant pressure

Measured heat is the enthalpy Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcalor) qwater = mcpDt qcal = CcalDt Reaction at Constant P ΔH = qrxn = - (qwater + qcalor) No heat enters or leaves! Insulated system! Pressure is constant. Measured heat is the enthalpy

Calorimetry

Constant Pressure Calorimetry: Problems (a) A 50.0 mL sample of 0.250 M HCl at 19.50 °C is added to 50.0 mL of 0.250 M NaOH, also at 19.50 °C, in a calorimeter. After mixing, the solution temperature rises to 21.21 °C. Calculate the heat of this reaction. Assume te volumes of the solutions are additive. Ignore the heat absorbed by the calorimeter. [-715 J] (b) Determine the value of ΔH that should be written in the equation for the neutralization reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ΔH = ? [-57.2 kJ]

Calorimetry: Problem 17. A quantity of 1.00 x 102 mL of 0.500M HCl is mixed with 1.00 x 102 mL of 0.500 M NaOH in a constant pressure calorimeter having a heat capacity of 335 J/ ºC. The initial temperature of the HCl and NaOH solutions is the same, 22.50 ºC, and the final temperature of the mixed solution is 24.90 ºC. Calculate the heat change for the neutralization reaction and the molar heat of the neutralization reaction. Assume s = 4.18 J/gºC for the solution. [-2.81 kJ; -56.2 kJ/mol]

Fuel Values of Foods and Other Substances Chemistry in Action: Fuel Values of Foods and Other Substances C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol 1 cal = 4.184 J One food Calorie: 1 Cal = 1000 cal = 4184 J

Standard Enthalpy of Formation Absolute value of the enthalpy of a substance cannot be measured. We need a frame of reference. Standard states for elements and compounds has to be established. Standard state for a solid or liquid substance is the pure element or compound at 1 atm at a given temperature. Standard state for gas: pure gas behaving as an ideal gas at 1 atm and temperature of interest. Standard enthalpy of formation: the enthalpy change for a reaction in which the reactants in their standard state yield products in their standard states. Standard state denoted with a superscript (º)

Standard Enthalpy of Formation Standard enthalpy of formation ΔHºf (also called heat of formation enthalpy change that occurs in the formation of 1 mole of the substance from its elements when both products and reactants are in their standard states (f stands for formation). Reference forms: standard enthalpy of formation of a pure element in its reference form is 0. standard enthalpy of formation (ΔH0) as a reference point for all enthalpy expressions is the formation of H2(g) = 0 kJ/mol

Standard Enthalpy of Formation Standard enthalpy of formation (DH0f) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm and 25 ºC . The standard enthalpy of formation of any element in its most stable form is zero. DH0 (O2) = 0 f DH0 (C, graphite) = 0 f DH0 (C, diamond) = 1.90 kJ/mol f DH0 (O3) = 142 kJ/mol f

standard

Determination of Heat of Formation 1. Direct method: From elements when it can be easily measured. C(graphite) + O2(g) →CO2(g) ΔHrxn = -393.5 kJ ΔHrxn= ΔHoCO2 – ΔH0O2 – ΔHgraphite = -393.5 kJ ΔH0CO2 = -393.5 kJ 2. Indirect method: Using the heat of reaction and heats of formations Given CH4(g) +2O2(g) → CO2(g) + 2H2O(g) ΔHrxn=-890.3 kJ Calculate the heat of formation of CH4 [-74.8kJ/mol]

Standard Enthalpy of Reaction The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm and 25ºC. rxn aA + bB cC + dD DH0 rxn dDH0 (D) f cDH0 (C) = [ + ] - bDH0 (B) aDH0 (A) DH0 rxn nDH0 (products) f = S mDH0 (reactants) -

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) Benzene (C6H6) 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. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ -5946 kJ 2 mol = - 2973 kJ/mol C6H6

Standard Enthalpy of Reaction Synthesis gas is a mixture of carbon monoxide and hydrogen that is used to synthesize a variety of organic compounds, such as methanol. One reaction for producing synthesis gas is shown here: 3CH4(g) + 2H2O(l) + CO2(g) → 4CO(g) + 8H2(g) ΔHº = ? [747.5 kJ] The combustion of isopropyl alcohol, common rubbing alcohol, is represented by this equation: 2(CH3)2CHOH(l) + 9O2(g) → 6CO2(g) + 8H2O(l) ΔHº = -4011 kJ use this equation and data from Thermodynamic tables to establish the standard enthalpy of formation for isopropyl alcohol. [ -318 kJ/mol]

Heat of Formation: Examples 20. Pentaborane-9, B5H9, is a colorless liquid. It is highly reactive substance that will burst into flame or even explode when exposed to oxygen: 2B5H9(l) + 12O2 → 5B2O3(s) + 9H2O(l) It was considered a potential rocket fuel as it release large amount of heat per gram. Calculate the kJ of heat released per gram of the compound reacted with oxygen. The standard enthalpy of formation of B5H9 is 73.2 kJ/mol [-71.58 kJ/gB5H9]

Heat of Reactions (Formations): Hess’s Law of Constant heat of Summations Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

Hess’ Law: Example Given the following reactions: CO(g) + ½ O2(g) → CO2(g) ΔH = -283.0 kJ C(graphite) + O2(g) → CO2(g) ΔH = -393.5 kJ Calculate the enthalpy for the formation of CO from its elements C(graphite) + ½ O2(g) → CO(g) ΔH = ?

Hess’s Law: Example

Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ rxn S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJ rxn CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ 2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1x2 kJ rxn + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJ rxn C(graphite) + 2S(rhombic) CS2 (l) DH0 = -393.5 + 2(-296.1) + 1072 = 86.3 kJ rxn 6.6

Hess’ Law: Examples 21. Given C(graphite) + O2 (g)→ CO2(g) ΔHºrxn= -393.5 kJ H2(g) + ½ O2(g) → H2O(l) ΔHºrxn = -285.8 kJ 2C2H2(g) + 5O2(g) → CO2(g) + 2H2O(l) ΔHºrxn = -2598.8 kJ Calculate the heat of formation of C2H2(g) 2C(graphite) + H2(g) → C2H2(g) [52.3 kJ]

Hess’ Law: Conceptual Example Without performing a calculation, determine which of the two substances should yield the greater quantity of heat upon Complete combustion, on a per mole basis: ethane, C2H6(g), or ethanol, CH3CH2OH(l)

The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788kJ – 784kJ = 4 kJ/mol

The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hfinal soln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? Which substances could be used for hot pack?

Heat of Solution and Dilution The process: NaCl(s) → Na+(aq) + Cl-(aq) The steps: 1. Separate ions in the crystal into ions in gaseous state 2. Hydration of the gaseous ions Energy involved: Lattice Energy, U Hydration energy, ΔH hydration ΔH solution= U + ΔH hydration

Heat of Solution and Dilution Lattice energy: energy required to completely separate one mole of solid ionic compound into gaseous ions (endothermic process) U + NaCl(s) →Na+(g) + Cl-(g) Hydration of the gaseous ions Heat of hydration: energy released when ions interact with water molecules. Na+(g) + Cl-(g) → Na+(aq) + Cl-(aq) + Energy

Heat of Dilution The heat change associated with the dilution of a solution. Can be exothermic or endothermic

The End