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Equilibrium Notes

Equilibrium Notes. Part 3 Based on BLB 10 th ed Ch 15. Practice Problem. A flask is charged with 1.500 atm of N 2 O 4 and 1.000 atm NO 2 at 25 o C. When the following equilibrium is reached, the partial pressure of NO 2 is 0.512 atm. N 2 O 4(g) 2NO 2(g)

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Equilibrium Notes

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  1. Equilibrium Notes Part 3 Based on BLB 10th ed Ch 15

  2. Practice Problem A flask is charged with 1.500 atm of N2O4 and 1.000 atm NO2 at 25oC. When the following equilibrium is reached, the partial pressure of NO2 is 0.512 atm. N2O4(g) 2NO2(g) • What is the equilibrium partial pressure of N2O4 ? • Calculate the value of Kp for the reaction.

  3. N2O4(g) 2NO2(g)

  4. N2O4(g) 2NO2(g)

  5. N2O4(g) 2NO2(g)

  6. N2O4(g) 2NO2(g)

  7. Answers • What is the equilibrium partial pressure of N2O4 ? 1.744 atm • Calculate the value of Kp for the reaction. 0.150

  8. Applications of the Equilibrium Constant K>>1, the reaction proceeds far to the right (product favored) K<<1, the reaction proceeds far to the left (reactant favored)

  9. The equilibrium constantalso allows us to… • Predict the direction in which a reaction mixture will proceed to achieve equilibrium • Calculate the concentrations of reactants and products when equilibrium is reached

  10. Predicting the direction of reaction Say you have 2.00 mol/L of H2, 1.00 mol/L N2 and 2.00 mol/L NH3 for N2 + 3H2 2NH3 Kc = 0.105 at 472oC How will the mixture reach equilibrium? Will N2 and H2 react to form more NH3? Will more NH3 decompose to N2 and H2?

  11. [NH3]2 (2.00)2 = = [N2][H2]3 (1.00)(2.00)3 To answer the questions… • Substitute the starting concentrations into the equilibrium expression and compare to the Kc. 2.00 mol/L H2, 1.00 mol/L N2 & 2.00 mol/L NH3 N2 + 3H2 2NH3Kc = 0.105 at 472oC 0.500

  12. To answer the questions… • This calculated 0.500 is Q, the reaction quotient • It is a number obtained by substituting starting reactant and product concentrations (or partial pressures) into an equilibrium constant expression,

  13. The Reaction Quotient (Q) • To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. • Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

  14. If Q = K, the system is at equilibrium.

  15. If Q > K, there is too much product and the equilibrium shifts to the left.

  16. If Q < K, there is too much reactant, and the equilibrium shifts to the right.

  17. [NH3]2 (2.00)2 = = [N2][H2]3 (1.00)(2.00)3 0.500 = Q To answer the questions… In this case, Q is bigger than Kc, so the reaction must move to the left and NH3 decomposes • Substitute the starting concentrations into the equilibrium expression and compare to the Kc. 2.00 mol/L H2, 1.00 mol/L N2 & 2.00 mol/L NH3 N2 + 3H2 2NH3Kc = 0.105 at 472oC

  18. H2(g) + I2(g) 2HI(g) Let’s try another one… At 448oC the equilibrium constant Kc for the reaction below is 50.5 Predict which direction the reaction will proceed to reach equilibrium at 448oC if we start with 2.0 x 10-2 mol HI, 1.0x10-2 mol H2, and 3.0 x 10-2 mol I2 in a 2.00-L container.

  19. (1.0x10-2)2 [HI]2 = 1.3 = Q = (5.0x10-3)(1.5x10-2) [H2][I2] Get initial concentrations [HI] = 2.0 x 10-2 mol HI / 2.00 L = 1.0x10-2 M [H2] = 1.0x10-2 mol H2 / 2.00 L = 5.0x10-3 M [I2] = 3.0 x 10-2 mol I2 / 2.00 L = 1.5x10-2 M

  20. Practice Exercise At 1000 K the value of Kp for the reaction is 0.338. Calculate the value for Qp , and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures are Answer:Qp = 16; Qp> Kp, and so the reaction will proceed from right to left, forming more SO3.

  21. N2(g) + 3H2(g) 2NH3(g) Calculating Equilibrium Concentrations We can use Kc or Kp to find missing concentrations or partial pressures For the Haber process (above), Kp = 1.45x10-5 at 500oC, the partial pressure of H2 is 0.928 atm and N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture?

  22. N2(g) + 3H2(g) 2NH3(g) [NH3]2 x2 = = [N2][H2]3 [.438][.928]3 atm 0.438 0.928 x Kp = 1.45 x 10-5 x2 = 5.01 x 10-6 x = 2.24 x 10-3 atm = PNH3

  23. Real life… Many times we will know the value of the equilibrium constant and initial amounts of all species. We must then solve for the equilibrium amounts. This treats the change as a variable. Stoichiometry gives us the relative change.

  24. H2(g) + I2(g) 2HI(g) Attempting a problem… A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448oC. Kc = 50.5 What are the equilibrium concentrations of H2, I2 and HI in mol/L? ICE, ICE baby!

  25. Start an ICE table…

  26. Set a variable equal to x

  27. Set a variable equal to x

  28. (2x)2 (1.000 – x)(2.000 – x) [HI]2 = [H2][I2] Simplify the equilibrium expression… 4x2 = 50.5(x2 – 3.000x + 2.000) 46.5x2 – 151.5x + 101.0 = 0 Kc = = 50.5

  29. x = -b  √(b2 – 4ac) 2a x = -(-151.5)  √[(-151.5)2 – 4(46.5)(101.0)] 2(46.5) Apply the quadratic eqn ax2 – bx + c = 0 46.5x2 – 151.5x + 101.0 = 0 x = 2.323 or 0.935

  30. Substitute both x values in x = 2.323 or 0.935 [H2] = 1.000 – x [I2] = 2.000 – x [HI] = 2x x = 2.323 gets a negative number… Impossible! [H2] = 0.065 M [I2] = 1.065 M [HI] = 1.870M

  31. PCl5(g) PCl3(g) + Cl2(g) Kp = 0.497 at 500K PCl5 has an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl5, PCl3 and Cl2 at this temperature? 3 PPCl = 0.97 atmPPCl = PCl = 0.693 atm 2 5

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