 Equilibrium.  Equilibrium is NOT when all things are equal.  Equilibrium is signaled by no net change in the concentrations of reactants or products.

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Equilibrium

 Equilibrium is NOT when all things are equal.  Equilibrium is signaled by no net change in the concentrations of reactants or products.

 At equilibrium, the concentrations of reactants and products no longer change with time.  For equilibrium to occur, neither reactants nor products can escape from the system.  At equilibrium a particular ratio or concentration terms equals a constant.

 Consider the reaction:  aA + bB  cC + dD  At equilibrium the equilibrium constant (k eq ) is calculated as…

 Write the equilibrium constant expressions for the following equations:  2 O 3 (g)  3 O 2 (g)  2 NO(g) + Cl 2 (g)  2 NOCl(g)  Ag + (aq) + 2 NH 3 (aq)  Ag(NH 3 ) 2 + (aq)

ExperimentInitial [NO] Initial [NO 2 ] Equilibriu m [N 2 O 4 ] Equilibriu m [NO 2 ] K eq 10.00.020.00140.01720.21 20.00.030.00280.02430.21 30.00.040.004520.0310.21 40.020.00.004520.0310.21

 When all of the reactants and products of a chemical reaction are gasses we can write the equilibrium constant expression in terms of each components partial pressure.

 k eq and k p are often times numerically different even for the same reaction.  We can however use one of the two to calculate the other.

 In the synthesis of ammonia from nitrogen and hydrogen, N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  k eq = 9.60 at 300 o C. Calculate k p for this reaction at 300 o C.

 k eq values for different reactions can range from very large numbers to very small numbers.  Consider the reaction:  If k eq >> 1 the equilibrium lies to the right and products predominate.  If k eq << 1 the equilibrium lies to the left and the reactants predominate.

 Equilibrium can be approached from either direction.  Consider:

 We have already seen cases where the overall reaction is actually a series of other reactions.  Consider the following two reactions:

 A heterogeneous equilibrium is one where not all of the chemical species are in the same physical state.  Example:  Whenever a pure liquid or a pure solid is involved in a heterogeneous equilibrium, its concentration is not included in the k eq expression.

 Write the k eq expression for the following reactions:  CO 2 (g) + H 2 (g)  CO(g) + H 2 O(l)  SnO 2 (s) + 2 CO(g)  Sn(s) + 2 CO 2 (g)

 The Haber process is a reaction that was first put to use by Fritz Haber as a way to produce ammonia.  One problem he had was that this reaction was an equilibrium reaction.

 A chemists performs the Haber reaction at 472 o C. The equilibrium mixture of the gases is found to contain 7.38 atm H 2, 2.46 atm N 2, and 0.166 atm of NH 3. Calculate k p for the Haber process at this temperature.

 Often times we do not know the equilibrium concentrations of all of the chemical species in a reaction.  But if we know the equilibrium concentration of at least one species we can use stoichiometry to find the others (and eventually k eq or k p )  Follow these steps: 1. Tabulate all the known initial and equilibrium concentrations. 2. Use these to calculate the change in concentration of that chemical species. 3. Use stoichiometry to calculate the change in the other chemical species. 4. Use the initial concentrations with the change in concentration to calculate the equilibrium concentration.

 A closed system initially containing 1.0 x 10 -3 M H 2 and 2.0 x 10 -3 M I 2 at 448 o C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that [HI] = 1.87 x 10 -3 M. Calculate k eq at 448 o C.

 We have seen that the magnitude of k eq tells us to which side the equilibrium lies.  We can also use k eq to tell us which way a reaction will proceed if it isn’t already at equilibrium.  If Q = k eq the reaction is at equilibrium.  If Q > k eq the concentration of products is too large and the reaction will proceed to the left.  If Q < k eq the concentration of reactants is too large and the reaction will proceed to the right.

 At 448 o C the equilibrium for the reaction:  H 2 (g) + I 2 (g)  2 HI(g) is 50.5. Predict the direction the reaction will proceed if we start with 0.02 mol of HI, 0.01 mol of H 2 and 0.03 mol of I 2 in a 2.00-L container.

 For the Haber process at 500 o C the equilibrium constant is 1.45 x 10 -5. An analysis of an equilibrium mixture of the reaction shows that the partial pressures of the reactants are H 2 = 0.928 atm, N 2 = 0.432 atm. Calculate the partial pressure of NH 3.

 Many times we will not know the equilibrium concentrations of at least one species.  Example:  A 1.0 L flask is filled with 1.0 mol H 2 and 2.0 mol I 2. The k eq for this reaction at 448 o C is 50.5. What are the equilibrium concentrations of H 2, I 2, and HI?

 The Chatelier’s Principle states that:  If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position to counteract the change.  In other words, k eq, must remain constant.

 N 2 (g) + H 2 (g)  2 NH 3 (g)

 If the volume of a gaseous system is decreased what would happen to the pressure of the system?  Reducing the volume of a gaseous equilibrium will cause the system to shift in the direction that decreases the total moles of gas.  Consider:  N 2 O 4 (g)  2 NO 2 (g)

 Changes in concentrations, volume, or pressure cause equilibrium shifts without changing the value of k eq.  However changing the temperature of an equilibrium system will change the numeric value of k eq.  When the temperature of an equilibrium system is increase, the system reacts as if we added a reactant to the endothermic reaction. The equilibrium will shift a such to consume that reactant.

 If the forward reaction of an equilibrium is endothermic increasing the temperature increases the value of k.  If the forward reaction is exothermic increasing the temperature decreases the value of k.

 Consider the equilibrium: N 2 O 4 (g)  2 NO 2 (g) In which direction will the equilibrium shift when (a) N 2 O 4 is added. (b) NO 2 is removed. (c) The total pressure of the system is increased. (d) The volume is increased. (e) The temperature is decreased.

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