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Lec.4

Lec.4. Chemical equilibrium. Recall. Relation between Kc and Kp. Example 1: calculate the number of moles for the following reaction And determine the relation form between Kp&Kc 1) N 2 O 4 2 NO 2 K p = K c (RT) 2-1 = K c (RT)

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Lec.4

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  1. Lec.4 Chemical equilibrium

  2. Recall

  3. Relation between Kc and Kp

  4. Example 1:calculate the number of moles for the following reaction And determine the relation form between Kp&Kc 1) N2 O4 2NO2 Kp =Kc (RT) 2-1 = Kc (RT) 2) N2+ 3H2 2NH3 Kp =Kc (RT) 2- 4 = Kc (RT) – 2 3) H2 + I2 2 HI Kp =Kc (RT) 2- 2 = Kc(RT) 0 Note : Kp =Kc when ∆n = zero i.e., when number of moles of products = number of moles of reactants.

  5. Example 2:

  6. Kp = 4 α2 P = 4*(0.47)2* 456/760 = 0.68 {1- α2} 1-(0.47)2 2) Temp. Is constant & same reaction Kp is the same 0.68 = 4 α2 P = 4 α2 * 0.25 {1-α2} {1-α2} 0.68 = 4 α2 = 2.7 0.25 {1-α2} 2.7-2.7 α2 -4α2 = 0 2.7 = 6.7 α2 α2 =0.4059 α = +0.637 The negative value is ignored. i.e.: % dissociation= 63.7% 3) Kp =Kc(RT) ∆n 0.68 = Kc (0.082*(45+273)) 2-1 Kc = 0.026

  7. Example(3): • 2SO2 + O2 2SO3 at 728°C Kc was found to be 274 ,Calculate Kp. Answer Kp=Kc RT ∆n ∆n =2-3 = -1 Kp= 2.74 (0.082 * (273+728) -1 = 274 0.082 * 1001 Kp= 3.4 • For the reaction:

  8. Le Chatelier's Principle • If a change occurs in one of the factors controlling the equilibrium, the equilibrium becomes displaced in such a direction to minimize the effect of the change imposed. • Three ways can affect the outcome of the equilibrium: • Changing concentrations by adding or removing products or reactants to the reaction vessel. • • Changing partial pressure of gaseous reactants and products. • • Changing the temperature.

  9. 1-Concentration: For Reaction A+ B C +D • Increase in [A] will force the reaction to move to the right • Increase in [B] will force the reaction to move to the right • Increase in [C] will force the reaction to move to the Left • Increase in [D] will force the reaction to move to the Left • 2- Temperature: II-Exothermic(forward reaction):

  10. 3- Pressure:

  11. Application to industrial process: • Synthesis of ammonia: [ Exothermic] N2 (g) + 3H2 (g) 2NH3 + 24000 Cal (4 moles ) (2 moles) • (+)sign because it is an exothermic reaction • Increasing pressure favors forward reaction • The reaction is exothermic so temperature must decrease. (by decreasing the Temp., the velocity of the reaction will decrease as the collision between molecules will decrease and longer time will be needed, so it need a catalyst ) • For gases reactions , the reaction is slow so a catalyst such as [ iron oxide + chromium] is used. • pressure used = 200- 300 atm, optimum Temperature =550 – 600 ºC (not high &not low)

  12. 2)Synthesis of nitric oxide :[Endothermic ] N2(g) + O2(g) 2NO(g) - 43000 Cal (2 moles) ( 2moles) • (-)sign because it is an endothermic reaction •  The pressure has no effect ( moles of product=moles of reactants) • Increasing the temperature favors endothermic reaction. (By increasing the temperature the velocity of the reaction will increase as the collision between molecules will increase and shorter time will be needed ) • We can not use catalyst at high temperature.

  13. 3) Synthesis of sulphur trioxide:[Exothermic ] 2SO2+O2 2SO3+22600 cal (3moles) (2 moles) • (+)sign because it is an exothermic reaction • Increasing pressure favors forward reaction . • increasing the temperature favors back reaction • Excess oxygen as a catalyst from the air favors the forward reaction.

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