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1. Chemical Equilibrium Chapter 6 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/

2. Chemical Equilibrium – Ch. 6 1. Consider the following reaction: N2(g) + 3 Cl2(g) 2 NCl3(g) a. Write the equilibrium expression (Kcand Kp) b. What is the value of Kp if the equilibrium pressures for N2, Cl2 and NCl3 were 1.8 atm, 0.5 atm and 0.017 atm respectively. c. Does the reaction favor reactants or products?

3. Chemical Equilibrium – Ch. 6 2. Consider the following equilibrium: H2 + F2 2 HF Kp = 62 a. What is the Kp for ⇒ 2 HF H2 + F2 b. What is the Kp for ⇒ 3 H2 + 3 F2 6 HF

4. Chemical Equilibrium – Ch. 6 Since reversible reactions can be written in either direction or balanced with any multiple of coefficients you must consider how this affects the constant – K 1. If you flip a reaction take the reciprocal of the constant 2. If you multiply the reaction coefficients raise the constant to the same multiple 3. If you add two or more reactions take the product of the constants

5. Chemical Equilibrium – Ch. 6 3. Consider the following equilibria: H2 + ½ O2 H2O K1 = 16 C + O2 CO2 K2 = 54 C2H2 +5/2 O2 2 CO2 + H2O K3 = 83 What is the equilibrium constant for: 2 C + H2 C2H2

6. Chemical Equilibrium – Ch. 6 4. Which reaction does the Kp= Kc? ( Kp = K(RT)Δn ) a. P4(s) + 6 Cl2 (g) 4 PCl3 (g) b. H2O (l) H2O (g) c. H2 (g) + Cl2 (g) 2 HCl(g) d. C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)

7. Chemical Equilibrium – Ch. 6 5. Consider the following reaction: UO2(s) + 4HF (g) UF4(g) + 2H2O (g)Kp = 11 If you fill a 5 L container with 15 g of UO2, 1.2 atm of HF, 0.75 atm of UF4 and 5.9 atm of H2O, which way will the reaction shift in order to get to equilibrium?

8. Chemical Equilibrium – Ch. 6 Reactions spontaneously go to equilibrium Determining which direction the reaction must go can be achieved by calculating the reaction quotient (Q) and comparing it to the equilibrium constant (K) if K > Q the reaction will go forward to achieve equilibrium if K < Q the reaction will go backward to achieve equilibrium if K = Q the reaction is at equilibrium

9. Chemical Equilibrium – Ch. 6 6. Pure PCl5 is introduced into an evacuated flask at 27 °C with a pressure of 0.45 atm.  The PCl5 is sparked upon which it decomposes into solid P and gaseous Cl2.  2 PCl5 (g) 2 P (s) + 5 Cl2 (g) At equilibrium the total pressure is measured to be 0.93 atm.  What is the equilibrium constant (Kp) at 27 °C?

10. Chemical Equilibrium – Ch. 6 7. Consider the following reaction: N2(g)  +  O2(g)  2  NO (g)Kp = ? Initially the partial pressures of N2 and O2 are 1 atm and 3 atm respectively. At equilibrium the pressure of NO is 1.5 atm. What is the Kp for this reaction?

11. Chemical Equilibrium – Ch. 6 8. At a certain temperature the partial pressures of an equilibrium mixture of N2O4(g) and NO2(g) are 0.34 atm and 1.2 atm respectively. N2O4(g) NO2(g) The volume of the container is doubled. Find the new partial pressures when the equilibrium is re-established.

12. Chemical Equilibrium – Ch. 6 9. Consider the following reaction:  2 NOBr(g)2 NO (g) + Br2(g) Kp = 34 An unknown pressure of NOBr is put into a rigid container at 40 °C, equilibrium is reached when 86 % of the original partial pressure of NOBr has decomposed.  What was the initial pressure of the NOBr?  What is the total pressure in the flask at equilibrium?

13. Chemical Equilibrium – Ch. 6 10. At a particular temperature Kp = 5.7x10-8 for the following reaction:  2 C (s) + 3 H2(g) C2H6(g) What is the pressure of C2H6 at equilibrium if initially there's 5 g of C and 3 atm of H2?

14. Chemical Equilibrium – Ch. 6 11. Consider the following endothermic reaction at equilibrium:  CCl4(g) C (s) + 2 Cl2(g) a. Is heat absorbed or released as the reaction goes forward? b. Which way will the reaction shift if the temperature is decreased? c. What happens to K as the temperature is increased? d. Which way will the reaction shift if the partial pressure of CCl4is increased? e. Which way will the reaction shift if you add more C? f. Which way will the reaction shift if you remove Cl2? g. Which way will the reaction shift if you compress the system? h. Which way will the reaction shift if you add a catalyst?

15. Answer Key – Ch. 6 1. Consider the following reaction: N2(g) + 3 Cl2(g) 2 NCl3(g) a. Write the equilibrium expression (K andKp) b. What is the value of Kp if the equilibrium pressures for N2, Cl2 and NCl3 were 1.8 atm, 0.5 atm and 0.017 atm respectively.

16. Answer Key – Ch. 6 1. …continued c. Does the reaction favor reactants or products? Since Kp < 1 the reaction favors the reactants

17. Answer Key – Ch. 6 2. Consider the following equilibrium: H2 + F22HF Kp = 62 a. What is the Kp for 2HF H2 + F2 Reaction was flipped ⇒ take the reciprocal of the constant Kp = (62)-1 = 0.016 b. What is the Kp for 3H2 + 3F2 6HF Reaction coefficients was multiplied by 3⇒ raise the constant by 3 Kp = (62)3 = 2.4 x 105

18. Answer Key – Ch. 6 3. Consider the following equilibria: N2 + O2 2NO K = 16 2 NOCl 2 NO + Cl2 K = 54 What is the equilibrium constant for 2 N2 + 2 O2 + 2 Cl2 4 NOCl Multiply 1st reaction by 2⇒ 2 N2 + 2 O2 4 NO ⇒K = (16)2 = 256 Flip and multiply 2nd reaction by 2⇒ 4 NO + 2 Cl2 4 NOCl⇒K = (54)-2 = 3.43 x 10–4 Take the sum of the new reactions canceling out the NO⇒ 2N2 + 2O2 + 2Cl2  4NOCl ⇒K = (256)(3.43 x 10–4) = 0.088

19. Answer Key – Ch. 6 4. Which reaction does the Kp=Kc? ( Kp = K(RT)Δn ) a. P4(s) + 6Cl2 (g) 4PCl3 (g) b. H2O (l) H2O (g) c. H2 (g) + Cl2 (g) 2HCl (g) d. C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g)

20. Answer Key – Ch. 6 5. Consider the following reaction: UO2(s) + 4HF (g) UF4(g) + 2H2O (g)Kp = 11 If you fill a 5 L container with 15 g of UO2, 1.2 atm of HF, 0.75 atm of UF4 and 5.9 atm of H2O, which way will the reaction shift in order to get to equilibrium? Since K < Q the reaction will go backward

21. Answer Key – Ch. 6 6. Pure PCl5 is introduced into an evacuated flask at 27 °C with a pressure of 0.45 atm.  The PCl5 is sparked upon which it decomposes into solid P and gaseous Cl2.  At equilibrium the total pressure is measured to be 0.93 atm.  What is the equilibrium constant (Kp) at 27 °C? Ptotal = PPCl5 + PCl2 0.93 = (0.45 – 2x) + (5x) x = 0.16 PPCl5 = 0.45 – 2(0.16) = 0.13 PCl2 = 5(0.16) = 0.8 Kp = (0.8)5/(0.13)2 Kp = 19

22. Answer Key – Ch. 6 7. Consider the following reaction: N2(g)  +  O2(g)  2  NO (g)Kp = ? Initially the partial pressures of N2 and O2 are 1 atm and 3 atm respectively. At equilibrium the pressure of NO is 1.5 atm. What is the Kp for this reaction? Since the P of NO at equilibrium is 1.5 2x = 1.5 => x = 0.75 So the P of N2 and O2 are 0.25 atm and 2.25 atm respectively Kp=(1.5)2/(0.25)(2.25)

23. Answer Key – Ch. 6 8. At a certain temperature the partial pressures of an equilibrium mixture of N2O4(g) and NO2(g) are 0.34 atm and 1.2 atm respectively. The volume of the container is doubled. Find the new partial pressures when the equilibrium is reestablished. N2O4(g)  2NO2(g) Since we’re given the equilibrium pressures we can find Kp⇒ Kp = (1.2)2/(0.34) = 4.24 According to P1V1 = P2V2⇒ if the volume is doubled the pressures will be halved ⇒ PN2O4 = 0.17 atm and PNO2 = 0.6 Calculate Q to know the direction the reaction proceeds Q = (0.6)2/(0.17) = 2.1 since K > Q the reaction goes forward in order to reestablish equilibrium …continue to next slide

24. Answer Key – Ch. 6 8. …continued Use K to solve for x 4.24 = (0.6 + 2x)2/(0.17 - x) x = 0.053 So the new equilibrium pressures ⇒ PN2O4 = 0.17 – 0.053 PNO2 = 0.6 + 2(0.053) PN2O4 = 0.12 atm PNO2 = 0.71 atm

25. Answer Key – Ch. 6 9. Consider the following reaction:  2 NOBr(g) 2 NO (g) + Br2(g)Kp = 34 An unknown pressure of NOBr is put into a rigid container at 40 °C, equilibrium is reached when 86 % of the original partial pressure of NOBr has decomposed.  What was the initial pressure of the NOBr?  What is the total pressure in the flask at equilibrium? …continue to next slide

26. Answer Key – Ch. 6 9.…continued Use K to solve for x 34 = (0.86x)2(0.43x)/(0.14x)2 x = 2.2 the initial pressure of NOBr is The total pressure in the flask at equilibrium Ptotal = 0.14x + 0.86x + 0.43x Ptotal = 3.1 atm 2.2 atm

27. Answer Key – Ch. 6 10. At a particular temperature Kp = 5.7x10-8 for the following reaction:  2 C (s) + 3 H2(g) C2H6(g) What is the pressure of C2H6 at equilibrium if initially there's 5 g of C and 3 atm of H2? Since K is so small we can make the assumption that x is negligible with respect to a non-zero value 5.7x10-8= x/33 x = 1.5x10-6atm negligible

28. Answer Key – Ch. 6 11. Consider the following endothermic reaction at equilibrium:  CCl4(g)  C (s) + 2 Cl2(g) a. Is heat absorbed or released as the reaction goes forward? Since the reaction is endothermic heat is absorbed if the reaction goes forward and vice versa b. Which way will the reaction shift if the temperature is decreased? As the temperature is decreased the system is losing heat causing the reaction to shift left c. What happens to K as the temperature is increased? As the temperature is increased heat is being added causing the reaction to shift right producing more products thereby causing an increase in K d. Which way will the reaction shift if the partial pressure of CCl4 is increased? More reactant pushes the reaction to the right

29. Answer Key – Ch. 6 e. Which way will the reaction shift if you add more C? Adding more solid will have no effect on the equilibrium f. Which way will the reaction shift if you remove Cl2? Removing a product causes the reaction to shift to the right g. Which way will the reaction shift if you compress the system? Decreasing the volume of the system causes the reaction to shift to the side with fewer moles of gas – in this case the left h. Which way will the reaction shift if you add a catalyst? Catalysts will speed up the forward and reverse direction therefore having no effect on the equilibrium

30. Answer Key – Ch. 6

31. Answer Key – Ch. 6