Warm up How many calories are in 535 kJ? How many nutritional calories is that? Explain the difference between heat and temperature.

Thermodynamics

1st law of Thermodynamics Based on the law of conservation of energy States that energy can be converted from one form to another Cannot calculate total energy but can calculate changes in energy

System Surroundings The rest of the universe outside the system Specific part of the universe that is of interest to us The rest of the universe outside the system Heat transfer: the transfer of thermal energy between 2 bodies of different temperatures. Heat is a measure of a process with a number and a unit. (It is best thought of as a verb not a noun.) open closed isolated Exchange: mass & energy energy nothing

Thermochemistry Study of heat changes in a chemical or physical process Enthalpy—used to quantify the heat content into or out of a system that occurs at constant volume in an open system Symbol: H, cannot be measured Symbol ΔH is used in heat changes At constant volume q = ΔH Heat of reaction (ΔHrxn) is the change in enthalpy with respect to a chemical reaction

DHrxn = H (products) – H (reactants) Thermochemical equations shows the enthalpy changes, ΔH must specify states of matter ΔHrxn (most general form, sometimes leave off the rxn)— used to describe changes in H. More specific would be: ΔHcomb,ΔHsoln, ΔHvap, ΔHhydration ΔHfus DHrxn = H (products) – H (reactants) All chemical and physical changes release or absorb heat a. Endothermic: absorbs heat, q is positive, ∆H is positive b. Exothermic: releases heat, q is negative, ∆H is negative

DHrxn = H (products) – H (reactants) DHrxn = heat given off or absorbed during a reaction at constant pressure, 1atm and 25C Hproducts < Hreactants Hproducts > Hreactants DH > 0 DH < 0

Endothermic Exothermic releases heat to its surroundings (feels warm) Energyproducts < Energyreactants -q or -H absorbs heat from the surroundings (feels cold) Energyproducts > Energyreactants +q or +H

Thermochemical Equations for Endothermic Reactions Endothermic : H is positive; heat is on the reactants side of the equation (Heat absorbed) Three ways to write the same thing: NH4NO3(s)  NH4+(aq) + NO3-(aq) H = 27 kJ NH4NO3(s) + Heat  NH4+(aq) + NO3-(aq) NH4NO3(s) + 27kJ  NH4+(aq) + NO3-(aq)

Thermochemical Equations for Exothermic Reactions Exothermic : H is negative; heat is on the products side of the equation (Heat released) Three ways to write the same thing CaO(s) + H2O(l)  Ca(OH)2(s) H = -65.2 kJ CaO(s) + H2O(l)  Ca(OH)2(s) + Heat CaO(s) + H2O(l)  Ca(OH)2(s) + 65.2 kJ

3. Heat of Combustion, DHcomb Change the problem in NT a. How much heat is released by the combustion of 250.0 g of hydrogen gas? DHcomb = -286 kJ 2 H2 + O2  H2O 2 250.0 g H2 1 mole H2 -286 kJ = -17,698 kJ 2.02g H2 2 mole H2 = -1.770 x 104 kJ Or 1.770 x 104 kJ released How do you know when to use a table vs a formula? If you have quantity (g or mol) you will use a table. (When I say “grams,” you say, “table.”)

Stop

Standard Heat of Formation, DH f Day 2 We will be using the standard heat of formation, DH f, to calculate the Direct Method of the Standard heat of a reaction, DH rxn at 1 atm and 25⁰C

ii. Standard Heat of Formation, DH f The circle is said “not,” which means standard ii. Standard Heat of Formation, DH f Take out your homework and find the C-13 table to use to look up values for DH f

ii. Standard Enthalpy, DH The circle is said “not,” which means standard ii. Standard Enthalpy, DH What happens when the exchange occurs under normal conditions(1atm & 25C)? Standard heat of Formation( Hf ) is Forming one mole of a compound using only elements as the reactants Example: C(s) + O2(g)  1CO2(g) 1 mole or molecule of CO2(g) Hf = -393.509 kJ The ΔHf of any element in its most stable form is 0kJ ΔHf (O2) = 0 kJ ΔHf (C, graphite) = 0 kJ ΔHf (O3) = 142 kJ/mol ΔHf (C, diamond) = 1.90 kJ/mol

Write the DH f to make H2O(l) Step 1: write the elements and the substance H2(g)+ O2(g)  H2O(l) Step 2: Balance the equation 2H2(g)+ O2(g)  2H2O(l) Step 3: Divide by any coefficients that are on the substance to get 1 mole H2(g)+ ½ O2(g)  H2O(l) This is the Standard Heat of Formation of one mole of water, DH f

You try and make DH f of aluminum chloride Al(s) + Cl2(g)  AlCl3 2Al(s) + 3Cl2(g)  2AlCl3 Al(s) + 3/2 Cl2(g)  AlCl3 = DH f

ii. Standard Enthalpy DH f values

ii. Standard Enthalpy, DH cont… What happens when the exchange occurs under normal conditions (1atm & 25C)? Standard heat of Reaction( Hrxn ) a. Direct method-equation (today’s lesson) aA + bB  cC + dD Hrxn = [c Hf(C) + dHf(D)] –[aHf(A) + bHf(B)] Hrxn =  Hf(products )-  Hf(reactants) b. Indirect method-Hess’s Law (tomorrow’s lesson)

ii. Standard Enthalpy, DH cont… Direct method-equation examples Calculate the standard heat of reaction for the combustion of methane. Write the balanced Chemical Equation: Then use the Hf values to substitute into the equation. Be sure to use coefficients in the eqn!!! aA + bB  cC + dD Hrxn = [c Hf(c) + dHf(D)] –[aHf(A) + bHf(B)] Hrxn =  Hf(products )-  Hf(reactants) CH4(g) + 2O2(g)  CO2(g)+ 2H20(l)

ΔHrxn = ΣΔHƒ(product) − ΣΔHƒ(reactants) Use the standard enthalpies of formation to calculate ∆Hrxn CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ∆Hf(CO2(g))= -393.509kJ ∆Hf(H2O(l))= -241.818kJ ∆Hf(CH4(g))= -74.81kJ ∆Hf(O2(g))= 0kJ ΔHrxn = ΣΔHƒ(product) − ΣΔHƒ(reactants) = [-393.509+(2×-241.818)] – [-74.81+0] = -877.145 + 74.81 = -802.335  -802.34 kJ exothermic reactants products

ii. Standard Enthalpy, DH cont… Direct method-equation examples 2. Calculate the standard heat of reaction for the combustion of hydrogen sulfide gas. (hint: produces liquid water and sulfur dioxide gas) Write the balanced Chemical Equation: 2H2S(g) + 3O2(g)  2H2O (l)+ 2S02 (g) Hrxn = [2(-285.830kJ )+2(-296.830kJ)] – [2(-20.63kJ)+ 3(0kJ)] Hrxn = -1,124.06kJ or 1,124.06kJ evolved Exothermic Reaction

ii. Standard Enthalpy, DH cont… Direct method-equation examples 3. Calculate the standard heat of reaction for the combustion of nitrogen monoxide gas and oxygen gas. (hint: yields nitrogen dioxide gas) Write the balanced Chemical Equation: 2NO(g) + O2(g)  2NO2 (g) Hrxn = [2(33.18kJ)] – [2(90.25kJ)+ 0kJ] Hrxn = -114.14kJ or 114.14kJ released Exothermic Reaction

ii. Standard Enthalpy, DH cont… Direct method-equation examples 3. Calculate the standard heat of reaction for the combustion of carbon monoxide gas and oxygen gas. (hint: yields carbon dioxide gas) Write the balanced Chemical Equation: 2CO(g) + O2(g)  2CO2 (g) Hrxn = [2(-393.509kJ )] – [2(-110.525kJ)+ 0kJ] Hrxn = -664.968kJ or 664.968kJ evolved Exothermic Reaction

Next Day-Indirect Method Hess’s Law

Next Day-Indirect Method Hess’s Law Hess’s Law is a lot like solving systems of equations but there are some differences. You may use a multiple of an equation. This multiple may include positive, negative, and even fractions. Do not try to subtract. Always add the equations together to avoid errors.

Hess’s Law-Example 1 2S(s) + 3O2(g)  2SO3(g) ∆H = ? S(s) + O2(g)  SO2(g) ∆H = -297 kJ 2SO3(g)  2SO2(g) + O2(g) ∆H = 198 kJ

Hess’s Law-Example 2 2NO(g) + O2(g)  2NO2(g) ∆H = ? N2(g) + O2(g)  2NO(g) ∆H = 180.6 kJ N2(g) + 2O2(g)  2NO2(g) ∆H = 66.4 kJ

Hess’s Law-Example 3 2C (s) + O2(g)  2CO(g) ∆H = ? C(s) + O2(g)  CO2(g) ∆H = 393.5 kJ 2CO(g) + O2(g)  2CO2(g) ∆H = -566kJ

Hess’s Law-Example 4 C(s) + 2H2(g)  CH4(g) ∆H = ? C(s) + O2(g)  CO2(g) ∆H =-393.5 kJ 2H2(g) + O2(g)  2H2O(l) ∆H =-571.6 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ∆H=-890.8 kJ