ࡱ> CEB'` pbjbj .r d^/hN;t<q.s.s.s.s.s.s.1h.4Rs.s./&&&Nq.&q.&&:+,', P|=9$, U-./0^/, 4&4',',J4q,&s.s.& ^/ 4.5 4.5 4.54Equilibria 4.5a demonstrate an understanding of the term dynamic equilibrium as applied to states of matter, solutions and chemical reactions Chemical reactions which take place in both directions are called reversible reactions. Reactions like these however reach equilibrium. This is when chemical reactions are still occurring but the rates of the forward and reverse reactions are equal and so the concentration of each species remains constant. This type of system is said to be in dynamic equilibrium. Example H2(g) + I2(g) <=>2HI(g) (A dynamic equilibrium). The proportion of products to reactants in the equilibrium mixture is described as the equilibrium composition or as the position of equilibrium: aP + bQ <=> cR + dS If the conversion of P and Q into R and S is small the position of the equilibrium lies to the left. If the equilibrium mixture is largely composed of R and S, the position of the equilibrium lies to the right. The position of equilibrium is dependant on the equilibrium constant Kc and can also be altered by a change in external conditions such as concentration or pressure. The equilibrium constant is only affected by a change in temperature. Task 4.5a With reference to the reaction between hydrogen and iodine, ice and water at 0oC, water and lead iodide, water and sodium carbonate explain: (a) reversible reaction (b) equilibrium reaction (c) dynamic equilibrium (d) position of equilibrium (e) shifting the position of equilibrium to the right or left. 4.5b recall that many important industrial reactions are reversible In the Haber process a moderately high temperature of around 500oC is used to speed the rate at which equilibrium is reached. This temperature is chosen in spite of the fact that a lower temperature gives a higher yield. The manufacture of nitric acid involves 3 steps the first of which is the catalytic oxidation of ammonia to nitrogen monoxide. 4NH3(g) + 5O2(g) <=> 4NO(g) + 6H2O(g) /\ H = -950kJmol-1 This reaction is run over a platinum gauze at 7 atmospheres and 900oC. Low pressure favours the products as 9 molecules of reactant form 10 molecules of product. However a higher pressure also increases the concentration of gas molecules of reactant molecules at the start and helps increase the yield of nitrogen monoxide. Low temperature also favours products but a moderately high temperature is used so that equilibrium is reached quickly. Sulfuric acid is made in the Contact process. This takes sulfur dioxide (made by burning sulfur) and reacts it in an equilibrium process with oxygen. 2SO2(g) + O2(g) = 2SO3(g) DH= -192kJmol-1 The optimum conditions for this process are: a temperature of 430oC a pressure of 2 atmospheres a catalyst of vanadium pentoxide V2O5 Task 4.5b.1 (a) Explain why a higher temperature is not used to speed up this reaction. (b) Explain the choice of pressure given the position of equilibrium is far to the right. (c)Write an equation for making sulfur dioxide Ammonia is changed into nitric acid in the Ostwald process. This is another equilibrium process which starts with the reaction of ammonia and oxygen. 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(g) DH= -900 kJmol-1 The optimum (best) conditions for this process which give the greatest yield are a pressure of 10 atmospheres a temperature of about 900oC a catalyst of platinum and rhodium The nitrogen monoxide is then mixed with oxygen and flows up a column down which water flows. The result is a solution of 65% nitric acid. 4NO(g) + 3O2(g) +2H2O(l) ---> 4HNO3(aq) Task 4.5b.2 Ethanol is made from ethene and water in the gas phase in an equilibrium reaction. At 50 atmospheres the yield of ethanol is 30% at 320oC and 45% at 200oC. At 80 atmospheres the yield of ethanol is 45% at 320oc AND 60% at 200oC. Comment on the optimum conditions for this reaction. 4.5c use practical data to establish the idea that a relationship exists between the equilibrium concentrations of reactants and products which produces the equilibrium constant for a particular reaction, eg data on the hydrogen-iodine equilibrium. For the reaction aP + bQ <=> cR + dS The equilibrium constant Kc = [R]c[S]d/[P]a[Q]b So for the equilibrium H2(g) + I2(g) <=>2HI(g) Kc = [HI(g)]2/[H2(g)][I2(g)]  EMBED Excel.Sheet.8  Task 4.5c Calculate Kc for the remaining reactions 4.5d calculate a value for the equilibrium constant for a reaction based on data from experiment, eg the reaction of ethanol and ethanoic acid (this can be used as an example of the use of ICT to present and analyse data), the equilibrium Fe2+(aq) + Ag+ (aq) ( Fe3+(aq) + Ag(s) or the distribution of ammonia or iodine between two immiscible solvents Hexane and water are immiscible. Iodine dissolves in water if KI is present. If 500cm3 of aqueous iodine of initial concentration 2.54gdm-3 (0.0100M) is mixed with 50cm3 hexane titration with 0.0200M sodium thiosulfate can determine the equilibrium concentration of aqueous iodine. Calculation can reveal the iodine concentration in the hexane layer. Kc can then be calculated using Kc=[iodine in hexane]/[iodine in water] 4.5e construct expressions for Kc and Kp for homogeneous and heterogeneous systems, in terms of equilibrium concentrations or equilibrium partial pressures, perform simple calculations on Kc and Kp and work out the units of the equilibrium constants Kc can be obtained from the equation for an equilibrium reaction. aA(aq) + bB(s) ( cC(aq) + dD(l) Kc = [C]ceqm [D]deqm/[A]aeqm [B]beqm However concentrations of pure liquids and pure solids will also be constants. Kp is applied to equilibrium reactions involving gases. aA(g) + bB(g) ( cC(g) Kp = pCc/pAa pBb Where pC is the partial pressure of C. pC = mole fraction of C * total pressure of gas mixture. Task 4.5e write expressions for Kc and/or Kp for each of the following reactions. Ag+(aq) + Fe2+(aq) ( Fe3+(aq) + Ag(s) CH3COOH(aq) + C2H5OH(aq) ( CH3COOC2H5(aq) + H2O(l) H2(g) +I2(g) ( 2HI(g) N2O4(g) ( 2NO2(g) N2(g) + 3H2(g) ( 2NH3(g) 2SO2(g) + O2(g) ( 2SO3(g) NH4HS(s) ( NH3(g) + H2S(g) CaCO3(s) ( CaO(s) + CO2(g) Calculate Kp for the decomposition of CaCO3 if the pressure at equilibrium in 2atm. Calculate Kp for the oxidation of sulphur dioxide if at equilibrium there is 1 mol of oxygen, and 2 mol of both sulphur dioxide and sulphur trioxide and the overall pressure was 10 atm. 4.5f demonstrate an understanding that when Stotal increases the magnitude of the equilibrium constant increases S = RlnK Where S = entropy change, R = gas constant, K = equilibrium constant. Both S and K tell us the direction and extent of chemical change. A reaction proceeds far to the right if S is positive and K is large. Task 4.5f Put the correct S and K values into the table: K=(, 10-10, 1010, 102 to 10-2) S = (>+200, <-200, -40 to +40). 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Stotal = Ssystem + Ssurroundings Ssystem does not change much with temperature unless one of the substances involved changes state. Ssurroundings = -H/T So for exothermic reactions Surroundings=+ve but the higher the temperature the smaller is the value so the less is the effect. Task 4.5h For both of the following reactions work out if the reaction proceeds at 700K and explain the effect on of Stotal increasing the temperature. Mg(s) +1/2O2(g) ( MgO(s) Ssystem = H = -601.7kJmol-1 ZnCXXXXJXn``` $7$8$H$Ifgdbkd $$IflFT  t06    44 laytvJXKXLXXXZZZ$[R[T\V\]neeeeeeeeee 7$8$H$gdkkd $$IflFT  t06    44 laytv ZZZZ Z&Z(Z*Z8ZZZZ\Z^ZlZ$[&[H[J[[[[[T\V\ʸʸʏʸ~p~p~p_pOhDC5B*CJPJaJph!h]Zh]ZB* CJPJaJphZh]ZB* CJPJaJphZ!hkh]ZB* CJPJaJphZh B* CJPJaJphZh 6B* CJ]aJphZh B* CJaJphZ#hkh 6B* CJ]aJphZ!hkh B* CJPJaJphZ h hk5B* CJaJphZ&h hk56B* CJ]aJphZV\h\B]N]]]]]]]]]]]]^hhh hh&h(h2hBhXh\h^hbhdhhhܼʆsqܩܕܩʆ``s hDChDCB*CJH*aJphU$hDChDCB*CJH*PJaJphhDChDCB*CJaJph' jhDChDCB*CJPJaJph$hDChDCB*CJH*PJaJphhDCB*CJPJaJph#hDChDC6B*CJ]aJph!hDChDCB*CJPJaJph$hDChDC5B*CJPJaJph]]]]0hdhhhh\iippp 7$8$H$gd6J 7$8$H$gdDC 7$8$H$gdk O3(s) ( ZnO(s) + CO2(g) Ssystem = +175kJmol-1K-1 H = +71kJmol-1 Answer Stotal = Ssystem + Ssurroundings = -216.6kJmol-1K-1 - (-601.7kJmol-1)/700K = -215.74 kJmol-1K-1 This reaction is not feasible as Stotal is neghhhhhhhhhhhhh i i$i&i*i.i4iJi°°‡xgxgYM<!hDCh6JB*CJPJaJphh6JB*CJaJphh6JB*CJH*aJph hDCh6JB*CJH*aJphhDCh6JB*CJaJphh6JB* CJPJaJphZh6J6B* CJ]aJphZh6JB* CJaJphZ#hkh6J6B* CJ]aJphZ!hkh6JB* CJPJaJphZh6JB*CJH*aJphhDCB* CJPJaJphZ hDChDCB*CJH*aJphJiNiZixiiiiiiiiiiijp6p8pppppp΄sa_ssaN hDCh6JB*CJH*aJphU#hkho6B* CJ]aJphZ!hkhoB* CJPJaJphZhoB* CJPJaJphZhoB*CJH*aJph hDChoB*CJH*aJphhDChoB*CJaJphhoB*CJPJaJphh6JB*CJPJaJph!h6Jh6JB*CJPJaJph$hDCh6JB*CJH*PJaJphative. 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