Unit 3 AP Chemistry Test (Thermochemistry)

What is the First Law of Thermodynamics?

- Law of Conservation of Energy ∙ Energy can neither be created or destroyed. ∙ Energy is only transferred between the system and the surroundings.

Calculating Enthalpy using Bond Energies Calculate ΔH°rxn for the reaction: H₂ (g) + F₂ (g) → 2 HF (g) Given the following bond energies: ΔH°rxn = [1(432kJ/mol) + 1(154kJ/mol)] - [2(565kJ/mol)] ΔH°rxn = _______?

-544kJ/mol

Enthalpies of Formations (ΔH°f) Practice Problem Write the formation reaction for water.

1 H₂ (g) + 1/2 O₂ (g) → 1 H₂O (l)

Enthalpies of Formations (ΔH°f) Practice Problems 1. Calculate the standard enthalpy change for the combustion of 1 mol of benzene (C₆H₆) to CO₂ and H₂O (l). 2. Calculate the standard enthalpy change for the combustion of ethanol (C₂H₅OH).

1. Calculate the standard enthalpy change for the combustion of 1 mol of benzene (C₆H₆) to CO₂ & H₂O (l). ∙ C₆H₆ + 15/2 O₂ → 6 CO₂ + 3 H₂O (l) ∙ ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants) A. ΔH°rxn = [6(ΔH°f CO₂ (g)) +3(ΔH°f H₂O (l))] - [1(ΔH°f C₆H₆ (g)) + 15/2(ΔH°f O₂ (g))] B. ΔH°rxn = [6(-393.5kJ) +3(-285.8kJ)] - [1(+49.2kJ) + 15/2(0kJ)] C. ΔH°rxn = -3267. 6kJ/mol 2. Calculate the standard enthalpy change for the combustion of ethanol ∙ C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O (l) ∙ ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants) A. ΔH°rxn = [2(ΔH°f CO₂ (g)) +3(ΔH°f H₂O (l))] - [1(ΔH°f C₂H₅OH (g)) + 3(ΔH°f O₂ (g))] B. ΔH°rxn = [2(-393.5kJ) +3(-285.8kJ)] - [1(-278kJ) + 7(0kJ)] C. ΔH°rxn = -1366.4kJ/mol

Calculating using Calorimetry Data: Dissolving a Salt 1. When a 4.25 g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter, the temperature drops from 22.0°C to 16.9°C. Calculate the enthalpy for the solution process. Assume that the specific heat and density of the solution is the same as water. A. Is the reaction endothermic or exothermic?

1. ∙ Determine the mass of the solution: 4.25 g + 60.0 g = 64.25 g (the salt becomes part of the solution) ∙ Calculate ΔT = 16.9 °C - 22.0 °C = -5.1 °C A. Calculate qsolution = (64.25 g)(4.182 J/g°C)(-5.1 °C) = -1370.992 J = -1400 J ∙ Determine qrxn (equal in magnitude to qsolution but opposite in sign) = +1400 J - This makes sense because the temperature decreased indicating an endothermic reaction! ∙ The limiting reactant is the only reactant: Ammonium Nitrate ∙ NH₄NO₃ (s) → NH₄⁺ (aq) + NO₃⁻ (aq) ∙ 4.25 g NH₄NO₃ = .0531 mol NH₄NO₃ ∙ Calculate ΔH°rxn = qrxn/moles of Limiting Reactant ∙ ΔH°rxn = +1370.992 J/.0531mol = +25826.27 J/mol or +25.82627 kJ/mol ∙ ΔH°rxn = +26 kJ/mol (A. endothermic!)

Calculating Enthalpy using Stoichiometry ∙ For reactions that the enthalpy have been experimentally determined, the value of the heat released or absorbed by a specific situation can be calculated using stoichiometry. The stoichiometric ratios become important here. ∙ For the combustion of methane: CH₄ + 2 O₂ → CO₂ + 2 H₂O, ΔHrxn= -890kJ/mol 1. If 1 mole of CH₄ and 2 moles of O₂ are burned, then ______ of heat will be released. (The reaction is ______ because the ΔH is negative) 2. The equation could also be written: CH₄ + 2 O₂ → CO₂ + 2 H₂O + _____ 3. This also shows the reaction is ____ because heat is a ____.

1. 890kJ, exothermic 2. 890kJ 3. exothermic, product

Enthalpy Practice Problem 1. A chemical reaction that gives off heat to its surroundings is said to be endothermic/exothermic and has a positive/negative value of ΔH. 2. As an ice cube melts under atmospheric pressure, is the process endothermic or exothermic and what is the sign of ΔH?

1. A chemical reaction that gives off heat to its surroundings is said to be endothermic/exothermic and has a positive/negative value of ΔH. 2. Melting (changing from a solid to a liquid) requires heat to be absorbed by the ice from the surrounding and, therefore, is endothermic. The sign of ΔH is positive.

Measuring the Heat Flow 1. How is this used to determine standard change in enthalpy? 2. What do the figures in the formula below represent? ∙ qsolution = mCΔT

1. Because the heat is reactant dependent, if the amount of reactant is known and the heat absorbed/released is measured, then the change in enthalpy can be calculated. 2. q = heat (Joules or KiloJoules), m = mass of the solution (grams), C = specific heat capacity of the solution (Joules/g°C), ΔT = Tfinal - Tinitial (°C)

Calorimetry 1. What is Calorimetry? 2. To raise the temperature of 1.00 gram of water 1.0°C, 4.184 J of heat is required (that's the definition of one Calorie - what you know of for food). This is referred to as the ______ ______ or ______ ______. 3. What does a calorimeter do? ∙ There must be something in the calorimeter to act as the surroundings to absorb or release heat to the system. Water is usually the substance in a calorimeter. Because the heat capacity of water is known (C = 4.184J/g°C), the exchange of heat between the system and the surroundings can be determined.

1. Calorimetry is the measure of heat flow. In the lab, this is one way standard enthalpies are determined experimentally. 2. Heat Capacity or Specific Heat. 3. Measure heat flow. In calorimetry, to measure heat flow, we can measure the magnitude of the temperature change as the heat flow proceeds.

Potential Energy Diagrams Practice Problem 1. Given the following graph (on the other side of the flash card), is the reaction endothermic or exothermic? 2. Justify your answer with a calculation of ΔH. 3. Indicate what value is represented: (a), (b), and (c).

1. Endothermic because the PE of the products is greater than the reactants so heat must be added to get this reaction to work. 2. ΔH = 300 - 100 = +200 3. ∙ A: Activation Energy of the forward reaction (+300) ∙ B: Activation Energy of the reverse reaction (+100) ∙ C: Enthalpy of the forward reaction (+200)

Ways to Calculate Enthalpy from Data ∙ What are the five ways to calculate enthalpy?

1. Stoichiometry 2. Hess's Law of Summation 3. Summation Equation 4. Using Bond Energies 5. Calorimetry

Units of Enthalpy 1. What is the SI Unit for Energy? 2. What is the formula for the one SI unit? 3. What does enthalpy mean?

1. The SI unit for energy is the joule (J). A joule (J) is defined as the kinetic energy of a 2kg mass moving at a speed of 1 m/s. A joule is not a large amount of energy, and we will often use kilojoules (kJ). 2. Ek= ½ mv² = ½(2kg)(1m/s)² =1 kgm²/s² = 1 J 3. Enthalpy is an extensive property meaning ΔH is proportional to the amount of reactant consumed in the process. More reactant will generate more heat. Therefore, the ΔHrxn = ΔH/1mol of reaction and the units for ΔH are usually kJ/mol.

Using a Coffee Cup Calorimetry ∙ Because the solution in the coffee cup is the "surroundings", the q of the "system" is equal in magnitude but opposite in sign from the solution. ∙ qsolution = -qrxn 1. What is the formula for enthalpy in coffee cup calorimetry?

1. The amount of heat released or absorbed by the reaction is dependent of the limiting reactant and therefore ΔH°rxn = qrxn/moles of Limiting Reactant. (ΔH°rxn = qrxn/mol LR)

Enthalpy 1. What is Enthalpy (ΔH°)? 2. What is the formula for enthalpy? 3. What is enthalpy at constant pressure? 4. What is enxothermic and endothermic? 5. What is the sign for enthalpy in exothermic and endothermic reactions?

1. The thermodynamic function that accounts for heat flow. 2. ΔH = Hproducts - Hreactants 3. At constant pressure, the enthalpy is equal to the energy flow as heat. 4. ∙ Energy (Heat) lost by the system to surroundings = Exothermic ∙ Energy (Heat) gained from the surroundings to system = Endothermic 5. ∙ ΔH = + for endothermic reactions (Energy of Products > Reactants) ∙ ΔH = - for exothermic reactions (Energy of Reactants > Products)

Potential Energy Diagrams 1. What are PE Diagrams (or Enthalpy Diagrams) used for?

1. To determine if a reaction is endothermic or exothermic (as well as enthalpy of the forward reaction and reverse reaction, activitation energy of the reverse and forward reaction). ∙ By comparing the PE of the reactants to the products, one can determine the components with the most potential energy and determine if the net energy is positive or negative - meaning is energy as heat released or absorbed by the system ∙ Calculating the enthalpy of the PE Diagram in the picture: ΔH = 20kJ - 40kJ = -20kJ (Exothermic Reaction)

Hess's Law of Summation ∙ To use Hess's Law to compute enthalpy changes, it is important to understand two characteristics of ΔHrxn: ∙ If a reaction is reversed, the sign of ΔHrxn is also reversed. "FLIP" 1. If an exothermic reaction is reversed, then it becomes _______ - hence the sign change or flip. 2. The magnitude of ΔHrxn is directly _______ to the quantities of reactants and products in a reaction. ∙ If the coefficients in a balanced reaction are multiplied, the value of ΔHrxn is multiplied by the same value. "MULTIPLY"

1. endothermic 2. proportional

Bond Energies and Enthalpy 1. Bond Energy values can be used to calculate the approximate ______ for reaction. 2. In order to break a bond, energy must be _______ to the system (endothermic). ∙ Why? Atoms are at their lowest potential energy levels when bonds are formed and therefore more stable. To separate two atoms (break a bond), energy must be ______. 3. Conversely, when bonds are formed, atoms return to a stable state with the lowest potential energy and the excess energy is released as heat (exothermic).

1. enthalpies 2. added ∙ added 3. released Energy is required to break bonds, energy is released when bonds form.

Enthalpies of Formations (ΔH°f) 1. Tables of the experimentally determined values called the enthalpy of formation (ΔH°f) can be used to determine the _______ of a reaction. ∙ It is important to note that a formation reaction is the formation of 1 mol of a compound from its elements in their standard states at standard conditions. 2. The " ° " on the (ΔH°f) indicates the standard conditions exist - pressure is ________, solutions are ________ and temperature is ________ or ________. 3. By definition, the standard enthalpy of formation of the most stable form of any element is ______. ∙ Example of a formation reaction: C (s) + 2 H₂ (g) + ½ O₂ (g) → CH₃OH (l), ΔH°f = -239kJ/mol Note: ∙ The elements are in there standard states: C is solid at room temperature, H₂ is diatomic, and a gas at room temp, etc. ∙ The equation is balanced for 1 mole of the compound which is why the coefficient for the O₂ is ½.

1. enthalpy 2. 1 atm, solutions are 1 M, and temperature is 298K or 25°C 3. Zero

Hess's Law of Summation 1. Hess's Law states if a reaction is carried out in a series of steps, ΔHrxn for the overall reaction ________ the sum of the enthalpy changes for the individual steps. The reaction of nitrogen and oxygen to produce nitrogen dioxide can be written in one step: ∙ N₂ + 2 O₂ → 2 NO₂, ΔHrxn= 68kJ/mol Or a series of steps: ∙ N₂ + O₂ → 2 NO, ΔH₁= 180kJ/mol ∙ 2 NO + O₂ → 2 NO₂, ΔH₂ = -112kJ/mol Net reaction: ∙ ______________________________ (and enthalpy of the reaction)

1. equals 2. N₂ + 2 O₂ → 2 NO₂, ΔHrxn= ΔH₁ + ΔH₂ = 68kJ/mol

Calculating with Enthalpies of Formations (ΔH°f) ∙ The formation reactions can be used with Hess's Law to determine the enthalpy of a reaction or a simplified calculation can be used: ΔH◦rxn = ΣΔH°f(products) - ΣΔH°f(reactants) ∙ The symbol Σ (sigma) means "the sum of" and the stoichiometric coefficients cannot be forgotten. - Calculate the standard enthalpy change for the following reaction: C₃H₈ (g) + 5 O₂ (g) → 3 CO₂ (g) + 4 H₂O (l) ∙ Use the equation given and looking up the enthalpy of formation (a table will usually be given). ∙ ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants)

1. ΔH°rxn = [3(ΔH°f CO₂ (g)) +4(ΔH°f H₂O (l))] - [1(ΔH°f C₃H₈ (g)) + 5(ΔH°f O₂ (g))] 2. Substitute in the values from the table: ΔH°rxn = [3(-393.5kJ) +4(-285.8kJ)] - [1(-103.85kJ) + 5(0kJ)] 3. ΔH°rxn = -2219.85kJ/mol ∙ Note: The Stoichiometry coefficients are used, mathematical orders of operations are followed and the Enthalpy of oxygen is 0 because it is an element in its standard state!

Stoichiometry Practice Problem #2 ∙ CH₄ + 2 O₂ → CO₂ + 2 H₂O, ΔHrxn= -890kJ/mol ∙ How much heat is released if 12.3 g of oxygen is burned in the presence of methane gas? ∙ Is the reaction exothermic or endothermic?

12.3 g O₂ x 1mol O₂/32.0 g x -890kJ/2 mol O₂ = -171kJ (or 171kJ of heat is released) ∙ Exothermic, heat is a product *Watch the stoichiometry! The reaction requires 2 moles of O₂ to produce 890kJ of heat.

Hess's Law of Summation Practice Problem #2 ∙ Calculate ΔH for the reaction (given the following chemical equations and their respective enthalpy changes): - 2 C (s) + H₂ (g) → C₂H₂ (g) ∙ C₂H₂ (g) + 5/2 O₂ (g) → 2 CO₂ (g) + H₂O (l), ΔHrxn = -1299.6kJ/mol ∙ C (s) + O₂ → CO₂, ΔHrxn= -393.5kJ/mol ∙ H₂ (g) + ½ O₂ (g) → H₂O (l), ΔHrxn= -285.8kJ/mol

2 CO₂ (g) + H₂O (l) → C₂H₂ (g) + 5/2 O2(g) ) ΔHrxn= +1299.6kJ/mol FLIP 2 C (s) + 2 O₂ → 2 CO₂, ΔHrxn= 2(-393.5kJ/mol) MULITPLY H₂ (g) + ½ O₂ (g) → H₂O (l), ΔHrxn= -285.8kJ/mol No change 2 C (s) + H₂ (g) → C₂H₂ (g), ΔHrxn= +226kJ/mol

Stoichiometry Practice Problem #1 ∙ CH₄ + 2 O₂ → CO₂ + 2 H₂O, ΔHrxn= -890kJ/mol ∙ Given the reaction above, how much heat is released if 8.2g of methane gas are burned in a constant pressure system? ∙ Is the reaction exothermic or endothermic?

8.2 g CH₄ x 1mol CH₄/16.0 g x -890kJ/1 mol CH₄ = -460kJ or 460kJ of heat is released ∙ Exothermic, heat is a product

What is Thermochemistry?

The branch of chemistry concerned with the quantities of HEAT evolved or absorbed during chemical reactions.

Bond Energies Practice Problem Using bond energies, calculate the enthalpy change for the combustion of ethanol (C₂H₅OH).

Using bond energies, calculate the enthalpy change for the combustion of ethanol. C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O 1. ΔH°rxn = ΣBond Energy (broken) - ΣBond Energy (formed) 2. ΔH°rxn = [1(C-C) + 5(C-H) +1(C-O) + 1(O-H) + 3(O=O)]- [4(C=O) + 6(O-H)] 3. ΔH°rxn = [1(347) + 5(413) +1(358) + 1(467) + 3(498)]- [4(799) + 6(467)] 4. ΔH°rxn = [4731]- [5998] = -1267kJ/mol *Note - this is the same as problem #6 so why are the answers different? Bond Energies are measured for the GASEOUS state and so the water produced is a gas. Heat is required to make the water for liquid to gas (endothermic) therefore less heat is release from the reaction overall.

Stoichiometry Practice Problem #3 ∙ 2 NO + O₂ → 2 NO₂, ΔHrxn= -112kJ/mol ∙ If 35.0 g NO reacts with 42.0 g of O₂, how much heat will be produced? ∙ Is the reaction exothermic or endothermic?

∙ 2 NO + O₂ → 2 NO₂, ΔHrxn= -112kJ/mol ∙ Use the Limiting Reactant to determine the amount of heat produced! ∙ 35.0 g NO x 1 mol NO/30.01 g x -112kJ/2 mol NO = -65.3kJ ∙ 42.0 g O₂ x 1 mol O₂/32.00 g x -112kJ/1 mol O₂ = -147kJ ∙ Because the NO is the limiting reactant, only 65.3kJ of heat is produced. ∙ The statement is produced because the reaction is exothermic (heat is a product).

Calculating using Calorimetry Data: A reaction 1. 1.00L of 1.00M Barium Nitrate solution at 25.0°C is mixed with 1.00 L of 1.00 M Sodium Sulfate solution at 25.0°C in a calorimeter. A solid forms and the temperature of the mixture increases to 28.1°C. Assuming the calorimeter absorbs only a negligible quantity of heat, that the specific heat of the solution is 4.184J/g°C, and that the density of the solution is 1.0g/ml, calculate the standard enthalpy change. A. Is the reaction exothermic or endothermic?

∙ Determine the mass of the solution: 2.00L total x 1.0g/ml = 2000g. ∙ Calculate ΔT = 28.1°C - 25.0°C = +3.1°C ∙ Calculate qsolution = (2000g)(4.182J/g°C)(+3.1°C) = +25940.8J = +26000J ∙ Determine qrxn (equal in magnitude to qsolution but opposite in sign) = -26000J ∙ This makes sense because the temperature increased indicating an exothermic reaction! ∙ Now the limiting reactant must be determined: Ba(NO₃)₂ + Na₂SO₄ → BaSO₄ + 2 NaNO₃ ∙ 1.00 mol of Barium Nitrate, 1.00 mol of Sodium Sulfate (either is the LR) ∙ Calculate ΔH°rxn = qrxn/moles of Limiting Reactant ∙ ΔH°rxn = -25940.8J/1.00mol = -25940.8J/mol or -25.9408kJ/mol ∙ ΔH°rxn = -26kJ/mol (A. exothermic!)

Hess's Law of Summation Practice Problem #1 Calculate ΔHrxn for 2 NO + O₂ → N₂O₄ using the following information: ∙ Reaction 1: N₂O₄ → 2 NO₂, ΔH₁ = +57.9kJ/mol ∙ Reaction 2: 2 NO₂ + O₂ → 2 NO₂, ΔH₂ = -112.0kJ/mol

∙ Reaction 1: 2 NO₂ → N₂O₄, ΔH₁ = -57.9kJ/mol FLIP Reaction 1 must be flipped to make N₂O₄ a product and therefore the ΔH₁ becomes -57.9kJ/mol. ∙ Reaction 2: 2 NO + O₂ → 2 NO₂, ΔH₂ = -112.0kJ/mol ∙ Reaction 2 has all reactants and products on the correct side with the correct coefficients. ∙ When the two reactions are added together, the 2 NO₂ cancel out and you are left with the correct overall reaction. ∙ ΔHrxn = -57.9kJ/mol + -112.0kJ/mol = -169.9kJ/mol ∙ The ΔHrxn for the overall reaction is -169.9kJ/mol.

Calorimetry Practice Problem: Calculating heat (q) 1. How much heat is needed to warm 300.0 g of water from 22°C to 98°C? ∙ qsolution = mCΔT

∙ qsolution = (300.0 g)(4.184J/g°C)(+76°C) = +95395.2J ∙ Therefore, 95,395.2J (or 95.3953 kJ) must be added to heat the water to 98°C.