Thermochemistry Mrs. Stoops Chemistry

Thermochemistry or Thermodynamics reactions involving heat. Abbreviations to know – rxn = reaction E = Energy  = delta = change Thermo Day 1

Kinetic Energy vs Potential Energy Potential – stored energy Kinetic – movement energy Units of Energy – Joule or J also KJ calories 1 cal = 4.184 J PE KE Thermo Day 1

System vs Surroundings We will focus on the system – the system changes depending on what we are looking at. Could be a beaker, or the room, or the Earth, or the solar system Thermo Day 1

Work & Heat Do you work? Work = force x distance moved Thermo Day 1

Heat Heat is energy transferred. Heat is transferred from HOT to COLD. Energy is the capacity to do work or transfer heat. NEW symbol  = delta – means change T means change in temperature = Tfinal - Tinitial Thermo Day 1

1st Law of Thermo 1st law of Thermodynamics = Energy is CONSERVED. E = Ef - Ei E is positive means Ef  Ei. Also Energy has moved from surroundings into the system E is negative means Ef  Ei. Energy has moved from system out to surroundings. Thermo Day 1

System vs Surroundings Another way to solve for E E = q + w = heat + work. Sign direction issues! q + Heat entering q – Heat leaving System W – Do work Thermo Day 1

Example problem: Calculate the change in energy on the system when 140 J of heat are absorbed and 85 J of work are done. E = (+140J) + (- 85J) = 55 J Thermo Day 1

Exothermic & Endothermic Cue words are important!!! Absorbs heat = goes INTO the system Endothermic = q + Releases heat = EXITS the system Exothermic = q – When something feels cold – the heat is being taken away and used, leaving the object “feeling” cold. Thermo Day 1

State Functions State functions – depend on the current conditions only, not the path taken. E, average temperature Thermo Day 1

Homework Day 1 Page 203: 25, 29 Thermo Day 1

Enthalpy - H Heat gained or lost **yeah, and q is too! State function H = H products – H reactants

Enthalpy - H Sign indicates the direction of heat transfer only!!! +H = gained heat (into system), Endothermic, REACTANT -H = lost heat (out of system), Exothermic, PRODUCT Hrxn – heat of reaction - we will calculate this later, for now it is given. Thermo Day 2

Thermochemical equation a reaction shown with the H value Ex: 2H2 + O2 → 2H2O H = -483.6 KJ What this means: Since H is negative, the enthalpy is a _____________ (which side of rxn?) 2H2 + O2 → 2H2O + 483.6 KJ *** the sign tells the side of reaction ONLY! product Thermo Day 2

Rules of using Enthalpy: 1. Enthalpy is an extensive property. The coefficients matter! Ex: CH4 + 2 O2 → CO2 + 2 H2O H = -890 KJ This is for 1 mole CH4 = based on the stoichiometry For 4 moles CH4 the H becomes -890 x 4 = -3560 KJ Thermo Day 2

Rules of Enthalpy 2. The enthalpy sign changes when the reaction is reversed CH4 + 2 O2 → CO2 + 2 H2O H = -890 KJ But if we switch products for reactants CO2 + 2 H2O → CH4 + 2 O2 H = +890 KJ Thermo Day 2

Rules of Enthalpy 3. Enthalpy depends on the states of the reactants and the products. The states have been listed in the reactions – you need to pay attention to. (s) solid, (l) liquid, (g) gas, & (aq) aqueous – dissolved in water H2 (g) + O2 (g) → H2O (g) Thermo Day 2

Using these ideas for stoichiometry . . . If you have a Thermochemical equation, you can be asked to solve a problem to find the heat transferred. Ex: CH4 + 2 O2 → CO2 + 2 H2O H = -890 KJ If you have 50 KJ of heat, how many grams of oxygen can be reacted? 50KJ x 2 mol O2 x 32 g O2 = 3.6 g O2 890 KJ 1 mol O2 *** would the answer be negative? What does the sign tell us? Thermo Day 2

Calorimetry Measures heat flow by measuring the temperature changes Heat capacity = amount of heat needed to raise the temperature by 1 (degree) Kelvin Molar heat capacity = amount of heat needed to raise the temperature by 1 degree for 1 mole Specific heat = heat capacity for 1 gram Thermo Day 2

To find specific heat or use specific heat Heat transferred = mass (in grams) x specific heat x change in temperature Equation form: q = m c t Values for c are on page 181 of text, given, or you are asked to solve for them Thermo Day 2

Homework Day 2 page 204: 37a, 41, 43, 51, 53 Thermo Day 2

Hess’s Law If the reaction is carried out in a series of steps, the H for the overall reaction will be equal to the sum of the H for all the steps. Generic example: Overall reaction A  D H = ? Steps: 1 A  B H = x 2 B  C H = y 3 C  D H = z Overall A  D H = x + y + z Thermo Day 3

Example: Find the H for the reaction: 2 C (s) + O2 (g)  2 CO (g) Steps: 1 C (s) + O2 (g)  CO2 (g) H = -393.5 KJ 2 CO (g) + ½ O2 (g)  CO2 (g) H = - 283.0 KJ Work: 2 C (s) + O2 (g)  2 CO (g) 2C (s) + 2O2 (g)  2CO2 (g) H x 2 = -393.5 x 2 = -787 KJ 2CO2 (g)  2CO(g) + O2 (g) H x 2, R = +283.0 x 2 = 566 KJ **Cancel** 2 C(s) + O2 (g)  2 CO(g) -221 KJ Thermo Day 3

Enthalpy of Formation Hf Depend on the state (s, l, g, aq) Standard enthalpy = Hº Standard enthalpy of formation = Hfº = are listed for 298 K Units are KJ/mol Listed in the text book = appendix C (page 1019) ***Element values are zero*** Thermo Day 3

Enthalpy of Formation Hrxn =  n Hfºproducts -  m Hfºreactants What are these?  - sigma, Greek letter, means sum of n & m – coefficients from balanced equation Thermo Day 3

Example Find the Hrxn for C2H5OH(l) + 3 O2 (g)  2 CO2(g) + 3 H2O (l) Look up values: C2H5OH(l) -277.7 KJ O2 (g) 0 KJ CO2(g) -393.5 KJ H2O (l) -285.83 KJ Plug into the formula: [(2*-393.5 KJ) + (3*-285.83 KJ)] – [(1*-277.7 KJ) + (3*0)] = H = -1366.79 KJ PRODUCTS - REACTANTS Thermo Day 3

Homework Day 3 P 205: 52, 61, 63, 69, 71b d, Thermo Day 3