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Understanding Gaseous Chemical Equilibrium Dynamics

Explore how to interpret equilibrium constants, reaction quotients, and predict reaction directions using Masterton and Hurley's Chapter 12 concepts. Learn to calculate concentrations and partial pressures in equilibrium systems.

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Understanding Gaseous Chemical Equilibrium Dynamics

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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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