Termokimia Part 2

8/8/2019 Termokimia Part 2

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THERMOCHEMISTRY

Created by Stephen LourensTanihaha



Berdasarkan jenis interaksi antara sistem denganlingkungan: Sistem terbuka: terjadi pertukaran materi dan energi

Sistem tertutup: hanya terjadi pertukaran energi Sistem terisolasi: tidak terjadi pertukaran materi

maupun energi.

Kalor reaksi yang berlangsung pada tekanantetap disebut perubahan entalpi ( H).

Pada umumnya, reaksi berlangsung padatekanan yang tetap sehingga kalor reaksiumumnya dinyatakan sebagai perubahanentalpi ( H).



Reaksi eksoterm & endoterm



Uji kepahaman Jelaskan perbedaan pokok antara reaksi eksoterm dan

endoterm dan beri contoh masing-masing? Manakah diantara peristiwa berikut ini yang merupakan

reaksi eksoterm?

kalor mengalir dari sistem ke lingkungan.

entalpi sistem bertambah.

berlangsung pada suhu rendah.

suhu lingkungan naik.

perubahan entalpi reaksi bertanda negatif.

Diketahui dua jenis reaksi sebagai berikut:

C(s) + O2 (g) CO2 (g) H = -393,5 kJ

2 NH3 (g) N2 (g) + 3 H2 (g) H = + 92 kJ

Gambarlah diagram tingkat energi untuk kedua reaksitersebut!-



Persamaan Termokimia

Persamaan reaksi yang mengikutsertakan perubahanentalpinya.

Jumlah mol zat yang terlibat dalam reaksi sama dengankoefisien reaksinya.

Entalpi bergantung pada keadaan zat (wujud) yang terlibatdalam reaksi, maka wujud zat harus dinyatakan.

CH4(g) + 2O2(g) CO2(g) +2H2O (l) H= -890.5 kJ

CH4(g) + 2O2(g) CO2(g) +2H2O (g) H= -802.3 kJ



Rules of Thermochemistry

1. H is directly proportional to the amount of reactant

or product

H2(g) + ½ O2(g)H2O(l) H= -286 kJ

2H2(g) + O2(g) 2H2O(l) H= -572 kJ



soal

1. H2(g) + Cl2(g) 2HCl (g) H= -185kJ

Calculate H when:

(a). 1.00 mol of HCl is formed

(b). 1.00 g of Cl2(g) reacts



Rules of thermochemistry

2. H for a reaction is equal in magnitude but

opposite in sign to H for the reverse reaction

H2(g) + ½ O2(g) H2O(l) H= -286 kJ

H2O(l) H2(g) + ½ O2(g) H= 286 kJ



Question

Write down the thermochemistry equation for:

1. Reaction between 1 mol Fe2O3 with Carbonforms Fe and CO. the H of the reaction is

621 kJ/mol of Fe2O3.2. The reaction between 1 mol of solid sulfur

(S) with O2 gas that forms SO2 gas. The H

of the reaction is -297 kJ/mol of sulfur.

3. The formation of 2 mol of NH4Cl solid from 1 mol N2 gas, 4 mol of H2 gas and 1 mol of Cl2, given that H NH4Cl = -314.4 kJ/mol



Standard Enthalpy Change ( Hº)

Defined as enthalpy change of the reaction measuredunder standard conditions (298.15K (25ºC) and 1

atm)

Standard Enthalpy Change of Formation ( Hºf )

Standard Enthalpy Change of Combustion ( Hºc)

Standard Enthalpy Change of Decomposition ( Hºd)



Standard Enthalpy Change of

Formation ( Hºf )

Enthalpy change of formation of one mole

subtance from its elements under standard

conditions.

For example, Hºf for the formation of 1

mol of methane gas (CH4) from grafite andhydrogen gas is -74.8 kJ/mol

C(s,graphite) + 2H2(g) CH4(g) Hº = -74.8 kJ



Question

1. Hºf NO2 is 33.84 kJ/mol. Write down the

thermochemical equation for the formation

of NO2 gas under standard conditions.2. On the formation of 10 grams MgSO4(s) , the

energy of heat about 107 kJ was released.

Write down the thermochemical equation for

the formation of MgSO4(s) . (Ar O=16;Mg=24; S=32)




Decomposition Hºd) Enthalpy change of decomposition of one

mole subtance into its elements under

standard conditions.

For example, Hºf for the decomposition of 1

mol of H2O(l)

is +286 kJ/mol

H2O(l) H2(g) + ½ O2(g) Hº = 286kJ




Combustion ( Hºc) Enthalpy change of complete combustion

of one mole subtance under standard

conditions.

For example, Hºf for the combustion of

1 mol of C(s,graphite) is -393.5 kJ/mol2C(s,graphite) + 2O2(g) 2CO2(g) Hº = -393.5 kJ



Holiday Homework

1. Find the definition of:

Standard enthalpy change of fusion

Standard enthalpy change of vaporizationAnd give the example!



Calculating H of Reaction

Calorimeter

Hesss Law (Law of Constant Heat

Summation) Enthalpy formation

Bond Energy

Termokimia Part 2

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