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Chemistry 1011

Chemistry 1011. TOPIC Gaseous Chemical Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 12. 12.4 Applications of the Equilibrium Constant. YOU ARE EXPECTED TO BE ABLE TO: Determine the potential for a reaction to occur from the magnitude of the equilibrium constant.

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Chemistry 1011

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  1. Chemistry 1011 TOPIC Gaseous Chemical Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 12 Chemistry 1011 Slot 5

  2. 12.4 Applications of the Equilibrium Constant YOU ARE EXPECTED TO BE ABLE TO: • Determine the potential for a reaction to occur from the magnitude of the equilibrium constant. • Distinguish between a reaction quotient and the equilibrium constant. • Predict the direction in which a reaction is proceeding given the value of the reaction quotient. • Calculate the concentration (partial pressure) of a component of an equilibrium system, given the equation, the value of the equilibrium constant, and the partial pressures of other components. Chemistry 1011 Slot 5

  3. Application #1 Interpreting the Magnitude of K • The magnitude of the equilibrium constant can give an indication of whether a reaction is likely to occur • If K is very small, then the concentration of products at equilibrium would be very low Eg N2(g) + O2(g) 2NO(g) Kp = 1.0 x 10-30 at 25oC • Alternatively, if K is >1 then the equilibrium mixture will contain product Eg N2(g) + 3H2(g) 2NH3(g) Kp = (PNH3)2 = 6.0 x 105 at 25oC PN2 x (PH2)3 Chemistry 1011 Slot 5

  4. Application #2 The Reaction Quotient • For aA(g) + bB (g) cC (g) + dD (g) K = (PC)c x (PD)d (PA)a x (PB)b • If the system is NOT at equilibrium, the actual pressure ratio, known as the reaction quotient, can have any value at all Q = (PC)c x (PD)d (PA)a x (PB)b Chemistry 1011 Slot 5

  5. Using the Reaction Quotient • Comparison of a reaction quotient to the equilibrium constant will indicate whether the reaction is at equilibrium • If not, it will indicate which direction the reaction will proceed in Chemistry 1011 Slot 5

  6. Q and the Direction of Reaction • If Q is LESS than K, then the reaction will proceed from left to right. The value of Q will increase until it becomes equal to K • If Q is GREATER than K, then the reaction will proceed from right to left. The value of Q will decrease until it becomes equal to K • If you start with pure reactants, the value of Q will initially be zero • If you start with pure products, the value of Q will initially be infinity Chemistry 1011 Slot 5

  7. The NOBr Equilibrium • Given that Kp for the NOBr equilibrium at 350oC is 2.8 x 10-2, will any net reaction occur if 1.00atm of NOBr, 0.80atm of NO and 0.40atm of Br2 are mixed? • If so, will NO be formed or consumed? 2 NOBr(g) 2NO(g) + Br2(g) Q = (PNO)2 x PBr2= (0.80)2 x (0.40) = 2.6 x 10-1 (PNOBr)2 (1.00)2 • Q > Kp • The reaction will move towards the reactants. • NO will be consumed Chemistry 1011 Slot 5

  8. Application #3 Determining Equilibrium Partial Pressures • Given a balanced equation, initial partial pressures or concentrations and the value for the equilibrium constant, it is possible to determine the equilibrium partial pressures of reactants and products • Sometimes a quadratic equation will have to be solved Chemistry 1011 Slot 5

  9. Determining Equilibrium Partial Pressures • Write a balanced equation for the equilibrium • Write an expression for the equilibrium constant • Distinguish equilibrium from initial partial pressures • Use ‘x’ for the unknown partial pressures. Express the equilibrium partial pressures of all species in terms of ‘x’ • Input equilibrium partial pressures into expression for Kp • Calculate ‘x’ Chemistry 1011 Slot 5

  10. Determining Equilibrium Partial Pressures • Consider the system: PCl5(g) PCl3(g) + Cl2(g) • Initially, a system contains PCl5 only, at a pressure of 3.00atm at 300oC. • The value of the equilibrium constant Kp at this temperature is 11.2 • Find • PPCl5 at equilibrium • PPCl3 at equilibrium • PCl2at equilibrium Chemistry 1011 Slot 5

  11. Determining Equilibrium Partial Pressures PCl5(g) PCl3(g) + Cl2(g Initially 3.00atm 0.00atm 0.00atm At equilibrium ? ? ? Let the equilibrium partial pressure of PCl3 be x atm DP - x atm +x atm +x atm At equilibrium 3.00 - x atm x atm x atm Kp =PPCl3 xPCl2 =xxx=11.2 PPCl5 (3.00 - x) x = 2.46atm • PPCl3 = 2.46atm; PCl2 = 2.46atm; PPCl5 = 0.54atm Chemistry 1011 Slot 5

  12. Determining Equilibrium Partial Pressures • Hydrogen cyanide can be made by the reaction: C2N2(g) + H2(g) 2HCN(g) • At a certain temperature, Kp = 64 • Calculate the partial pressures of all species at equilibrium at this temperature if the initial partial pressures of the reactants are 0.50atm Chemistry 1011 Slot 5

  13. Determining Equilibrium Partial Pressures C2N2(g) + H2(g) 2HCN(g) Initially 0.50atm 0.50atm 0.00atm At equilibrium ? ? ? Let the equilibrium partial pressure of HCN be 2x atm DP - x atm -x atm +2x atm At equilibrium (0.50 – x)atm (0.50 – x)atm 2x atm Kp = (PHCN)2= (2x)2=64 PC2N2x PH2 (0.50 – x)2 x = 0.40atm • PHCN= 0.80atm; PC2N2 = 0.10atm; PH2 = 0.10atm Chemistry 1011 Slot 5

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