Chapter 4 Chemical equilibria, entropy and Gibbs function X 4.1 The standard equilibrium constant X 4.2 The application of the standard equilibrium constant X 4.3 The shift of a chemical equilibrium X 4.4 Spontaneous reactions and entropy x§4.5 Gibbs function
Chapter 4 Chemical equilibria, entropy and Gibbs function §4.3 The shift of a chemical equilibrium §4.2 The application of the standard equilibrium constant §4.1 The standard equilibrium constant §4.5 Gibbs function §4.4 Spontaneous reactions and entropy
4.1 The standard equilibrium constant 4.1.1 The basic features of a chemical equilibrium 4.1.2 The standard equilibrium constant expression 4.1.3 Experimental measurement of the standard equilibrium constant
§4.1 The standard equilibrium constant 4.1.3 Experimental measurement of the standard equilibrium constant 4.1.2 The standard equilibrium constant expression 4.1.1 The basic features of a chemical equilibrium
4.1.1 The basic features of a chemical equilibrium Most reactions are reversible,an example is ca-LH,(g+l,(g=2Hg”x10°x10 mol.L.s- t/s 0 0.01000.0100 0 7.60 0 2000 0.003970.003970.01211.20 2.04 4850 0.002130.002130.0157 0.345 3.43 85000.002130.002130.01570.3453.43
4.1.1 The basic features of a chemical equilibrium 0 0.0100 0.0100 0 7.60 0 2000 0.00397 0.00397 0.0121 1.20 2.04 4850 0.00213 0.00213 0.0157 0.345 3.43 Most reactions are reversible, an example is t/s 1 / mol L − c 6 + 10 7 − 10 1 1 mol L s − − H (g) I (g) 2HI(g) 2 + 2 8500 0.00213 0.00213 0.0157 0.345 3.43
Initially,concentrations of H2 and I2 are large,only the forward reaction occurs;As time passes,the concentrations of H,and I, decrease and in result the forward reaction slows down.The concentrations of HI increase and in result the reverse reaction speeds up. The reaction mixture is at equilibrium until the forward and reverse reactions go at the same rate
Initially, concentrations of H2 and I2 are large, only the forward reaction occurs; As time passes, the concentrations of H2 and I2 decrease and in result the forward reaction slows down. The concentrations of HI increase and in result the reverse reaction speeds up. The reaction mixture is at equilibrium until the forward and reverse reactions go at the same rate
0正 I-Tlow/5 HI H2, t/s t/s H,(g+I,(g=2HΠ(g) HI 2HⅢ(g)÷H2(g)+2(g) H2,k t/s
正 逆 正 = 逆 H (g) I (g) 2HI(g) 2 + 2 2HI(g) H2 (g) + I2 (g) Conc
A chemical equilibrium: A reversible chemical reaction can be at equilibrium under certain conditions: U正=D逆≠0 The basic features of chemical equilibrium: (1)the composition of an equilibrium mixture undergoes no further change with time. (2)chemical equilibria are dynamic. (3)the equilibrium composition is independent of the path of approach
A chemical equilibrium： 正 = 逆 0 A reversible chemical reaction can be at equilibrium under certain conditions: The basic features of chemical equilibrium： (1) the composition of an equilibrium mixture undergoes no further change with time. (2) chemical equilibria are dynamic. (3) the equilibrium composition is independent of the path of approach
4.1.2 The standard equilibrium constant expression Equilibria involving gases H2(g+I2(g)=2HΠ(g) K- [p(HⅡ)/po]2 p=100kpa p(H2)/pe][p(I2)/pe] Equilibria in aqueous solution: Sn2+(aq)+2Fe3(aq)-Sn4+(aq)+2Fe2*(aq) ko_[c(Snce )Ic(Fe2c)2 [c(Sn2+c )]Ic(Fe3+c)]2 c=1 mol.l-1
4.1.2 The standard equilibrium constant expression ☆Equilibria in aqueous solution： H (g) I (g) 2HI(g) 2 + 2 ☆Equilibria involving gases ： Sn2+(aq)+2Fe3+(aq) Sn4+ (aq)+2Fe2+(aq) [ (H )/ ][ (I )/ ] [ (HI)/ ] 2 2 2 p p p p p p K = p = 100kpa 2 3 2 4 2 2 [ (Sn )][ (Fe )] [ (Sn )][ (Fe )] c /c c /c c /c c /c + + + + K = c = 1 mol.l-1
For a more general chemical reaction: aA(g)+bB(aq)+cC(s)=xX(g)+yY(aq)+zZ(1) o-lox)ip() p=100kpa Lp(A)/pe]ac(B)/c c=1 mol.I-1 *The value of a constant ke depends on temperature and is independent of concen- trations and partial pressures of the reaction components..Kis a dimensionless quantity(无量纲，google) *The form of a standard equilibrium constant expression depends on exactly how a stoichio- metric chemical equation is written
For a more general chemical reaction： aA(g) + bB(aq) + cC(s) xX(g) + yY(aq) + zZ(l) K ( ) ( ) ( ) ( ) b c c a p p y c c x p p A / B / X / Y / = The value of a constant depends on temperature and is independent of concentrations and partial pressures of the reaction components. K The form of a standard equilibrium constant expression depends on exactly how a stoichiometric chemical equation is written. K is a dimensionless quantity (无量纲,google) p = 100kpa c = 1 mol.l-1
H2(g+I2(g=2HΠ(g 足 = [p(H)/po12 [p(H2)/pe][p(L2)/pe] H2(g+号12(g)-Hg 月 2 2 K月 [p(HI)/pe] [pH,)/p9T2pL,/p9=(9n 2HI(g)=H2(g)+I2(g) k3-pH,/p91Lp)'p-(9 [p(HI)/pe
H (g) I (g) 2HI(g) 2 + 2 [ (H )/ ][ (I )/ ] [ (HI)/ ] 2 2 2 p p p p p p = 2 I 2 (g) HI(g) 2 1 H (g) 2 1 + ( )1/2 = = 1/ 2 2 1/ 2 2 [ (H )/ ] [ (I )/ ] [ (HI)/ ] p p p p p p 2HI(g) H (g) I (g) 2 2 + =( )-1 [ (H ) / ][ (I ) / ] [ (HI) / ] 2 2 2 p p p p p p = K1 K1 K3 K3 K1 K2 K1 K2
The principle of multiple equilibria Example:Equilibrium constants for the following reactions have been determined: ①BrCI(g)=1/2Cl2(g+1/2Br2(g)K,o=0.67 ②L2(g+Br2(g)=21IBr(g)K9=0.051 Using the above information,calculate the standard equilibrium constant for the following reaction: 32BrCl(g)+12(g)--2IBr(g)+Cl2(g) Answer:①x2+②=③ 2BrCl (g)+12(g)=2IBr(g)+Cl(g) K月=()29=0.672×0.051=0.023
Example：Equilibrium constants for the following reactions have been determined: The principle of multiple equilibria Answer：①x2 + ② = ③ ②I2 (g)+Br2 (g) 2IBr(g) K 2 = 0.051 Using the above information, calculate the standard equilibrium constant for the following reaction: ①BrCl(g) 1/2Cl2 (g)+1/2Br2 (g) K1 = 0.67 2BrCl (g)+ I2 (g) 2IBr(g)+ Cl2 (g) ③2BrCl (g)+ I2 (g) 2IBr(g)+ Cl2 (g) K3 = · = 0.67 K3 K1 K2 2×0.051=0.023 2 ( )