LEARNING MATERIALS

Unit Education: SMA Negeri 1 UngaranClass / Semester: XI IPA / 1Subjects: ChemistryTopic: Law of conservation of energy, understanding the system and the environment, as well as understanding endothermic exothermic reaction.Time Allocation: 4 x 45 minutes

A. Learning objectives :1. Through the playing of video, student can explain the law of conservation of energy2. Through the playing of video and group discussion, student can distinguish between sistem and environment correctly3. Through the playing of video and group discussion, students can distinguish the reaction that release heat (exothermic) and reaction that receive heat (endothermic)4. Students can know the differences between exothermic and endothermic recation based on experimentB. Instructions Learning Materials:1. Use this resource to brief information about the Law of Conservation of Energy materials, definition of system and environment as well as the material of exothermic and endothermic reactions. If the condition does not need to be completed worksheets allow studentsC. Learning MaterialsThermochemical1. Law of conservation of energyThe law of conservation of energy is one of the basic laws of physics and therefore governs the microscopic motion of individual atoms in a chemical reaction. The law of conservation energy states:In a closed system, i.e., a system that isolated from its surrounding, the total energy of the system is conserved.In SI units, energy has units of Joules. 1 Joule = 1 kg m2s-2

Some forms of energy:1. Kinetic energy - energy of motion.2. Potential energy - energy of ``location'' with respect to some reference point.3. Chemical energy - energy stored in chemical bonds, which can be released in reactions.4. Electrical energy - energy created by separating charges; energy stored in a battery, for example.5. Thermal energy - energy given off as heat, such as friction.(Source: http://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_2/node4.html)

2. System and EnvironmentIf a piece of magnesium ribbon we enter into a hydrochloric acid solution, then the magnesium ribbon will soon dissolve or react with HCl with heat release which causes the beaker and its contents into heat. Mixture of magnesium ribbon and the HCl solution we call the system. While the beaker and the surrounding air we call the Environment. Thus, the system is part of the universe that has become the center of attention. Another part of the universe that we call the system interacts with the environment. Chemical system is a mixture of reagents which are being studied as in Figure 1.

Figure 1. System A mixture of magnesium and hydrochloric acid solutionIn general, a system is much smaller than its surroundings. In nature this happens many natural events or changes so that the system contains an infinite number, there are large (such as solar), small size (a human and a machine), and the small size (such as a cell and one atom). As a result, a small system can be in a large system, or the system is an environment for other systems. However, if a system is summed with the environment, will be as big as the other system is summed with the environment, called the universe.The universe is a coupled system environment (Figure 2) Therefore, the universe is only one, not both. The interaction between the system and the environment can be either exchange or exchanges matter and energy. In connection with that, the system can be divided into three, namely open systems, closed systems and isolated systems.

Figure 2. System and Environment

System is said to be open if the system and the environment can experience the exchange of matter and energy. Exchange material means no reaction products can leave the system (reaction vessel), for example, gas, or there is something of an environment that can enter the system.The system in Figure 1 belong to the open system. Furthermore, if the system is said to be closed between the system and the environment can not be an exchange of material, but it can be an exchange of energy. In an isolated system, no exchange of matter and energy with their environment (see Figure 3 below).

Figure 3. a) An open system, b) closed and c) isolated

Energy exchange between the system and the environment can be either heat (q) or other forms of energy that collectively we call work (w). The existence of the energy exchange will change the amount of energy contained in the system. Work is a form of energy exchange between the system and the environment outside heat. (Lewis, 2009)

LEARNING MATERIALS

Unit Education: SMA Negeri 1 UngaranClass / Semester: XI IPA / 1Subjects: ChemistryTopic: Law of conservation of energy, understanding the system and the environment, as well as understanding endothermic exothermic reaction.Allocation of Time: 6 x 45 minutes

A. Learning objectives :1. Students can know how to determined H of reaction based on calorimetri data2. Student can know how to determined H of reaction based on Hesss Law3. Student can solved the example of problem to determined H of reaction based on calorimetri data and Hesss Law4. Student can know how todetermined entalpi of eaction based on experiment

B. Tips Instructional Materials:1. Use this resource to brief information about the material enthalpy of reaction calorimetry is based on the data, Hess law, the standard enthalpy change of formation and bond energy data.

C. Teaching MaterialsEnthalpy of a system analogous to the contents of the wallet. Although we do not know the contents of one's wallet, we can determine the change purse to observe the change. If the person receives a sum of money, the purse will be increased. Conversely, if the issuing of money, the purse will be reduced. So is the chemical reaction, the energy of a substance can not be calculated, but we can calculate the enthalpy change. Enthalpy change of the reaction can be determined by experiment, Hess law, the standard enthalpy of formation of some substances and energy ties some substance.In this teaching material will be described materials on the reaction enthalpy determination based on the data calorimetry, with Hess's Law, standard enthalpy of formation and bond energy data based on several substances.In this teaching material will be described materials on the reaction enthalpy determination based on the data calorimetry, with Hess's Law, standard enthalpy of formation and bond energy data based on several substances.

A. Determination of H Through ExperimentEnthalpy change of the reaction can be determined by using a device called a calorimeter (gauges heat). In the calorimeter, a substance which will be put in place reacted reaction. The place is surrounded by the water of known mass. The heat of reaction that is released is absorbed by the water and the water temperature will rise. Changes in water temperature is measured with a thermometer. Calorimeter was placed in an insulated container filled with water to prevent the escape of heat.

Figure of Calorimeter

Based on the research results, to raise the temperature of 1 kg of water at 10C by 4.2 kJ of heat required or 1 kcal. Required to heat 1 gram of water by 4.2 A or 1 cal. The amount of heat is called the specific heat of water with the symbol c.The amount of heat that is absorbed into the water is calculated by multiplying three factors: the mass of water in the calorimeter (grams), changes in water temperature (0C), and the specific heat of water.The formula is written:Q = m.c.Tq = heat released or absorbedm = mass of water (g)c = specific heat of water (4.2 Jg-1 .)T = change in temperature ()(source:Problems Example:In the calorimeter are substances that react endothermic. These reactions lead to 1 kg of water contained in the calorimeter decreased 5 temperature. Determine the heat of reaction of the reaction!Solution:q = m.c.T= 1000 g. 4.2 Jg-1 .5 = 21000 J = 21 kJ

B. Determination of H with Hess's LawHess's Law is a law that is used to determine the enthalpy change of a reaction. In Hess law, the value of the enthalpy change is expressed as a function of the state (H). According to this law, because the change in enthalpy is a state function, the change in the chemical reaction will be worth the same in spite of the steps necessary to produce different reaction products. In other words, the change in enthalpy of a reaction is determined only by the initial state and the final state of the reaction and does not depend on the course of the reaction.

Figure 2. The process produces a reaction

Enthalpy change of a chemical reaction is the difference between the value of the enthalpy of formation enthalpy of formation of the reaction products and reactants. Mathematically it can be written as follows:

H = Hf reaction results - Hf reagent

To produce a product of the reaction, sometimes passing through several steps such as image 1 above. To produce C, there are two ways: from A to C or from A to B and then from B to C. For such a process, then apply:

H = H1 + H2

Problems Example:If known: C2H5OH + 3O2 2CO2 + 3H2O H = -327 kcalC + O2 CO2 H = -94 kcalH2O H2 + O2 H = -68,4kkalCalculate the enthalpy change that occurs in the formation of the elements C2H5OH according to the equation: 2C + 3H2 + O2 C2H5OH

Solution:All three of the above reaction we arrange such that when added together will produce C2H5OH formation reaction, and its H is the sum of the three reaction H.(C + O2 CO2) H = -188 kJ x2 (H price multiplied by 2)(H2 + O2 H2O) x3 H = -205.2 kJ (H price multiplied by 3)2CO2 + 3H2O C2H5OH + 3O2 H = -327 kJ2C + O2 + 3H2 C2H5OH H = -66.2 kJSo the enthalpy change that occurs in the formation of C2H5OH is -66.2 kJ

C. Determination of H Through Establishment Data StandardsChanges in energy (heat) in the formation of one mole of a substance directly from its elements are called enthalpy of formation. If measurements are made at standard conditions (298 K, 1 atm) and all the elements in the standard form, the enthalpy change is called standard enthalpy of formation (Hfo). example:C (s, graphite) + 2 H2 (g) CH4 (g) Hfo = -74.8 kJ / mol

Table 1. The standard enthalpy of formation of several substances

Chemical FormulaChemical EquationHfo(kJ mol-1)

H2O(l)H2(g) + O2(g) H20(l)-285,85

H2O(g)H2(g) + O2(g) H2O(g)-241,8

C(graphite)C(graphite) C(graphite)0

C(diamond)C(graphite) C(diamond)+1,88

C(g)C(graphite) C(g)+718,4

CO(g)C(graphite) + O2(g) CO(g)-110,5

CO2(g)C(graphite) + O2(g) CO2(g)-393,5

C2H5OH(l)2 C(grafiit) + 3 H2(g) + O2(g) C2H5OH(l)-277,7

NaCl(s)Na(s) + Cl2(g) NaCl(s)-410,9

C2H2(g)2 C(graphite) + H2(g) C2H2(g)+226,7

Table 1.1 Standard Enthalpy of Formation of Several SubstancesSubstancesHfo (kJ/mol)SubstancesHfo (kJ/mol)

H2(g)0H2O(l)-285,85

O2(g)0H2O(g)-241,8

N2(g)0CO(g)-110,5

C(grafit)0CO2 (g)-393,5

Cl2(g)0NaCl(s)-410,9

The elements have a value of zero standard enthalpy of formation in free form.

How to calculate H reaction based on the data standard enthalpy of formation:Hreaksi = Hfoproduk - Hforeaktan

d. Determination of the Enthalpy of Reaction by the Energy InstituteChemical reactions occur because of termination of the old ties and the formation of new bonds. At the termination of the bond required energy while in the formation of energy released.Bond energy is the energy required to break the chemical bonds in 1 mole of a compound in the gaseous state into a standard gas atoms. Average bond energy is the average energy that it need to decide on a bond in a compound.

Table 2. Energy Different Types of AssociationNo.BondingBonding Energy (kJ/mol)BondingBonding Energy (kJ/mol)BondingBonding Energy (kJ/mol)BondingBonding Energy (kJ/mol)

1.Single Bonding

C-H413Si-C301H-Br366S-S266

C-C348Si-O368H-I299FF155

C-N293N-H391O-H463Cl-F253

C-O358N-N163O-O146Cl-Cl242

C-F485N-O201O-F190Br-F237

C-Cl328N-F272O-Cl203Br-Cl218

C-Br276N-Cl200O-I234Br-Br193

C-I240N-Br243S-H339I-Cl208

C-S259H-H436S-F327I-Br175

Si-H323H-F567S-Cl253I-I151

Si-Si226H-Cl431S-Br218

2.Double Bonding

C=C614CN891N=N418S=O323

CC839C=O799NN941S=S418

C=N615CO1072O=O495

How to calculate H reaction by the bond energy:Hreaction = bond breaking energy - bond formation energyHreaction = bond breaking energy - bond formation energy

Summary

1. In a closed system, ie, a system that isolated from its surroundings, the total energy of the system is conserved.2. The system is part of the universe that has become the center of attention.3. Environment is another part of the universe that interact with our system.4. Changes in the reaction enthalpy can be determined by using a device called a calorimeter (gauges heat).5. Hess's Law is a law that is used to determine the enthalpy change of a reaction.6. In Hess's law, the value of the enthalpy change is expressed as a function of the state (H).7. In the Hess law enthalpy change of a reaction is determined only by the initial state and the final state of the reaction and does not depend on the course of the reaction.8. The enthalpy change during the formation of one mole of a substance directly from its elements is called the enthalpy of formation.9. The enthalpy change during the formation of 1 mole of a substance directly from the elements and performed at standard conditions (298 K, 1 atm) is called the standard enthalpy of formation (Hfo).10. The bond energy is the energy required to break the chemical bonds in 1 mole of a compound in the gaseous state into a standard gas atoms.11. The average bond energy is the average energy that it need to decide on a bond in a compound.

Exercise

A. OPTIONS PLURALPut a cross (X) in a, b, c, d or e to the best answer!1. Based on the interaction between the system with the environment, there are three kinds of systems in thermochemical one of which is ..A. The system is openB. Distributed SystemsC. System tetahanD. Separate SystemsE. System neglected

2. Heat required to break bonds by the gas molecules into atoms or groups in a gaseous state called ..A. Energy decompositionB. The standard enthalpy of formationC. The enthalpy of decompositionD. enthalpy of combustionE. Energy ties

3. How much heat is required to raise the temperature of 50 g of water from 25 C to 60 C? Known water specific heat of 4.18 J / gk.A. 7.215 kJB. 7.315 kJC. 7.415 kJD. 7.515 kJE. 7.815 kJ

4. The heat generated from the combustion of 15 grams of ethane (C2H6) (Ar C = 12 and H = 1) according to the reaction:2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (l) H = -3084 kJ is ....A. -385.5 kJB. -771 kJC. -1542 kJD. -1850.4 kJE. -885.5 Kj

5. The characteristics of exothermic reaction is ....A. environments absorb heat from the systemB. The system absorbs heat from the surroundingsC. systems and environments have the same heatD. heating system and the environment when added equal to zeroE. system and the environment has not changed

6. Diketahui persamaan termokimia berikut :H2 (g) + O2 (g) H2O (g)H = p kJH2 (g) + O2 (g) H2O (l)H = q kJH2 (g) + O2 (g) H2O (s)H = r kJKalor sublimasi es (padat menjadi gas) dalam kJ/mol adalah ...A. pB. r - pC. p - rD. p - q rE. r7. If 0.92 grams of ethanol (C2H5OH) with Mr = 46 burned perfectly and generated heat can raise the temperature to 500 cm3 of water of 250C becomes 380C. Enthalpy change of combustion heat of ethanol is ... (the specific heat of water = 4.18 kJ / mol)A. 27.17 kJ / molB. 135.9 kJ / molC. 271.7 kJ / molD. 1359 kJ / molE. 2717 kJ / mol

8. Given the enthalpy change following diagram:N2 (g) + 2O2 (g) 2 NO2 (g)H1 = -66 kJ

H2 = -114 kJ

H3 = x kJ2 NO (g) + O2 (g)

Heat of formation of 33.6 L (STP) NO gas is ....A. 36 kJB. 48 kJC. 72 kJD. 96 kJE. 24 kJ

9. The following thermochemical equation is the standard enthalpy change of formation (Hfo), except ....A. C (s) + 2 Cl 2 (g) CCl4 (l) Hf = -134 kJB. Ca (s) + Cl2 (g) CaCl2 (s) Hf = -795.8 kJC. Fe2 (s) + 3/2 O2 (g) Fe2O3 (s) Hf = -822.2 kJD. Na (s) + H2 (g) + C (s) + 3/2 O2 (g) NaHCO3 (s) Hf = -947.7 kJE. C (s) + 2 Cl 2 (g) of CCL2 (l) Hf = -134 kJ

10. Given the following thermochemical equation.H2 (g) + O2 (g) 2 H2O (l) H = -571 kJ2Ca (s) + O2 (g) 2CaO (s) H = -1269 kJEnthalpy of formation of Ca (OH) 2 is ...A. -694 kJmol-1B. -994 kJmol-1C. -684 kJmol-1D. -984 kJmol-1E. +684 kJmol-111. Dari data energi ikatan diketahui :C-C : 348 kJmol-1O=O : 500 kJmol-1C-H : 414 kJmol-1C=O : 740 kJmol-1O-H : 464 kJmol-1Besarnya H pembakaran 100 mol propane, C3H8 adalah .A. -1,644 x 105 kJB. -1,544 x 105 kJC. -1,444 x 105 kJD. -1,344 x 105 kJE. -1,344 x 105 kJ

15. A piece of metal inserted into a solution of hydrochloric acid to determine the reactivity of hydrochloric acid on the metal. In this case the so-called system is ....A solution of hydrochloric acidB. MetalC. solvent waterD. hydrochloric acid and metalsE. aquadest

B. ESSAYAnswer the following questions correctly on the answer sheet that has been provided!1. Know Hf H2CO3 (aq) = -699.65 kJ / mol. Calculate the magnitude of the change in enthalpy at 496 grams decomposition H2CO3 (Ar H = 1, C = 12, O = 16) and write the equation thermochemical decomposition of H2CO3.

2. Given the reaction:CaCO3 (s) CaO (s) + CO2 (g) H = + 130 kJ3 CaO (s) + 2 H3PO4 (l) Ca3 (PO4) 2 (s) + 3 H2O (l) H = - 512 kJCalculate the magnitude of H of the reaction:CaCO3 (s) + 2 H3PO4 (l) Ca3 (PO4) 2 (s) + 3 H2O (l) + 3 CO2 (g)

3. Know CCl4 (g) C (g) + 4CL (g) Ho = 1303.58 kJDetermine the average bond energy CCl in CCl4.

4. Using the bond energy table, determine the enthalpy change of the following reaction:a. CH4 (g) + Cl 2 (g) CH 3 Cl (g) + HCl (g)b. H O H C OH (g) + O2 H C OH (g) + H2O(g) Hc. CH2CH2(g) + HCl(g) CH3CH2Cl(g) O d. CH3CH (g) + H2(g) CH3CH2OH(g)

5. Given:Hf CH4 (g) = -75 kJ / molheat of vaporization C (s) = 714 kJ / molbond energy of H2 (g) = 437 kJ / molCalculate the magnitude of the average bond energy C - H in the reaction:CH4 (g) C (s) + 4 H (g)