CLASS XI SEMESTER 2 SMKN 7 BANDUNG THERMOCHEMISTRY CLASS XI SEMESTER 2 SMKN 7 BANDUNG

Isi dengan Judul Halaman Terkait Thermo chemistry Be part of chemistry studying dynamics or change of chemical reaction by observing its(the temperature is only. In thermo chemistry there is system term, area. System is some of the universe we are studying, outside system is area. Relation between system and area; Hal.: 2 Isi dengan Judul Halaman Terkait

1.Enthalpy (H) and enthalpy change (ΔH) Reaction of happened in calorimeter bomb stays at volume which is permanent because canister bomb cannot be big or minimizes. Means if(when gas formed at reaction of here, pressure will be big hence accentual at alterable system. Because in the situation volume which permanent hence heat of reaction measured with calorimeter bomb called as heat of reaction at constant volume. Coffee cup calorimeter correlates with air and if (there are) any reaction yielding gas, its(the gas earns evaporates into the air and accentual at system earns remain to constant. Hence change of energy is measured with coffee cup calorimeter is temperature reaction of at permanent pressure. Hal.: 3 Isi dengan Judul Halaman Terkait

Reaction exothermic and endothermic reaction Exothermic reaction At exothermic reaction happened heat transfer from system kenvironmental or at the reaction is released [by] temperature.At exothermic reaction of price ? H = negativity ( -) Example : C(s) + O2(g) CO2 (g) + 393.5 kJ ΔH = -393.5 kJ Exothermic reaction At exothermic reaction happened heat transfer from system kenvironmental or at the reaction is released [by] temperature.At exothermic reaction of price ? H = negativity ( -) Example : C(s) + O2(g) CO2 (g) + 393.5 kJ ΔH = -393.5 kJ Hal.: 4 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait Endothermic Reaction Endothermic reaction At reaction of happened heat transfer from area of to system or at the reaction is required by temperature. At endothermic reaction of price Δ H = positive (+) Example : CaCO3(s)  CaO(s) + CO2(g) - 1785 kJ ; ΔH = + 1785 kJ Hal.: 5 Isi dengan Judul Halaman Terkait

Standard enthalpy change (ΔH) Enthalphy change reaction of measured at temperature 298°K and accentual 1 atm agreed on as standard enthalpy, expressed with symbol ΔH° Thermochemical equation is equation of reaction equiped with the price of enthalphy change (ΔH). Thermochemical equation besides express number of moles reaktan and number of product moles also express number of heats freed or absorbent at the reaction. To express level of enthalphy change happened at chemical reaction, applied set of kJ . Enthalphy change inmolar applied set of molar kJ /mol ( kJ mole-1) Example : Reaction of 1 methane gas mole with 2 oxygen gas mole frees heat 802,3 kJ at temperature 298°K and pressure 1 atm. Equation its (the thermochemistry can be written as follows : CH4(g) + 2O2 CO2(g) + 2H2O ΔH°= -802,3 kJ /mol Hal.: 6 Isi dengan Judul Halaman Terkait

standard enthalpy type of change type 1.Standard Forming Enthalphy Change ( ΔHf ): (ΔHf = standard enthalpy of formothion) ΔH to form 1 compound mole directly from its the elements measured by 298 K and pressure 1 atm.That need to be gave attention to standard forming enthalphy change is a. Because compound formed only 1 mole, equation may contains fraction coefficient. Example : H2(g)+ 1/2 O2 ? H2O ; ΔH = - 285,85 kJ Enthalpy forming in a state of standard is null Example: Al(s) ΔHf = 0,0 kJ Hal.: 7 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait 2. Change of enthalpy decomposition of standard (ΔHd):(ΔHd=Standar enthalpi of decomposition) ΔHd from decomposition of 1 direct compound mole becomes elements(=reverse from ΔH forming) Example : 1. Decomposition enthalpy change of water is 285,85 kj H2O (l) H2 (g) +1/2O2(g) ; ΔHd = +285,85 kj 2. Decomposition enthalpy change of gas NO is -90,4 kj/mole NO1/2 N2 +1/2 O2 ; ΔHd =-90,4 kj/mole Hal.: 8 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait 3. Combustion Enthalphy Change of Standard ( ΔHc): (ΔHc= standard enthalpy of combustion) ΔHc to burn 1 compound mole with O2 from the air which measured by 298 K and pressure 1 atm. Example: Combustion enthalphy change of gas CH4 is -802 kj/mol CH4(g) +2O2(g) CO2(g)+ 2H2O : ΔHc=-802 kJ 2. Combustion enthalphy change CH3OH is - 638 kJ /mol CH3OH(g) + 3/2O2(g) CO2 (g)+ 2H2O ΔHc= - 638 kJ/mol 4. Reaction Enthalpy: ΔH from an equation reaction of where matters which there is in quation reaction of expressed in set of mole and equation coefficients reaction of simple integer. Example : 2Al + 3H2SO4  Al2 (SO4)3 + 3H2 ; ΔH =- 1468 kJ Hal.: 9 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait 5. Neutralisation Enthalpy ΔH yielded ( always exotherm) at reaction of sour neutralizing acid or alkaline. Example: NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) ; ΔH = - 890,4 kJ/mol 6. Hukumlavoisier-laplace "Number of heats discharged at forming of 1 matter mole from Its the elements equal to number of heats needed to elaborates the matter becomes its the former elements." Mean : If reaction is returned [by] hence heat sign formed also returned from positive become negativity or on the contrary. Example : N2(g) + 3H2  2NH3 ΔH =-112 kJ 2NH3(g)  N2(g) + 3H2(g) ; ΔH = +112 kJ Hal.: 10 Isi dengan Judul Halaman Terkait

Law Hess about hot amounts Example of problem: 1. Known cycle diagram as follows We return equation by changing reaktan and product this thing means reaction to run from left of dextrorse heat of reaction cycle diagram hence its the reaction can be described as following Hal.: 11 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait Because enthalpy is state function, hence magnitude ΔH from reaction chemistry do not be depended from trajectory experienced by reactant to form result of reaction. To see the importance of Iesson about temperature from reaction this, we see change which have been recognized that is evaporation from water at Oits(the boiling point. Especially, we pay attention to change of 1 dilution mole of water, H2O at 100 C becomes 1 gaseous water mole, H2O(g) this absorbsi will 41 kJ , hence ΔH = + 41 kJ . Entirety change can be written with equation : H2O(l)  H2O (g) ΔH = +41 kJ Hal.: 12 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait initial state  situation of transition-1, ΔH2 situation of transition-1 situation of transition-2, ΔH3 Situation of transition-2 final situation, ΔH4(+) ------------------------------------------------------------------------------------------------------------- 4. Initial state  Final situation, ΔH1 So, ΔH1 =ΔH2 + ΔH3 + ΔH4 Hal.: 13 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait situation of transition might possibly more than two Cycle diagram Hessss Hal.: 14 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait Example of problem Reaction determines ΔH (ΔH4) Its the Cycle diagram create and energy level diagram! Reply : There is 4 reaction step : Initial state : C2H4(g) + 6 F2 g) situation of End : 2CF4(g) + 4 HF ( g) Phase reaction of 1,2 and 3 is arranged its the situation (right/ left) and its the coefficient checked off ( multiplied with a number) that if it is summed up result of his(its becomes reaction phase 4.4. Hal.: 15 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait Reaction phase phase Hal.: 16 Isi dengan Judul Halaman Terkait

Standard forming enthalpy Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait In doing this calculation, we assume Δ Hf0 for every element in the situation pure his, its is most stable at temperature 250C and 1 atm = 0. The price of heat of formation of standard for every element, ΔHf0 = 0 Hal.: 18 Isi dengan Judul Halaman Terkait

Heat Capacity and specific heat Heat capacity is number of temperatures needed to changes temperature an object equal to 10C. Specific heat is number of temperatures needed to boosts up temperature 1 matter gram equal to 10C. Example; specific heat ( specific heat) water 4,18 j/g0C, iron specific heat 0,452 j/g0C Formula; q = m.c.ΔT Description; q: number of heats ( joule) m : mass substance (gram) C : specific heat ΔT : temperature change ( t final –t initial) Hal.: 19 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait Calorimeter Calorimeter is equipment applied to measure heat of reaction Hal.: 20 Isi dengan Judul Halaman Terkait

Binding Energy and reaction enthalpy Binding energy is energy needed to decides chemical bond in 1 compound mole Chemical reaction between molecules requires tying resolving the and forming of new tying with structured atoms in differs in. The chemist has developed method for studies species between a real reactive, that is species which its(the tying has broken and has not lapped over again. For example, atom hydrogen can be taken away from methane molecule, By leaving two fragment, which both no has stable valence electron structure in image of elektron-titik Lewis. Both would continuously react acurately with molecule or other fragment, then form of stable reaction result. However, during brief existence of the reactive species, many in character earns measured. Hal.: 21 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait An important amount measured is enthalpy change when a tying breaks in gas phase so-called tying enthalpy. This enthalpy always positive because heat must be given into gathering of molecules stable to break its the tying. For example, enthalpy change tying for C-H in methane is 438 kJ/mole standard measured to react : Where one tying moles C-H is broken, one for every methane molecule. Rather tying enthalpy differs in out of one other compound compounds. Hal.: 22 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait Binding energy calculate : If forming enthalpy NH3 is -46 kJ, binding energy H-H =436 kJ , binding energy N-H = 390 kJ , how much is it binding energy NΞN? Solution : ΔHreaction={( binding energy NΞN)+3 ( binding energy H-H))-{2(3 binding energy N-H)}-92 kJ = {(binding energy NΞN)+3(436)-{2(3 x 390) }-92 kJ = {(binding energy NΞN)+1308)-{2340} So binding energy NΞN = -92 + 2340 – 1308 kJ = 940 kJ Hal.: 23 Isi dengan Judul Halaman Terkait

Isi dengan Judul Halaman Terkait Thanks Hal.: 24 Isi dengan Judul Halaman Terkait