GEN-CHEM-CHAPTER-3.pptx yht7jj9uuhyggh77

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About This Presentation

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Slide Content

THERMOCHEMISTRY THERMOCHEMISTRY

THERMOCHEMISTRY THERMOCHEMISTRY a branch of chemistry that describes the energy changes that occur during chemical reactions .

THERMOCHEMISTRY THERMOCHEMISTRY Scientist classified everything that exist in the universe into 2 broad categories: matter

THERMOCHEMISTRY THERMOCHEMISTRY matter Matter is anything that has mass and volume (takes up space).  Can be thought as the “stuff” Scientists define energy as the ability/capacity to do work. That which moves the “stuff”

How can we move?

Solar energy photosynthesis Primary source of enrgy Chemical The chemical energy in our food is then converted to other forms of energy needed to sustain life. ENERGY How can we move?

ENERGY Can be transferred in two ways: cause motion change temperature

3.1 Energy Changes in Chemical Reactions Change in matter( converting into different product during chemical rxn ) are usually accompanied by: Absorption of energy evolution of energy

Energy change observed is actually a consequence of breaking or forming bonds.

Energy change observed is actually a consequence of breaking or forming bonds. Energy is absorbed to break bonds 2NH 3 + heat  3H 2 + N 2 BINARY COMPOUND DECOMPOSITION

Energy change observed is actually a consequence of breaking or forming bonds. When bonds are formed energy is released

Energy change observed is actually a consequence of breaking or forming bonds.

SYSTEM & SURROUNDING SYSTEM Portion of the universe single out for study SURROUNDING Everything else outside the system BOUNDARY OR WALL surface that encloses the system and separates it from its surroundings

SYSTEM & SURROUNDING

SYSTEM & SURROUNDING An  open system  can exchange both matter and energy with its surroundings A  closed system  can exchange energy but not matter with its surroundings An  isolated system  exchanges neither energy nor matter with the surroundings.

SYSTEM & SURROUNDING

SYSTEM & SURROUNDING

Exothermic & Endothermic Processes

Exothermic & Endothermic Processes

3.2 First Law of Thermodynamics The first law of thermodynamics, also known as  the law of conservation of energy   It states that energy can neither be created nor destroyed, but it can be changed from one form to another.

3.2 First Law of Thermodynamics

3.2 First Law of Thermodynamics CHEMICAL ENERGY ELECTRICAL ENERGY RADIANT ENERGY RADIATION THERMAL

3.2 First Law of Thermodynamics According to this law, energy can also be transferred back and forth between the system and the surroundings in the form of heat and work.   Δ  E = q + w

3.2 First Law of Thermodynamics   Δ  E = q + w

3.2 First Law of Thermodynamics   Δ  E = q + w   Δ  E ( + ), internal energy increases Δ  E ( - ), internal energy decreases

3.2 First Law of Thermodynamics Calculate Δ E if 100 kJ of heat energy is absorbed by the system and 30 kJ of work is done on the surrounding.

3.2 First Law of Thermodynamics Calculate Δ E if 100 kJ of heat energy is absorbed by the system and 30 kJ of work is done on the surrounding. Given: Q = +100 kJ (absorbed) W = -30 kJ (done by the system) Required: Δ E (internal energy) Solution: Δ E = q + w Δ E = 100 kJ – 30 kJ Δ E = 70 kJ

3.2 First Law of Thermodynamics (b) What is the change in internal energy of a system when a total of 150.00 J of heat transfer occurs out of (from) the system and 159.00 J of work is done on the system? (answer: 9 J )

3.2 First Law of Thermodynamics (a) Suppose there is heat transfer of 40.00 J to a system, while the system does 10.00 J of work. Later, there is heat transfer of 25.00 J out of the system while 4.00 J of work is done on the system. What is the net change in internal energy of the system? (answer: 9 J )

3.3Enthalpy of Chemical Reaction ENTHALPY : a thermodynamic quantity equivalent to the total heat content of a system. It is from Greek word “ enthalpein ”, meaning to warm. Δ H = H product – H reactant

3.3Enthalpy of Chemical Reaction Δ H = H product – H reactant ENDOTHERMIC: q = Δ H > 0 ( Δ H product > Δ H reactant ) q = + Δ H EXOTHERMIC: q = Δ H < 0 ( Δ H product < Δ H reactant ) q = - Δ H

Thermochemical Equation The enthalpy relation between reactants and products is usually shown in a thermochemical equation. A thermochemical equation is a balance equation in which the value of Δ H, with the appropriate sign, is usually writte n at the right side. CH 3 OH(g)  CO 2 (g) + H 2 (g) Δ H=+90.7 kJ(endothermic) H 2 (g) + F 2 (g)  2HF(g) Δ H=-557 kJ(exothermic)

Extensive Property of Enthalpy Enthalpy is an extensive property, meaning, the value of Δ H is dependent on the amount of reactants consumed. EXAMPLE: C 3 H 6 O(l) + 4O 2 (g)  3CO 2 (l) + 3H 2 O(l) ) Δ H=-1790kJ 1 mole C 3 H 6 O(l) = 1790kJ (released) 2 moles C 3 H 6 O(l) = 3580kJ (released) 58 g C 3 H 6 O(l) = 1790kJ (released) Since 58g C 3 H 6 O = 1 mole C 3 H 6 O

Extensive Property of Enthalpy C 3 H 6 O(l) + 4O 2 (g)  3CO 2 (l) + 3H 2 O(l) ) Δ H=-1790kJ Calculate th e amount of heat released during the combustion of: 0.30 moles of C 3 H 6 O 100g of C 3 H 6 O

3.4 CALORIMETRY Calorimetry is concerned with then measurement of heat transferred to or from a substance. Calorimeter is a device used to measure the amount of heat involved in a chemical or physical process.

3.4 CALORIMETRY What happens to a body when it absorb/release heat?

3.4 CALORIMETRY When a body absorb or release heat, its temperature changes. This temperature change depends on the capacity of the material used. Heat Capacity, C, is defined as the amount of heat require d to raise the temperature of 1 mole of substance by 1 * C.

3.4 CALORIMETRY Specific Heat, c, is defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 * C.

3.4 CALORIMETRY Amount of heat transferred is calculated using the following formula: q=mc( deltaT )

3.4 CALORIMETRY q=mc( Δ T) q is amount of heat absorbed or released in joule or calorie c specific heat (for water c=4.18J/g C or c=1cal/g C) m mass of solution in g Δ T change in temp( T final – T initial ) in C

3.4 CALORIMETRY What quantity of heat (in J) is necessary to raise 3.00 L of water ( D water =1.00 g/mL) from 22.0°C to 63.0°C?

3.4 CALORIMETRY q=mc Δ T m = 3L( ) = 3000g c = 4.18 Δ T = T final – T initial = 63 ° C - 22.0°C = 41°C q = (3000g)( 4.18 )( 41°C) = 515,000 J  

3.4 CALORIMETRY Molar Heat of rxn , Δ H is calculated using the equation: Δ H=  

3.4 CALORIMETRY What is the Δ H (in J/mole) required to raise 3.00 L of water ( D water =1.00 g/mL) from 22.0°C to 63.0°C?

3.4 CALORIMETRY q=mc Δ T m = 3L( ) = 3000g c = 4.18 Δ T = T final – T initial = 63 ° C - 22.0°C = 41°C q = (3000g)( 4.18 )( 41°C) = 515,000 J  

3.4 CALORIMETRY q = (3000g)( 4.18 )( 41°C) = 515,000 J Δ H= = Moles water = 3000g( ) = 166.67moles Δ H= = 3090.49 Every mole of water absorbs 3090.49 J  

3.5 STANDARD ENTHALPY OF FORMATION AND REACTION Enthalpies of formation can be used to calculate standard enthalpy of reaction. The standard enthalpy of reaction, Δ H , is the enthalpy change when all reactants and products are in their standard states. The standard enthalpy change for a given reaction is equal to: Δ H ° =  

3.5 STANDARD ENTHALPY OF FORMATION AND REACTION Elements in their standard state that are involved in the reaction are usually omitted since their heats of formation are equal to zero.( Δ H f  

3.5 STANDARD ENTHALPY OF FORMATION AND REACTION Al(s) + Fe 2 O 3 (s)  Fe(s) + Al 2 O 3 (s) Calculating standard enthalpy change for the reaction: Δ H -  

Sample Problem: Given the standard enthalpy of formation for the following: Calculate the standard enthalpy of reaction for combustion of hexane COMPOUNDS Hexane(l) -151.9 kJ/mol CO 2 (g) -393.5 kJ/mol H 2 O(l) -285.8 kJ/mol COMPOUNDS Hexane(l) -151.9 kJ/mol CO 2 (g) -393.5 kJ/mol H 2 O(l) -285.8 kJ/mol

1. Create a balanced chemical equation: 2C 6 H 14 (l) + 19O 2 (g)  12CO 2 (g) + 14H 2 O(l) 2. Calculate Δ H Δ H = [12(-395.5kJ/mol)+14(-285.8kJ/mol)] – [2(-151.9kJ/mol)] Δ H = -8419.4 kJ   COMPOUNDS Hexane(l) -151.9 kJ/mol CO 2 (g) -393.5 kJ/mol H 2 O(l) -285.8 kJ/mol COMPOUNDS Hexane(l) -151.9 kJ/mol CO 2 (g) -393.5 kJ/mol H 2 O(l) -285.8 kJ/mol

Hess’s Law Hess’s law is a very convenient way to obtain Δ H values for reactions that are difficult to carry out in a calorimeter . It provides an indirect method of determining heat of reaction using as basis related chemical reactions with previously determined values of Δ H f  

Hess’s Law Calculate the standard enthalpy of formation, Δ H 4 , produced from graphite and hydrogen gas according to the following reaction: C(graphite) + 2H 2 (g)  CH 4 (g)  

Hess’s Law Given the thermochemical equations of different reactions calculate Δ H C(graphite) + O 2 (g)  CO 2 (g) Δ H =-393.5 kJ H 2 (g) + ½ O 2 (g)  H 2 O(l) Δ H =-285.8 kJ CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2H 2 O(l) Δ H =-890.4 kJ  

Hess’s Law C(graphite) + O 2 (g)  CO 2 (g) Δ H =-393.5 kJ 2H 2 (g) + O 2 (g)  2H 2 O(l) Δ H =2(-285.8 kJ) CO 2 (g) + 2H 2 O(l)  CH 4 (g) + 2 O 2 (g) Δ H =890.4 kJ C(graphite) + 2H 2 (g)  CH 4 (g) Δ H =-74.7 kJ  

Decomposition of a compound is an example of forming/breaking of bond. Therefore this rection an Exothermic/Endothermic reaction

Differentiate the following and give one example for each system.

Find the change in internal energy of the system when 175 J of work is done on a system that evolves 50 J of heat to the surroundings

The change in internal energy for a reaction is 760 J. As the piston moves up, the system absorbs 839 J of heat from its surroundings. Is work done on the system? How much work was done?

Write the thermochemical equation for the combustion of hexane, with 3923.9 kJ of heat.

3.3Enthalpy of Chemical Reaction Δ H = H product – H reactant if Δ H is a positive value, what can we conclude about the system? EXOTHERMIC or ENDOTHERMIC

2Cl(g)  Cl 2 (g) Δ H=-243.4 kJ Which has higher enthalpy, 2Cl(g) or Cl 2 (g)?

How many joules are required to raise the temperature of 450 ml of water fro m 30C to 100C?

The specifi c heat of copper metal is 0.385 J/g-K. How many kJ of heat are needed to raise the temperature of 1.1 kg block of copper from 27.5C to 50C?

Calculate the standard enthalpy change, Δ H °, for the following reaction: 2CH 3 OH + 3O 2  2CO 2 + 4H 2 O Given: CH 3 OH Δ H f °=-238.7 kJ/mol CO 2 Δ H f °=-393.5 kJ/mol H 2 O Δ H f °=-285.8 kJ/mol
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