Energy Chapters 3,4,5,6

Thermodynamics: The branch of science that deals with the study of energy and its transformations Thermochemistry : The study of heat changes that occur in a chemical reaction

What is Energy ? It is the capacity of a system to do work Can cause a change in matter (change of state) Plays a significant role in chemical reactions It is neither created nor destroyed, it can only transform Energy changes are measured in joules (J) 1 joule = 1 N m 1N = 1 kg x m/s2

Energy in Reactions Most chemical reactions either release or absorb energy usually in the form of heat. Energy is always conserved amongst two parts: i) the system – the chemical reaction ii) the surroundings – everything outside the system The total energy is : system + surroundings

1. Exothermic Reactions Occurs when there is a release of energy When atoms combine to form a compound Example: combustion burning of a substance Energy (in kJ) written on the products side of equations System surroundings 2H2 (g) + O2 (g)  2 H2O (g) + 483.6 kJ heat

2. Endothermic Reactions Occurs when energy is absorbed during the reactions When a compound breaks apart into individual atoms (energy required to break bonds) Energy (in kJ) written on reactants side of equations Surroundings system H2O (s) + 6.02 kJ  H2O (l) 2H2O (l) + 285.8 kJ  H2 (g) + ½ O2 (g) heat

Experimental Observations for Energy Change Exothermic Heat loss by system to surroundings Container feels warm Endothermic Heat gained by system from surroundings Container feels cool Measuring heat Exothermic reaction: heat given off and temperature of water rises Endothermic reaction: heat taken in and temperature of water drops

Enthalpy (H) Def’n: is a measure of the energy content of a system (At a constant pressure) Molar enthalpy: measures energy for exo/endo of one mole of a substance The heat content of a substance. Heat cannot be measured directly so we measure the change of enthalpy (delta H).

Enthalpy changes (ΔH) Def’n : is a measure of the heat change in a reaction HR = enthalpy of the reactants HP = enthalpy of the products Reactants (HR)  Products (HP) .

If ΔH > 0 , the reaction is endothermic (heat is absorbed). Energy lies on the left of the equation Writing the equation: 2NH3 (g) + 92.4 KJ  N2(g) + 3H2 (g) 2NH3 (g)  N2(g) + 3H2 (g) ΔH = + 92.4 kJ/mol

If ΔH < 0, the reaction is exothermic (heat is released by system). Energy lies on the right of the equation Writing the equation: N2 (g) + 3H2 (g)  2NH3 (g) + 92.4 KJ N2 (g) + 3H2 (g)  2NH3 (g) ΔH = - 92.4 kJ/mol

Writing the equation: N2 (g) + 3H2 (g)  2NH3 (g) + 92.4 kJ N2 (g) + 3H2 (g)  2NH3 (g) ΔH = - 92.4 kJ/mol

Forms of Energy Kinetic: Energy due to motion Potential: Stored energy due to position of the atoms within a molecule. It is stored in the bonds that hold atoms together.

Vibrational Energy : the atoms move towards and away from the center of mass Rotational Energy : the molecule rotates around its center of mass Translational Energy: the molecule moves from place to place

kinetic energy kinetic energy kinetic energy

Reaction Pathway Shows the change in energy during a chemical reaction Activation energy ? ΔH=

2H2(l) + O2(l)  2H2O(g) + energy A. Exothermic Reaction reaction that releases energy products have lower Potential Energy than reactants energy released 2H2(l) + O2(l)  2H2O(g) + energy

B. Endothermic Reaction reaction that absorbs energy reactants have lower Potential Energy than products energy absorbed 2Al2O3 + energy  4Al + 3O2

Types of Systems Open system : is a system that freely exchanges energy and matter with its surroundings. Closed system: is a system that exchanges only energy with its surroundings, not matter. Isolated system: isolated system does not exchange energy or matter with its surroundings.

Calorimetry and the calorimeter Calorimetry is used to experimentally determine the quantities of heat involved in transformation Calorimeter: instrument that experimentally determines the heat energy (ΔH) absorbed or released during a reaction. Reaction in a sealed container representing a closed system Determines the enthalpy of a substance undergoing a chemical change

Don’t copy (Can be found on pg 132 ans 143 of textbook)

Calorimeter with Styrofoam Two Styrofoam cups One thermometer - Insulated container that contains a thermometer

Q= m x c x ΔT Thermal Energy Q = Quantity of heat energy, expressed in joules (J) m= mass of the water in the calorimeter in g c= the specific heat of water or what ever is acting as the environment (J/(g °C)) Specific heat: the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius (pg 134) c for water = 4.19 J/(g °C) ΔT = Temperature change (Tf – Ti) expressed in °C PAGE 136 137 138

Calculate the quantity of thermal energy absorbed by a 5 kg block of concrete to raise its temperature from 17.1°C to 35.5°C. m=5kg = 5000g c= 2.10 J/(g °C) specific heat Ti = 17.1 °C Tf = 35.5 °C ΔT = ? Q = ?

A) ΔT = ? ΔT = Tf - Ti ΔT = 35.5 °C - 17.1 °C ΔT = 18.4 °C B) Q= m x C x ΔT = (5000 g)x (2.10 J/(g °C)) x (18.4 °C ) = 193 200 J

Heat Transfer between two systems Heat travels from hot  cold -Q1 = Q2 -Qlost = Qgained Therefore -(m1c1ΔT1)= m2c2ΔT2 ΔT1 = temp. change in system 1 ΔT2 = temp. change in system 2 Nov 3) !!!!

Calculate the mass of cold water at 10°C needed to cool to 30°C a 10-g piece of glass at 95°C. c1 = 0.84 J/(g °C) c2=4.19 J/(g °C) Data m1= 10 g Ti1 = 95 °C Tf1= 30 °C m2= ? Pg 138

1) Calculate T1 ΔT1 = Tf1 – Ti1 = 30 °C - 95 °C = - 65 °C 2) Calculate T2 ΔT2 = Tf2 – Ti2 = 30 °C - 10 °C = 20 °C 3) -(m1c1ΔT1)= m2c2ΔT2 m2= -(m1c1ΔT1)= -(10 x 0.84 x -65 ) = 6.5g of water c2ΔT2 4.19 x 20

Temperature of the two systems Tf = m2c2Ti2 + m1c1Ti1 m1c1 + m2c2 Tf = Final temperature of the two systems in °C Tf = m2Ti2 + m1Ti1 m1 + m2

Calculate the final temperature when 200g of water at 20°C is mixed with 400g of water at 60°C. Heat loss = Heat gained -Q= Q -(m x c x ΔT) = m x c x ΔT -( 400x 4.19 x (χ-60)) = (200)x(4.19)x ( χ- 20) -(1676 (χ-60)) = 838 (x-20) -1676χ + 100560 = 838χ – 16760 117320 = 2514 χ 2514 2514 χ = 46.7 °C

Tf = mHTH + mCTC Tf = m2Ti2 + m1Ti1 mH + mc m1 + m2 Tf = Final temp. in °C mH= mass of hot water mc= mass of cold water TH = temp. of hot water Tc= temp. of cold water Tf = (400x 60) + (200 x 20) 400 + 200 = (24000 ) + (400) 600 = 46.7 °C

Molar Enthalpy and Thermal Energy Q= m x c x ΔT n= mass/Molar Mass ΔH = -Q/n (kJ/mol) Nov 8 ΔH= (+)  Endothermic ΔH= (-)  Exothermic

Find the molar heat of LiOH when you dissolve 4. 0g of LiOH in 200 Find the molar heat of LiOH when you dissolve 4.0g of LiOH in 200.0 g of water causing the water temperature to increase from 25.0°C to 31.5°C LiOH m= 4.0g Water m= 200.0 g Ti = 25 °C Tf= 31.5 °C c=4.19 J/(g °C)

Step 1: Q= m x c x ΔT = 200x 4.19 x (31.5 – 25) = 5447 J = 5.447 kJ Step 2: n= mass/Molar Mass = 4.0g / 24.0 g/mol = 0.167 mol Step 3: ΔH = -Q/n = - 5.447 / 0.167 = - 32.6 kJ/ mol

In a calorimetry experiment, 4 In a calorimetry experiment, 4.24g of LiCl(s) is dissolved in 100g of water at an initial temperature of 16.3 °C. The final temperature of the solution is 25.1 °C. Calculate the molar enthalpy ΔH, for LiCl. LiCl m= 4.24g Water m= 100.0 g Ti = 16.3 °C Tf= 25.1 °C c=4.19 J/(g °C)

Step 1: Q= m x c x ΔT = 100x 4.19 x (25.1 – 16.3) = 3687.2 J = 3.69 kJ Step 2: n= mass/Molar Mass = 4.24g / 42.39 g/mol = 0.10 mol Step 3: ΔH = -Q/n = - 3.69 / 0.1 = - 36.9 kJ/ mol

Percent error Percent error = (experimental value) - (theoretical value) x 100 theoretical value

Find the molar heat of LiOH when you dissolve 4. 0g of LiOH in 200 Find the molar heat of LiOH when you dissolve 4.0g of LiOH in 200.0 ml of water causing the water temp to increase from 25.0°C to 31.5°C Step 1: Q= m x c x ΔT = 200x 4.19 x (31.5 – 25) = 5447 J = 5.447 kJ Step 2: n= mass/Molar Mass = 4.0g / 24.0 g/mol = 0.167 mols Step 3: ΔH = -Q/n =- 5.447 / 0.167 = - 32.6 kJ/ mol

Percent error Percent error = (experimental value) - (theoretical value) x 100 theoretical value

Making and breaking bonds. Pg 156-157 But examples of bond energies

ΔH = H bonds broken - H bonds formed (reactants) (products) Each type of chemical bond has a characteristic bond energy. Breaking bonds requires energy. It is endothermic. Making new bonds gives out energy. It is exothermic. The energy associated for different bonds can be found on page 156 of your textbook. To calculate ΔH use the formula: ΔH = H bonds broken - H bonds formed (reactants) (products) Breaking bonds absorbs energy – It is endo Forming bonds releases energy – it is exothemic

Can be found on page 156 of textbook (Don’t Copy ) Bond Energy (kJ/mol) H-H 436 H-O 460 H-Cl 432 C=C 607 0=0 358 N=O 631 N Ξ N 418

H2(g) + Br2(g) 2HBr(g) Calculate the DH for the above reaction. DH = H bonds broken - H bonds formed DH =(436+192) - (368+368) DH = - 108 kJ/mol Bonds Broken: H-H 436 kJ/mol Br-Br 192 kJ/mol Bonds Formed: H-Br 368 kJ/mol

Calculate the enthalpy change for the following reactions H2 (g) + Cl2 (g)  2 HCl (g) CH4(g) + 2O2(g)  CO2 (g) + 2H2O(g)

CH3COOH(aq) + 5/2 O2(g)  2CO2(g) + 3H2O(g) H C C H O H