Chemical Equilibrium Chapter 14 Equilibrium occurs when there

Chemical Equilibrium Chapter 14

� Equilibrium occurs when there is a constant ratio between the concentration of the reactants and the products. Different reactions have different equilibriums. Some may appear to be completely products, however, all reactions have some reactants present. A reaction may look "finished" when equilibrium is reached, but actually the forward and reverse reactions continue to happen at the same rate. A reverse reaction is when the written reaction goes from right to left instead of the forward reaction which proceeds from left to right. This is why equilibrium is also referred to as "steady state. ”

The following factors which are effective in changing the equilibrium: a. temperature b. concentration c. pressure d. catalyst Except for the last one, each of these factors affects the state of equilibrium in a system. Catalysts have no effect on the final equilibrium state since they affect the rates of both the forward and reverse reactions by lowering the activation energy requirement for both processes.

�The Equilibrium Constant, Kc, relates to a chemical reaction at equilibrium. It can be calculated if the equilibrium concentration of each reactant and product in a reaction at equilibrium is known. �Kc=molar concentration of products/molar concentration of reactants with each concentration raised to the power of the corresponding stoichiometric coefficient. The equilibrium constant

�Page 651 � 14. 7 and 14. 8 all Home work

�The equilibrium constant for a chemical reaction that has the form 2 A C+D is Kc=[C][D]/[A]² �Suppose that experiments show that both the forward reaction and the reverse reaction are elementary second order reactions, with the following rate laws: A+A C+D Rate =k[A]² �C+D A+A Rate =k’[C][D] Rates and Equilibria

�At equilibrium these two rates are equal, k[A]²=k’[C][D] �[C][D]/[A]²=k/k’ �Kc=k/k’ �The equilibrium constant is equal to the ratio of the rate constants for the forward and reverse reaction.

�Equilibria in systems having more than one phase are called heterogenous equilibria. �Pure liquids and solids are ignored when writing expressions for equilibrium constants. �Ni(s) +4 CO(g)↔Ni(CO)4(g) �Kc=[Ni(CO)4]/[CO]4 Heterogenous Equilibria

�An example of an equilibrium that involves gases is thermal decomposition of calcium carbonate: �Ca. CO 3(s)↔Ca. O(s)+CO 2(g) Kc=[CO 2] �According to the gas law equation PV=n. RT; for gas x Px. V=nx. RT concentration nx/V=Px/RT �If we express the equilibrium constant in terms of partial pressures instead of molar concentrations, the result is still a constant. The equilibrium constant in terms of partial pressures is denoted as Kp. Gaseous Equilibria

�Write the expression for Kp for the reaction N 2(g)+3 H 2(g)↔ 2 NH 3(g) �If K is large then products are favored at equilibrium; if K is small, then reactants are favored. Class Practice

�The equilibrium constant for the gas phase reaction PCl₅(g)↔PCl 3(g)+Cl 2(g) is Kp=25 at 298 K. The partial pressures of PCl₅ and Cl 2 at equilibrium are 0. 0021 and 0. 48 atm, respectively. What is the equilibrium partial pressure of PCl 3? Using K to determine a partial pressure

�Le Chatelier’s principle gives a qualitative idea of an equilibrium system's response to changes in reaction conditions. If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. Response of equilibria to changes in the conditions

�To illustrate what happens when we change the concentration of one of the reactants or products of a reaction at equilibrium, let's consider the following system at 500 o. C. � N 2(g)+3 H 2(g)- 2 NH 3(g) Kc = 0. 040 Initial: 0. 100 M 0 M �Equilibrium: 0. 100 -∆C 0. 100 -3∆C 2∆C �We obtain the following results when we solve this problem. �[NH 3] = 2∆C 0. 0020 M �[N 2] = 0. 100 -∆C 0. 099 M �[H 2] = 0. 100 -3∆C 0. 097 M Adding and removing reagents

�What would happen if we add enough N 2 to increase the initial concentration by a factor of 10? �The reaction can't be at equilibrium any more because there is far too much N 2 in the system. Adding an excess of one of the reactants therefore places a stress on the system. The system responds by minimizing the effect of this stress by shifting the equilibrium towards the products. The reaction comes back to equilibrium when the concentrations of the three components reach the following values.

� NH 3] = 2∆C 0. 0055 M � [N 2] = 1. 00 ∆C 1. 00 M � [H 2] = 0. 10 3∆C 0. 092 M � By comparing the new equilibrium concentrations with those obtained before excess N 2 was added to the system, we can see the magnitude of the effect of adding the excess N 2. � Before After � [NH 3] 0. 0020 M [NH 3] 0. 0055 M � [N 2] 0. 099 M [N 2] 0. 99 M � [H 2] 0. 097 M [H 2] 0. 092 M

�Increasing the amount of N 2 in the system by a factor of 10 leads to an increase in the amount of NH 3 at equilibrium by a factor of about 3. Adding an excess of one of the products would have the opposite effect; it would shift the equilibrium toward the reactants.

�The effect of changing the pressure on a gas-phase reaction depends on the stoichiometry of the reaction. We can demonstrate this by looking at the result of compressing the following reaction at equilibrium. �N 2(g) + 3 H 2(g)↔ 2 NH 3(g) �Let's start with a system that initially contains 2. 5 atm of N 2 and 7. 5 atm of H 2 at 500 o. C, where Kp is 1. 4 x 10 -5, allow the reaction to come to equilibrium, and then compress the system by a factor of 10. When this is done, we get the following results. Compressing a reaction mixture

� Before Compression After Compression � PNH 3 = 0. 12 atm PNH 3 = 8. 4 atm � PN 2 = 2. 4 atm PN 2 = 21 atm � PH 2 = 7. 3 atm PH 2 = 62 atm � Before the system was compressed, the partial pressure of NH 3 was only about 1% of the total pressure. After the system is compressed, the partial pressure of NH 3 is almost 10% of the total. Reaction at equilibrium was subjected to a stress an increase in the total pressure on the system. The reaction then shifted in the direction that minimized the effect of this stress. The reaction shifted toward the products because this reduces the number of particles in the gas, thereby decreasing the total pressure on the system.

�Changes in the concentrations of the reactants or products of a reaction shift the position of the equilibrium, but do not change the equilibrium constant for the reaction. �Similarly, a change in the pressure on a gasphase reaction shifts the position of the equilibrium without changing the magnitude of the equilibrium constant. �Changes in the temperature of the system, however, affect the position of the equilibrium by changing the magnitude of the equilibrium constant for the reaction. Temperature and equilibrium

� Chemical reactions either give off heat to their surroundings or absorb heat from their surroundings. � If we consider heat to be one of the reactants or products of a reaction, we can understand the effect of changes in temperature on the equilibrium. � Increasing the temperature of a reaction that gives off heat is the same as adding more of one of the products of the reaction. It places a stress on the reaction, which must be alleviated by converting some of the products back to reactants

�The reaction in which NO 2 dimerizes to form N 2 O 4 provides an example of the effect of changes in temperature on the equilibrium constant for a reaction. This reaction is exothermic. � 2 NO 2(g) N 2 O 4(g) Ho = -57. 20 k. J �Thus, raising the temperature of this system is equivalent to adding excess product to the system. The equilibrium constant therefore decreases with increasing temperature.

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