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Chapter 6 Chemical equilibrium

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6 Chemical equilibrium 6.1 Equilibrium condition & affinity of chemical reaction 6.2 Isothermal of chemical reaction 6.3 Equilibrium constant and chemical equation 6.4 Chemical equilibrium of heterogeneous reaction 6.5 Calculation of equilibrium constant 6.6 Standard forming Gibbs free energy 6.7 Calculation of equilibrium const form Q function 6.8 Effect of T, P, indifferent gas on chemical equilibrium 6.9 Simultaneous equilibrium 6.10 Coupling reaction

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6.1 Equilibrium condition and affinity of chemical reaction Chemical reaction system Extent of reaction Equation of thermodynamics Direction and limit of chemical reaction Affinity of chemical reaction

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6.1.1 Chemical reaction system Dekonder defined extent of reaction ξ v B: negative for reactant; positive for resultant.

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6.1.1.1 Extent of reaction At any moment of the reaction, we can use any reactant/resultant to stands for the extent of reaction, the values are all the same. The changing value of every substance must satisfy:

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6.1.1.2 Equation of thermodynamics When T, P certain, When ξ=1 mol:

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6.1.1.3 Conditions of equation (1) The chemical reaction of isothermal, isobaric and it do not export work; (2) Chemical potential μ B of every substances keep unchangeable. Formula (a): the micro change happens in the limit system; Formula (b): extent of reaction is 1 mol in the abundant system which. The concentration/ μ B is also unchangeable.

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6.1.2 Direction and limit of chemical reaction can be used to judge chemical direction. (Δ r G m ) T,P <0, reaction process to the right spontaneously; (Δ r G m ) T,P >0, reaction process to the left spontaneously; (Δ r G m ) T,P =0, reaction equilibrium. and

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6.1.2.2 Judgment The slope of the curve.. Because change is small, extent of reaction is in 0~1 mol. right spontaneously left spontaneously equilibrium

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6.1.2.3

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6.1.2.4 Mixing Gibbs free energy D+E=2F R, G maximum, D and E no mixing; P, G maximum, after D and E mixing; T, G maximum of all the substances in equilibrium, containing mixing G; S, G of pure resultant F.

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6.1.2.5 van ’ t Hoff equilibrium box In order to make the reaction process to the end, process must be in the van ’ t Hoff equilibrium box (point S from point R directly), prevent any mixture between the reactants or between the reactant and the resultant.

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6.1.2.6

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6.1.3 Affinity of chemical reaction In 1922, Belgium expert, De donder first brought in the concept of chemical reaction affinity, A, a state function. or

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A has the properties of “ potential ” : A>0 reaction process towards the positive direction A<0 reaction process towards the converse direction A=0 reaction gets equilibrium 6.1.3 A used as judgment

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6.2 Isothermal of chemical reaction Chemical potential of gas (B) Isothermal of chemical reaction Equilibrium constant Judgment of reaction direction

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6.2.1 Chemical potential of gas (B) If the gas is ideal gas, f B =p B. Put chemical potential expression into the calculation formula of (Δ r G m ) T,p : Go

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Make: 6.2.2.2 Chemical potential of gas (B)

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6.2.3 Isothermal of chemical reaction For a reaction, Q f is called fagasity quotient. Δ r G $ m (T) is also can be worked out. thereby, Δ r G m can be calculated.

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6.2.4 Equilibrium constant In equilibrium, Δ r G m =0, therefore: K $ f, thermodynamics equilibrium constant, or standard equilibrium constant. It is only temperature function.

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6.2.5 Judgment of reaction direction For ideal gas Chemical reaction isothermal formula is also can be denote by: K $ p >Q p Δ r G m <0 reaction process to the right spontaneously K $ p 0 reaction process to the left spontaneously K $ p =Q p Δ r G m =0 reaction gets to equilibrium

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T he changing value of standard Gibbs free energy; Obviously, in the chemical reaction equation, the computation coefficient assumes the multiple relation, the value of Δ r G $ m (T) is also assumes multiple relation, while the value of K $ f assumes exponential relation. （1）（1） （2）（2） 6.3 Equilibrium constant and chemical equation

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K p by pressure 6.3.1 K p When Σv B =0, the units of K p is 1. Experimental equilibrium constant: Use the real pressure, mole fraction or concentration of reactant/resultant.

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6.3.2 K x For the ideal gas, it accord with the Dalton partial pressure law, About mole fraction equilibrium constant.

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6.3.3 K c For the ideal gas, About mole concentration equilibrium constant

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6.3.4 K a Liquid reaction use activity to stands for the equilibrium constant K a therefore Because

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6.4 Chemical equilibrium of Heterogenous reaction The reaction covering a gas phase and a liquid/solid phase is called heterogenous chemical reaction. If the agglomerate phase is pure state, the chemical potential of pure state is its standard state chemical potential. therefore, K p =exp(p j p /p i r ).

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There is a reaction, gas can be seen as ideal gas: CaCO 3 =CaO(s)+CO 2 (g) K $ p =p(CO 2 )/p $ p(CO 2 ) is called the dissociation pressure of CaCO 3 (s). Example

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6.4.2 Dissociation pressure When certain substance dissociates at some gas pressure it produces, this gas pressure is called dissociation pressure. This pressure has fixed value in certain T If the gas it produces is more than one kind, therefore, the summation of all the gas pressure is called dissociation pressure.

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NH 4 HS(s)=NH 3 (g)+H 2 S(g) Dissociation pressure p=p(NH 3 )+p(H 2 S) Therefore, the thermodynamics equilibrium constant: example

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6.5 Calculation of equilibrium constant 6.5.1 Equilibrium conversion rate It is called theory conversion rate, is the percentage of reactant turning into the resultant, after getting equilibrium,

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6.5.2 Industrial conversion rate Because the reaction is not in equilibrium, the real conversion rate is smaller than equilibrium conversion rate. 6.5.3 Measuring of equilibrium constant 6.5.3.1 Physical method 6.5.3.2 Chemical method

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6.6 Standard forming Gibbs free energy standard mole reaction Gibbs free energy changing value Mole standard forming Gibbs free energy The ionic mole standard forming Gibbs free energy The use of numerical value of Δ f G $ m

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6.6.1 Standard forming Gibbs free energy Under the standard pressure, the Gibbs free energy changing value cause by the steady single substance forming 1 mol compound Δ r G $ m ( compound, substance state, temperature ) It is usually can be check out in the table at 298.15K.

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6.6.2.1 Calculate the thermodynamics equilibrium constant 6.6.2 The application of Δ r G $ m

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6.6.2.2 Calculation of Special K Some of equilibrium constant can not be measured easily. (1)-(2) we can get (3)

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6.6.2.3 E stimating reaction feasibility can be used to judge the direction of the reaction approximately.

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6.7 Calculating equilibrium constant from partition function Sharing zero-point energy value of chemical equilibrium Sharing zero-point energy value of chemical equilibrium Free energy function Heat content function

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6.7.1 Sharing zero-point energy value of chemical equilibrium The energy zero selection of the same substance particle will not effect the calculation of its energy changing value. When both of the quantum number of transition and vibrate are zero (J=0,v=0), their energy level is fixed as zero.

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6.7.1.2 Sharing energy expression of chemical equilibrium As for many substances, the energy zero point of every substance is different. The lowest energy level at 0K is chosen, the difference of energy from energy zero to sharing zero is ε 0. The partition function expression of F, G, H, U has an item U 0 (U 0 =Nε 0 ) while S, C v and p are unchangeable.

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6.7.1.2

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6.7.2 Free energy function Free energy function is also free energy function At 0K, U 0 =H 0 Therefore, is called free energy function Therefore Because

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When N=1 mol, NK=R, suppose at the condition of standard state Free energy function can be worked out from q. The free energy function of every substance at different temperature can be checked out in the table. 6.7.2.2 Free energy function

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6.7.2.3 Calculation of equil. Const. D + E = G + H the second item is the change value of reaction thermodynamics energy at 0K.

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6.7.2.4 Calculation of Δ r U $ m (0) value 6.7.2.4.1. From K $ value and the value of every substance free energy function, inversely.

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6.7.2.4.2 From G definition Both Plus one, minus one Δ r U $ m (0), then we can get:

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6.7.2.4.3 From Kirchhoff formula

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6.7.2.4 From molecule dissociation energy D

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6.7.2.5 From heat content function The heat content change of reaction and the value of heat content function are already known, so we can get the value of Δ r U $ m (0).

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6.7.3 Heat content function The value of heat content function can be worked out by q. When T is 298.15K, the value of H $ m (298.15K)-U $ m (0) can be checked out in the table.

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Make use of the value of heat content function to calculate the heat content change of the chemical reaction: 6.7.3.2 Calculation of Chem. Heat

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6.7.3.3 Calculation of K K N is the equilibrium constant which is denoted by the molecule number, q is the total partition function after disparting the zero energy. Suppose D+E=G

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If further disparting the V in the transition partition function, therefore, use f to denote the partition function: 6.7.3.3 Work out the f of every partition function, we can get the equilibrium constant value K c.

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6.8 Effect of T, p and indifferent gas on chemical equilibrium Effect of T Effect of p Effect of indifferent gas

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6.7.2.3 Calculation of equil. Const. D + E = G + H the second item is the change value of reaction thermodynamics energy at 0K.

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6.7.2.4 Calculation of Δ r U $ m (0) value 6.7.2.4.1. From K $ value and the value of every substance free energy function, inversely.

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6.7.2.4.2 From G definition Both Plus one, minus one Δ r U $ m (0), then we can get:

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6.7.2.4.3 From Kirchhoff formula

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6.7.2.4 From molecule dissociation energy D

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6.7.2.5 From heat content function The heat content change of reaction and the value of heat content function are already known, so we can get the value of Δ r U $ m (0).

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6.7.3 Heat content function The value of heat content function can be worked out by q. When T is 298.15K, the value of H $ m (298.15K)-U $ m (0) can be checked out in the table.

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Make use of the value of heat content function to calculate the heat content change of the chemical reaction: 6.7.3.2 Calculation of Chem. Heat

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Example 1/ 3227 计算甲醇反应 CO 2 ＋ 2H 2 ＝ CH 3 OH(g) 在 1000 K 时平衡常数 K p 。 已知下列数据（ T=1000 K 时） ： -{[G (T) － U(0K)]/T}/J K -1 mol -1 Δ f U (0 K)/kJ mol -1 CO 204.054 -113.813 H 2 136.984 0 CH 3 OH(g) 257.651 -190.246

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Answer-1 Δ {[G (T)-U (0 K)]/T} =220.371 J · K -1 · mol -1 Δ U (0 K)= Δ [ Δ f U (0 K)] =-76.433 kJ · mol -1 -RlnK = Δ {[G (T)-U (0 K)]/T} + Δ r U (0 K)/T =143.938 K =3.03 × 10 -8

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Exampl-2/3205 试由下面数据计算反应 N 2 (g) + 3H 2 (g) ＝ 2NH 3 (g) 在 1000 K 时的平衡 常数 K 。 已知 : 1000 K N 2 (g) H 2 (g) NH 3 (g) {(G (T) -H)/T} / J K mol -1 -197.9 -137.0 -203.5 298 K (H (T)- H )/ kJ mol-1 8.669 8.468 9.920 Δ f H /kJ mol -1 -46.10

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- RlnK = Δ r G /T = Δ [(G (T)-H )/T] 1000 K + Δ r H /T Δ [(G (T)- H )/T] 1000 K = 201.9 J · K -1 · mol -1 Δ r H (298K) = 2[ Δ H (NH 3 )]= -92.2 kJ · mol -1 Δ [H (T)- H ] 298 K = -14.23 kJ · mol -1 Δ r H = - Δ [H (T)- H ] 298 K + Δ r H (298 K) = - 77.97 kJ · mol -1 -RlnK= Δ r G /T = Δ [(G (T)-H )/T] 1000 K + Δ r H /T = 123.9 K = 3.381 × 10 -7

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Example 3/ 3152 已知 298 K 时下列数据： H 2 O(g) H 2 (g) O 2 (g) Δ f H/kJ mol -1 -241.83 0 0 S/J K -1 mol -1 188.74 130.58 205.03 试求 H 2 O(g) H 2 (g) ＋ (1/2)O 2 (g) 在 25 ℃时的 K 。

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Answer 3 Δ r H =- Δ f H (H 2 O) =241.83 kJ · mol -1 Δ r S = S (H 2 )+(1/2)S (O 2 )- S (H 2 O)= 44.355 J · K -1 · mol -1 Δ r G = Δ r H -T Δ r S =228 612 J · mol -1 K =exp(- Δ r G /RT) =0.84 × 10 -40

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Example 4/ 3165 在 448 ～ 688 K 的温度区间内，用分光光度 法研究了下面的气相反应： I 2 ＋环戊烯 2HI ＋环戊二烯 得到 K 与温度的关系为： lnK = 17.39 － (51 034/4.575) × (1/T) (1) 计算在 573 K 时，反应的 Δ r G ， Δ r S ， Δ r H (2) 若开始时用等量的 I 2 和环戊烯混合，温 度为 573 K ，起始总压为 101.325 kPa ， 试求平衡后 I 2 的分压；

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Answer Δ r G =-RTlnK =(92 743-144.58T/K)-- (1) 代以 T=573 K 得 Δ r G =9870 J · mol -1 将 Δ r G = Δ r H -T Δ r S 与 (1) 式比较, 得： Δ r H =92 743 J · mol -1, Δ r S =144.6 J · K -1 · mol -1 (2)K =exp(- Δ r G /RT) =0.1260 I 2 (g) + 环戊烯 (g) = 2HI(g) + 环戊二烯 (g) 平衡： (p /2)-p (p /2)-p 2p p K={[p 2 (HI)p]/[p(I 2 )p( 环 )]}(1/p ) =0.1260 解此三次方程得 ： p=0.1552p

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6.8.1 Effect of T For endothermic reaction, Δ r H m $ >0, T increases, K $ p increase too, it is good for the positive reaction. For exothermic reaction, Δ r H m $ <0, T increases the temperature, K $ p decreases. it is bad for the positive reaction. Gibbs-Helmholtz equation: Then

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6.8.1.2 Effect of T If the T cover is not so big, and Δ r H $ m can be considered as constant, so we can get the fixed integral formula is: If Δ r H $ m is relative with T, therefore, put the relative formula into the differential coefficient formula to integral, and use the table to work out the integral constant.

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6.8.1.3 For ideal gas reaction When ideal gas is denoted by concentration, because p=cRT, we can get This formula is useful in the gas reaction thermodynamics.

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6.8.2 Effect of p According to the principle of Le chatelier, increasing pressure, the reaction process to the direction of volume decreasing. Here, we can use the effect on the equilibrium constant by pressure to explain the principle from essential.

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K $ p is only temperature function 6.8.2.1 For the ideal gas

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6.8.2.1.2 therefore Because K $ c is only temperature function

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6.8.2.1.3

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6.8.2.1.4 K x is relative with P, The gas molecule numer decreases, adding pressure, the reaction process towards positive direction, it is true in opposite.

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6.8.2.2 For the agglomerate reaction ΔV * B >0, increasing P, K $ a decreases, it is bad for the reaction, it is true in opposite. When P is not too large, because ΔV * B is not so big, the effect by P can be ignored.

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6.8.3 Effect of indifferent gas Increases the indifferent gas, n increases, the item in the bracket decreases. Because K $ p is fixed value, therefore, will increase, the content of resultant will increase.

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6.9 Simultaneous equilibrium In one reaction system, if several reactions happen at the same time, when getting equilibrium state, this condition is called simultaneous equilibrium. We need to consider the change of every substance amount in every reaction. In every equilibrium equation, the amount of the same substance must keep identical.

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Example At 600K, CH 3 Cl(g) and H 2 O(g) form CHOH, then forming (CH 3 ) 2 O, two equilibrium coexist: (1) CH 3 Cl(g)+H 2 O(g)=CH 3 OH(g)+HCl(g) (2) 2CH 3 OH(g)=(CH 3 ) 2 O(g)+H 2 O(g) K $ p,1 =0.00154, K $ p,2 =10.6. The amount of CH 3 OH and H 2 O is the same in equilibrium, respectively. What is the equilibrium translation rate of CH 3 Cl.

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6.9.3 Example 2 Suppose the mole fraction of CH 3 Cl and H 2 O is 1.0, when getting equilibrium, the forming mole fraction of HCl is x, the forming (CH 3 ) 2 O is y. (1) CH 3 Cl(g)+H 2 O(g)=CH 3 OH(g)+HCl(g) 1-x 1-x+y x-2y x (2) 2CH 3 OH(g)=(CH 3 ) 2 O(g)+H 2 O(g) X-2y y 1-x+y

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6.9.4 Example 2 Because both of the reactions are zero. Therefore K $ p =K x Unite these two equations, then we can get x=0.048, y=0.009. The translation rate of CH 3 Cl is 0.148 or 4.8%.

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6.10 Coupling reaction Suppose two chemical reaction happen in the system, if the resultant of one reaction in another reaction is one reactant, therefore these two reactions are called coupling reaction.

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Use the reaction whose Δ r G $ m value is very negative to bring along the reaction whose Δ r G $ m is larger than zero. 6.10 example

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6.10.2 example 2 Therefore: (3) TiO 2 +C(s)+2Cl 2 (g)=TiCl 4 (l)+CO 2 (g) Δ r G $ m,3 =-232.44kJ.mol -1 After couple reaction (1) and (2), reaction (3) can take place smoothly. at 298.15K:

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Exercises P446/1,2,5 P447/9, 10 P447/1 P448/9

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P449/14 P450/16 P451/23 P453/33 P453/34

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