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

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

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 2

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 N 2(g) + O 2(g) 2NO (g) Kp = 1.0 x at 25 o C Alternatively, if K is >1 then the equilibrium mixture will contain product Eg N 2(g) + 3H 2(g) 2NH 3(g) K p = (P NH3 ) 2 = 6.0 x 10 5 at 25 o C P N2 x (P H2 ) 3 Chemistry 1011 Slot 5 3

4 Application #2 The Reaction Quotient For aa (g) + bb (g) cc (g) + dd (g) K = (P C )c x (P D ) d (P A ) a x (P B ) b If the system is NOT at equilibrium, the actual pressure ratio, known as the reaction quotient, can have any value at all Q = (P C )c x (P D ) d (P A ) a x (P B ) b Chemistry 1011 Slot 5 4

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 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 6

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

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 8

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 9

10 Determining Equilibrium Partial Pressures Consider the system: PCl 5(g) PCl 3(g) + Cl 2(g) Initially, a system contains PCl 5 only, at a pressure of 3.00atm at 300 o C. The value of the equilibrium constant Kp at this temperature is 11.2 Find P PCl5 at equilibrium P PCl3 at equilibrium P Cl2 at equilibrium Chemistry 1011 Slot 5 10

11 Determining Equilibrium Partial Pressures PCl 5(g) PCl 3(g) + Cl 2(g Initially 3.00atm 0.00atm 0.00atm At equilibrium??? Let the equilibrium partial pressure of PCl 3 be x atm P - x atm +x atm +x atm At equilibrium x atm x atm x atm Kp = P PCl3 x P Cl2 = x x x = 11.2 P PCl5 ( x) x = 2.46atm P PCl3 = 2.46atm; P Cl2 = 2.46atm; P PCl5 = 0.54atm Chemistry 1011 Slot 5 11

12 Determining Equilibrium Partial Pressures Hydrogen cyanide can be made by the reaction: C 2 N 2(g) + H 2(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 12

13 Determining Equilibrium Partial Pressures C 2 N 2(g) + H 2(g) 2HCN (g) Initially 0.50atm 0.50atm 0.00atm At equilibrium??? Let the equilibrium partial pressure of HCN be 2x atm P - x atm -x atm +2x atm At equilibrium (0.50 x)atm (0.50 x)atm 2x atm Kp = (P HCN ) 2 = (2x) 2 = 64 P C2 N x P 2 H (0.50 x)2 2 x = 0.40atm P HCN = 0.80atm; P C 2 N 2 = 0.10atm; P H2 = 0.10atm Chemistry 1011 Slot 5 13