Unit 2: Thermochemistry

Energy The capacity to do work or transfer heat Law of conservation of energy states that energy cannot be create nor destroyed but can change forms Energy lost by a system must be gained by the surroundings and vice versa

Types of Energy Radiant energy Thermal energy Chemical energy From electromagnetic radiation Thermal energy From heat (movement of particles) Chemical energy Stored in the bonds of chemical compounds, released in reactions

Heat and Thermochemistry Heat is the energy used to cause the temperature of an object to increase Thermodynamics is the study of energy and its transformations Thermochemistry is a portion of thermodynamics that studies chemical reactions and energy changes that involve heat

System Vs. Surroundings System is the portion of the universe we single out for study In chemistry the reactants and products are the system Surroundings are everything else besides the system

Open Vs. Closed System Open systems can exchange matter and energy with the surroundings Uncovered pot of boiling water on the stove Closed systems can exchange energy but not matter with the surroundings Piston with H2 and O2 reacting to produce H2O and energy in the form of heat and work

Exothermic Vs. Endothermic Exothermic process – the system loses heat to the surroundings (spontaneous) Combustion of gasoline Endothermic process – the system gains heat from the surroundings (not spontaneous) Melting ice cubes

Enthalpy (“H”) Internal energy plus the product of pressure and volume of the system 𝐻=𝐸+𝑃𝑉 Heat evolved (absorbed or released) by a reaction is the change in enthalpy (ΔH) ∆𝐻= 𝑞 𝑃 =∆𝐸+𝑃∆𝑉 qP = heat or energy evolved from a system at a constant pressure

Enthalpy of Reaction (ΔHrxn) “Heat of reaction” Unit are in Joules (J)/kilojoules (kJ) Enthalpy change associated with a chemical reaction -ΔH = exothermic = gives off heat +ΔH = endothermic = absorbs heat ∆𝐻= 𝑞 𝑛 𝑞 𝑟𝑥𝑛 = −𝑞 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Thermochemical Equations In your data sheets booklet these equations are given We must incorporate mol calculations into these questions

Thermochemical Equations Consider the following reaction: 2Mg(s) + O2(g)  2MgO(s) ΔH=-1204kJ Is the reaction exothermic or endothermic? Calculate the amount of heat transferred when 3.55g of Mg(s) reacts at a constant pressure. How many grams of MgO(s) are produced during an enthalpy change of -234kJ? How many kJ of heat are absorbed when 40.3g of MgO(s) is decomposed into Mg(s) and O2(g) at constant pressure?

Thermochemical Equations Exothermic (Δ𝐻=−) 3.55𝑔 𝑀𝑔 1𝑚𝑜𝑙 𝑀𝑔 24.3050𝑔 𝑀𝑔 −1204𝑘𝐽 2𝑚𝑜𝑙 𝑀𝑔 =−87.9𝑘𝐽 −234kJ 2𝑚𝑜𝑙 𝑀𝑔𝑂 −1204𝑘𝐽 40.3044𝑔 𝑀𝑔𝑂 1𝑚𝑜𝑙 𝑀𝑔𝑂 =15.7𝑔 𝑀𝑔𝑂 40.3𝑔 𝑀𝑔𝑂 1𝑚𝑜𝑙 𝑀𝑔𝑂 40.3044𝑔 𝑀𝑔𝑂 1204𝑘𝐽 2𝑚𝑜𝑙 𝑀𝑔𝑂 =+602𝑘𝐽

Calorimetry Measurement of heat flow Carried out in/with a calorimeter by measuring the magnitude of temperature change the heat flow produces In this course we do calculations using a specific type of calorimeter (coffee cup calorimeter) 𝑞=𝑚𝐶∆𝑇

Heat Capacity (“C”) Amount of heat required to raise its temperature by 1K (or 1℃) Greater C means more heat required top produce a temperature increase Sand vs. water

Specific Heat (“CS”) Heat capacity for pure substances is usually for a given amount Specific heat is the amount of energy needed to raise the temperature of 1g of a substance by 1K.

Calorimetry Calorimeters are vessels used to carry out experiments that allow us to analyze temperature changes This allows us to see the energy being created by a reaction

Bomb Calorimeter Vs. Coffee Cup Calorimeter Used to study combustion Constant volume Used to study principles of calorimetry Mostly solutions Constant pressure

Bomb Calorimeter Vs. Coffee Cup Calorimeter

Calorimetry Calculations For calculations we use: q = heat m = mass C = specific heat (or heat capacity) ΔT = change in temperature

Coffee Cup Calorimetry The specific heat of octane (C8H18) is 2.22J/g•K How many joules of heat are needed to raise the temperature of 80.0g of octane from 10.0 to 25.0℃? Which will require more heat, increasing the temperature of 1mol of C8H18 by a certain amount or increasing the temperature of 1mol of H2O by the same amount?

Coffee Cup Calorimetry 𝑞=𝑚𝐶∆𝑇= 80.0𝑔 2.22𝐽 𝑔∙𝐾 15𝐾 =2.66× 10 3 𝐽 1𝑚𝑜𝑙 𝐶8𝐻18 114.22852𝑔 𝐶8𝐻18 1𝑚𝑜𝑙 𝐶8𝐻18 2.22𝐽 𝑔∙𝐾 = 253.59𝐽 𝐾 1𝑚𝑜𝑙 𝐻2𝑂 18.01528𝑔 𝐻2𝑂 1𝑚𝑜𝑙 𝐻2𝑂 4.184𝐽 𝑔∙𝐾 = 75.376𝐽 𝐾 It will take more energy to increase one mol of C8H18 that one mol of H2O

Coffee Cup Calorimetry When 50mL of 1.0M HCl and 50mL of 1.0M NaOH is combine in a coffee cup calorimeter the temperature of the resultant solution increases from 21.0℃ to 27.5℃. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming that the calorimeter loses only negligible quantity of heat, that the total volume of the solution is 100mL, that its density is 1.0g/mL, and that its specific heat is 4.18J/g•K.

Coffee Cup Calorimetry 𝑚= 100𝑚𝐿 1.0𝑔 1𝑚𝐿 =100𝑔 ∆𝑇=27.5℃−21.0℃=6.5℃=6.5𝐾 𝑞 𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟 =𝑚𝐶∆𝑇= 100𝑔 4.18𝐽 𝑔∙𝐾 6.5𝐾 =2717𝐽=2.7𝑘𝐽=-ΔH (exothermic) 0.050𝐿 𝐻𝐶𝑙 1.0𝑚𝑜𝑙 𝐻𝐶𝑙 1𝐿 =0.050𝑚𝑜𝑙𝐻𝐶𝑙 ∆𝐻 𝑚𝑜𝑙 = −2.7𝑘𝐽 0.050𝑚𝑜𝑙 =−54𝑘𝐽/𝑚𝑜𝑙

Enthalpy Diagrams Used to give a visual representation of energy changes throughout a reaction

Enthalpy Diagrams Things to remember Enthalpy is an extensive property Changes with the number of mols being reacted Enthalpy stays the same magnitude for the reverse reaction but changes the sign Enthalpy change depends on the states of the reactants and products Enthalpy of gases>liquids>solids +ΔH = endothermic (products higher than reactants) -ΔH = exothermic (products lower than reactants)

Enthalpy Diagrams

2H2(g) + O2(g) 2H2O(g) ΔH = -483.6kJ Enthalpy Diagrams Draw the enthalpy diagram for the following equation: 2H2(g) + O2(g) 2H2O(g) ΔH = -483.6kJ Remember to include: Direction of increasing enthalpy Products (with proper coefficients and states) Reactants (with proper coefficients and states) Direction of reaction ΔH (heat of reaction) with proper sign “Exothermic” or “endothermic”

Hess’s Law No matter how many steps a reaction has the total enthalpy change is equal to the sum of all enthalpy changes ΔHtotal rxn = ΔHall steps

Hess’s Law Calculate ΔH for the reaction: 2C(s) + H2(g)  C2H2(g) given the following chemical equations and their respective enthalpy changes: C2H2(g) + 5 2 O2(g)  2CO2(g) + H2O(l) ΔH=-1299.6kJ C(s) + O2(g)  CO2(g) ΔH=-393.5kJ H2(g) + 1 2 O2(g)  H2O(l) ΔH=-285.8kJ

Hess’s Law 2CO2(g) + H2O(l)  C2H2(g) + 5 2 O2(g) ΔH=1299.6kJ 2C(s) + 2O2(g)  2CO2(g) ΔH=-787.0kJ H2(g) + 1 2 O2(g)  H2O(l) ΔH=-285.8kJ --------------------------------------- 2C(s) + H2(g)  C2H2(g) ΔH=226.8kJ

Hess’s Law Things to remember Products and reactants should be on same side in component equations as in net equation Mol ratio should add up to be the same in component and net equations Reversing an equation changes the sign of ΔH for that equation Add ΔH’s with appropriate signs When multiplying an equation by any number multiply the ΔH as well