TERMOKIMIA (Thermochemistry) PERTEMUAN KE 6 – KE 7 KIMIA FISIKA I TERMOKIMIA (Thermochemistry)

The study of the energy transferred as heat during the course of chemical reactions is called thermochemistry. Thermochemistry is a branch of thermodynamics because a reaction vessel and its contents form a system, and chemical reactions result in the exchange of energy between the system and the surroundings. Thus we can use calorimetry to measure the energy supplied or discarded as heat by a reaction, and can identify q with a change in internal energy (if the reaction occurs at constant volume) or a change in enthalpy (if the reaction occurs at constant pressure). Constant pressure; qp = H Constant volume ; qv = U

Conversely, if we know U or H for a reaction, we can predict the energy (transferred as heat) the reaction can produce. We have already remarked that a process that releases energy by heating the surroundings is classified as exothermic and one that absorbs energy by cooling the surroundings is classified as endothermic. Exothermic ∆H<0 Endothermic ∆H>0

Heat reaction is influenced by: - amount of matter - Physical state - Temperature - Pressure - Type of reaction (P constant or V constant) To writing the equation of chemical reaction need to write state (l,ag,s,), coeficient and experiment condition.. Example : Reaction of formation CO2 in1 atm and 298 K C(grafit)+ 2O2(g) CO2 (g) +393,515 kJ

Change of energy in constant pressure (atmosferic pressure), so  H = qp In Reaction : aA + bB cC + dD + x kJ If enthalpy of reactant = H1 Enthalpy of product = H2 So : H1 = H2 + x kJ H2-H1 = -x kJ  H = -x kJ

Standard enthalpy changes Changes in enthalpy are normally reported for processes taking place under a set of standard conditions. In most of our discussions we shall consider the standard enthalpy change ∆Ho, the change in enthalpy for a process in which the initial and final substances are in their standard states The standard state of a substance at a specified temperature is its pure form at 1 bar.

For example, the standard state of liquid ethanol at 298 K is pure liquid ethanol at 298 K and I bar; the standard state of solid iron at 500 K is pure iron at 500 K and 1 bar. The standard enthalpy change for a reaction or a physical process is the difference between the products in their standard states and the reactants in their standard states, all at the same specified temperature. As an example of a standard enthalpy change, the standard enthalpy of vaporization, ∆Hovap , is the enthalpy change per mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar, as in

Enthalpies of physical change The standard enthalpy change that accompanies a change of physical state is called the standard enthalpy of transition and is denoted ∆trans Ho The standard enthalpy of vaporization, ∆Vap Ho is one example. Another is the standard enthalpy of fusion, ∆Fus Ho , the standard enthalpy change accompanying the conversion of a solid to a liquid, as in

Enthalpies of chemical change We consider enthalpy changes that accompany chemical reactions. There are two ways of reporting the change in enthalpy that accompanies a chemical reaction. One is to write the thermochemical equation, a combination of a chemical equation and the corresponding change in standard enthalpy: Or 2A + B 3C + D the standard reaction enthalpy is

Hess's law Standard enthalpies of individual reactions can be combined to obtain the enthalpy of another reaction. This application of the First Law is called Hess's law The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided. The individual steps need not be realizable in practice: they may be hypothetical reactions, the only requirement being that their chemical equations should balance.

The thermodynamic basis of the law is the path-independence of the value of ∆rHo and the implication that we may take the specified reactants, pass through any (possibly hypothetical) set of reactions to the specified products, and overall obtain the same change of enthalpy. The importance of Hess's law is that information about a reaction of interest, which may be difficult to determine directly, can be assembled from information on other reactions.

Example 2.8:

Latihan Jika entalpi pembakaran standar 1-heksena dan heksana masing2 adalah -4003 kJ mol-1 dan -4163 kJ mol-1 dan entalpi pembentukan air adalah -285,83 kJ mol-1. Hitung entalpi hidrogenasi standar dari 1-heksena menjadi heksana.

Standard enthalpies of formation ∆Hof The standard enthalpy of formation, ∆Hof , of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference states. The reference state of an element is its most stable state at the specified temperature and 1 bar. Standard enthalpies of formation are expressed as enthalpies per mole of molecules or (for ionic substances) formula units of the compound Formulation of reaction entalphy

Exercise 2.14, 2.15 2.16, 2.17 and 2.20

Contoh soal 2.20 0,3212 gram glukosa dibakar dalam kalorimeter bom yang kapasitas kalornya 641 J K-1, temperatur naik 7,793 K. Hitung entalpi molar pembakaran standar, energi dalam pembakaran standar dan entalpi pembentukan glukosa standarnya. Jawab: Tentukan perub entalpi dari nilai q (q=C∆T) (Dicari dari hub U dan H (apakah ada perub mol gas) Tulis reaksi pembakarannya, kemudian gunakan perub entalpi standar pembentukan dari tabel untuk menentukan entalpi pembentukan standar glukosa

Macam-macam Panas /Perub entalpi Panas atomisasi : Panas yang diperlukan untuk menghasilkan 1 mol zat dalam bentuk gas dari keadaan yang paling stabil pada keadaan standar . Contoh : C grafit C(g)  H = 716,68 kJ Panas penguapan standar : panas yang diperlukan untuk menguapkan 1 mol zat cair menjadi upanya pada keadaan standar contoh : H2O(l) H2O(g)  H=44,01 kJ Panas peleburan standar : panas yang diperlukan atau dilepas pada peleburan . Contoh : H2O(s) H2O(l)  H = 6,0 kJ Panas pelarutan integral: Panas yang timbul atau diserap pada pelarutan suatu zat dalam suatu pelarut. Besarnya tergantung jumlah zat pelarut dan zat terlarut.

Panas pengenceran integral : panas yang timbul atau diserap jika suatu larutan dengan konsentrasi tertentu diencerkan lebih lanjut dengan menambahkan pelarut Panas pelarutan diferensial = panas yang timbul atau diserap jika 1 mol zat terlarut ditambahkan ke dalam sejumlah besar larutan tanpa mengubah konsentrasi larutan. Panas Pengenceran diferensial : Panas yang timbul atau diserap jika 1 mol pelarut ditambahkan ke dalam sejumlah larutan tanpa mengubah konsentrasi larutan tersebut. Panas netralisasi : panas yang diserap atau dilepaskan jika 1 mol ekivalen asam kuat tepat dinetralkan oleh 1 mol ekivalen basa kuat. Panas Hidrasi : panas yang timbul atau diperlukan pada pembentukan hidrat. Contoh : CaCl2 (s) + 2H2O (l) CaCl2 .2H2O (s)  H = -7960 kal

ENERGI IKATAN Energi disosiasi ikatan : energi yang diperlukan untuk memutuskan satu buah ikatan pada suatu molekul . Contoh : H2 (g) 2H (g) H 298 K=435,9 kJ/mol Energi ikatan : rata-rata dari energi disosiasi ikatan total Data energi ikat dapat digunakan untuk meramalkan panas reaksi pada pembentukan molekul sederhana C2H2 (g) + 2H2 (g) C2H6 (g)  H= ? Hr = ∑ H pemecahan + ∑ H pembentukan dengan: H pemecahan selalu (+) H pembentukan selalu (-) Contoh 2.7

Pengaruh suhu terhadap H Untuk Cp konstan Untuk reaksi : A + B C + D

Sehingga Hukum Kirchhoff

Contoh 2.16

Berapa perubahan entalpi untuk penguapan air pada suhu 0 oC dengan asumsi bahwa pada suhu 0 – 25 oC Cp tdk tergantung pada T

Persamaan Kirchoff Jika kapasitas panas pada tekanan tetap sebagai fungsi suhu, maka: Hubungan Cp dengan suhu dapat ditulis: Cp = a+bT+c/T2+d/T3 a,b,c,d adalah tetapan Cp = a+ bT+ c/T2+ d/T3

Exercise Cp of n-butane/JK-1mol-1= 19,41+ 0,233 T/K Calculate q for increasing of 1 mole n-butane from 25 oC to 300 oC in constant pressure. Answer: Use formula:

Tentukan kalor yang diperlukan untuk menaikkan suhu 1 mol N2 dari 25 oC menjadi 250 oC pada tekanan tetap, jika nilai kapasitas panasnya tergantung pada temperatur. (Gunakan perumusan Cp sebagai fungsi suhu) Penyelesaian: Tulis pers hukum kirchoff dg Cp fungsi suhu Integralkan persamaan itu Masukkan nilai a, b, dan c dari tabel 2.16 Hitung nilainya dengan teliti