Chapter+12,+Section+2


 * **Le Chatelier's Principle: What Happens to an Equilibrium When Conditions Change** ||
 * **Le Chatelier's Principle:** ||


 * || **When a stress is applied to a system at equilibrium the system will adjust to relieve the stress.** ||  ||


 * **Stresses applied to chemical equilibrium systems:** These changes are initial ones made to a system already at equilibrium. The amounts of reactant and product must follow the equilibrium constant for the reaction. ||


 * * [|concentration changes-amounts of reactants or products are changed]
 * [|temperature changes-reaction mixture is cooled or heated]
 * [|pressure changes-size of container is altered to change pressure]
 * [|addition of a catalyst-reduces activation energy BUT doesn't change relative amounts of reactant and product at equilibrium.] ||


 * ** Summary: ** ||


 * **1.** **An increase in concentration on one side of an equation favors or drives the reaction to the opposite side. adding reactants favors products** **adding products favors reactants** ||


 * **2.** **An increase in temperature favors or drives an endothermic reaction forward to products**. ||


 * **3. An increase in temperature drives an exothermic reaction backwards to reactants.** ||


 * **4. An increase in pressure drives a reaction toward the side with fewer molecules (moles) of gas. Increased pressure "forces" the reaction into a smaller volume. The gas volume is smaller with fewer gas molecules.** ||


 * **5.** **Adding a catalyst does not alter the relative amounts of reactant and product. Forward reaction happens more easily and so does the reverse reaction. Equilibrium is only reached faster.** ||


 * **Le Chatelier's Principle:** ||

|| 
 * The equilibrium system is balanced in terms of the forward and reverse reaction rates. Any "stress" that alters one of these rates makes the system "shift" . The "shift" occurs so that the two rates eventually equalize. The reaction rate for the forward reaction is typically high because the reactants are present in the mixture at the start. There normally are no products around at the start of a reaction. The reverse reaction will become more important as the amanita of product increases.

||  ||


 * Increasing the concentration of a substance in the equilibrium mixture creates a temporary increase the reaction rate on that side of the equilibrium. The illustration shows the effect of additional reactant on the forward rate. The forward reaction rate "jumps". The reverse reaction rate gradually climbs up to reestablish balance. The new equilibrium has an increased amount of product. The system shifted to relieve the stress. Addition of a reactant has resulted in an increase in the amount of product. [[image:http://www.800mainstreet.com/7/0007-007-jump.gif width="327" height="188" align="center"]]

The equilibrium system is balanced in terms of the forward and reverse reaction rates. Any "stress" that alters one of these rates makes the system "shift". The "shift" occurs so that the two rates gradually come back to be equality. ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">**Balance beam model** ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">An equilibrium can be viewed like a balance beam. The reactants are on the right hand side and the products are on the left hand side. When the system is balanced the forward and reverse rates are equal. The picture shows an equilibrium where K = 1. The balance arms are equal length. ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">[[image:http://www.800mainstreet.com/7/0007-007-bal-be.gif width="376" height="79" align="center"]]

||


 * <span style="font-family: Arial,Helvetica,sans-serif;">If reactants are added to the system the "balance" will be lost temporarily. The balance can be recovered when the system forms more product to bring back an equilibrium condition. ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">[[image:http://www.800mainstreet.com/7/0007-007-tem-imb.gif width="390" height="98" align="center"]]

||


 * <span style="font-family: Arial,Helvetica,sans-serif;">The added reactants cause an increase in the amount of products. The reactants will be converted to form the additional product. ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">**Example:** ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">** The added reactants cause an increase in the amount of products. ** ||

There is a ** MISLEADING ** statement in the text. All of the excess CO is not used up. The definition of excess is at issue. Some of the added CO will be converted to products, but not all. || <span style="font-family: Arial,Helvetica,sans-serif;">
 * <span style="font-family: Arial,Helvetica,sans-serif;">Let's look at the reaction between hydrogen and carbon monoxide. If the reaction is at equilibrium and we add more CO(g) to the mixture, then the forward reaction will occur faster than the reverse. More product will form and not be depleted because the reverse rate is the "slow" one from the earlier equilibrium. Gradually the reverse rate will increase because of the new products. The forward rate will gradually decrease as excess reactant is consumed. The concentration of products will continue to rise until the reverse rate and forward rates equalize. The amount of CH3OH, methanol, will rise.
 * 1 CO(g) + 2 H2 (g) <---> CH3OH (g)**

||  ||

||
 * <span style="font-family: Arial,Helvetica,sans-serif;">**The equilibrium can be recovered by the formation of new methanol, CH****3****OH(g)[[image:http://www.800mainstreet.com/7/0007-007-ad-cob.gif width="344" height="68" align="bottom"]]**
 * The equilibrium shifts to relieve the stress of added reactant CO.**


 * <span style="font-family: Arial,Helvetica,sans-serif;">** What would happen if product methanol was added to the equilibrium? ** ||

||
 * <span style="font-family: Arial,Helvetica,sans-serif;">** The equilibrium can be recovered by the formation of CO and H ** <span style="font-family: Arial,Helvetica,sans-serif;">**2. [[image:http://www.800mainstreet.com/7/0007-0007-met-ad.gif width="346" height="96" align="bottom"]]**
 * The addition of a substance to one side of an equilibrium results in an increase in materials on the opposite side.**

The decrease in methanol would cause CO and H2 to form methanol to replace the amount removed. The forward rate would temporarily be greater than the reverse rate. The formation of methanol would occur faster than its decomposition. The amount of product would go up to replace what was removed. ||
 * <span style="font-family: Arial,Helvetica,sans-serif;">** What would happen if product methanol was removed from the equilibrium? **


 * <span style="font-family: Arial,Helvetica,sans-serif;">**Effect of pressure changes:** ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">Usually only reactions involving gases are altered by pressure changes. Gas pressure results from gas molecules banging against the walls of the container. If you handle gases you know that gas molecules "seek" low pressure. Compressed gas "escapes" high pressure in a balloon or out of a tire valve stem. Low pressure also goes with fewer molecules of gas. If the pressure is on an equilibrium is increased the system will shift to lower the pressure. The lower number of gas molecules uses less space or volume. This favors the side of the reaction with fewer gas molecules. ||

<span style="font-family: Arial,Helvetica,sans-serif;">** 1 CO(g) + 2 H2 (g) <---> CH3OH (g) ** <span style="font-family: Arial,Helvetica,sans-serif;">The reactants side has three molecules of gas while the products side has only one molecule of gas. Increasing the pressure favors the side with lass volume. The products are favored because the fewer gas molecules occupy less volume. The forward rate would be greater than the reverse rate. The formation of methanol would occur faster than its decomposition. <span style="font-family: Arial,Helvetica,sans-serif;">**1 CO(g) + 2 H2 (g) <---> CH3OH (g)** ||
 * <span style="font-family: Arial,Helvetica,sans-serif;">** What would happen to the ** equilibrium if the pressure was raised?


 * <span style="font-family: Arial,Helvetica,sans-serif;">** Example: ** ||

<span style="font-family: Arial,Helvetica,sans-serif;">The reactants side has two molecules of gas while the products side has only one molecule of gas. Increasing the pressure favors the products side with less volume. The forward rate would be greater than the reverse rate. The formation of methane would occur faster than its decomposition. ||
 * <span style="font-family: Arial,Helvetica,sans-serif;">**What would happen to the 1 C(s) + 2 H2 (g) <---> CH4(g) equilibrium if the pressure was raised?**


 * <span style="font-family: Arial,Helvetica,sans-serif;">** Example: ** ||

The reactants side has no molecules of gas while the products side has only one molecule of gas. Increasing the pressure favors the reactants side with less volume. The reverse rate would be greater than the forward rate. The formation of CaCO3(s) would occur faster than its decomposition. ||
 * <span style="font-family: Arial,Helvetica,sans-serif;">** What would happen to the 1 CaCO3(s) <---> CaO(s) + CO2(g) equilibrium if the pressure was raised? **


 * <span style="font-family: Arial,Helvetica,sans-serif;">** Example: ** ||

<span style="font-family: Arial,Helvetica,sans-serif;">Both sides of the equation have two molecules of gas. The pressure increase would cause no change. ||
 * <span style="font-family: Arial,Helvetica,sans-serif;">**What would happen to the N2** ** <span style="font-family: Arial,Helvetica,sans-serif;">(g) ) + O 2(g) <---> 2 NO(g) equilibrium if the pressure was raised?  **


 * <span style="font-family: Arial,Helvetica,sans-serif;">**Effect of temperature changes:** ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">**Temperature changes can shift an equilibrium balance just as concentration changes. Raising the temperature for an exothermic reaction favors reactants.** ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">**Raising the temperature for an endothermic reaction favors formation of products.** ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">** Example: ** ||

<span style="font-family: Arial,Helvetica,sans-serif;"> ||
 * <span style="font-family: Arial,Helvetica,sans-serif;">** What happens to the ammonia equilibrium if the temperature is raised? **
 * 3 H** **<span style="font-family: Arial,Helvetica,sans-serif;">2 (g) + 1 N 2(g) <---> 2 NH 3 (g) + heat **


 * <span style="font-family: Arial,Helvetica,sans-serif;">[[image:http://www.800mainstreet.com/7/0007-007-hot-amon.gif width="374" height="73" align="center"]]

||


 * <span style="font-family: Arial,Helvetica,sans-serif;">The equilibrium balance can be reestablished if the amount of reactants increase. Raising the temperature for the ammonia reaction favors the reverse reaction and the formation of reactants. ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">** Summary: ** ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">** 1. ** **An increase in concentration on one side of an equation favors or drives the reaction to the opposite side.** ** <span style="font-family: Arial,Helvetica,sans-serif;">adding reactants favors products  **** <span style="font-family: Arial,Helvetica,sans-serif;">adding products favors reactants  ** ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">** 2. ** **An increase in temperature favors or drives an endothermic reaction forward to products**. ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">** 3. An increase in temperature drives an exothermic reaction backwards to reactants.** ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">** 4 . An increase in pressure drives a reaction toward the side with fewer molecules (moles) of gas. Increased pressure "forces" the reaction into a smaller volume. The gas volume is smaller with fewer gas molecules.** ||


 * <span style="font-family: Arial,Helvetica,sans-serif;">** 5. ** **Adding a catalyst does not alter the relative amounts of reactant and product. Forward reaction happens more easily and so does the reverse reaction. Equilibrium is only reached faster.** ||