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## Chemical Equilibrium

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**Equilibrium**• Initially all liquid • Gas only, produced • Balance of gas and liquid • production**Equilibrium**• When compounds react, they eventually form a mixture of products and unreacted reactants, in a dynamic equilibrium. • A dynamic equilibrium consists of a forward reaction, in which substances react to give products, and a reverse reaction, in which products react to give the original reactants.**Chemical Equilibrium**• H2 + I2< -- > 2HI • Initially only H2 and I2 arepresent. • The rxn. proceeds only • As HI concentration increases, some HI is able to decompose back into H2 and I2 • Rxn. proceeds < -- > • At some point • The rate of H2 + I2-- > 2HI equals • The rate of 2HI -- > H2 + I2 • Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal.**Chemical Equilibrium**• For example, the Haber process for producing ammonia from N2 and H2 does not go to completion. • It establishes an equilibrium state where all three species are present. (see Figure 15.3)**Chemical Equilibrium**• Chemical Equilibrium is a fundamentally important concept to master because most chemical reactions fail to go to completion.**Suppose we place 1.000 mol N2 and 3.000 mol H2 in a reaction**vessel at 450 oC and 10.0 atmospheres of pressure. The reaction is A Problem to Consider • Applying Stoichiometry to an Equilibrium Mixture. • What is the composition of the equilibrium mixture if it contains 0.080 mol NH3?**A Problem to Consider**• Using the information given, set up an ICE table. • The equilibrium amount of NH3 was given as 0.080 mol. Therefore, 2x = 0.080 molNH3 (x = 0.040 mol).**A Problem to Consider**• Using the information given, set up the following table. Equilibrium amount of N2 = 1.000 - 0.040 = 0.960 mol N2 Equilibrium amount of H2 = 3.000 - (3 x 0.040) = 2.880 mol H2 Equilibrium amount of NH3 = 2x = 0.080 mol NH3**The Equilibrium Constant**• Every reversible system has its own “position of equilibrium” under any given set of conditions. • The ratio of products produced to unreacted reactants for any given reversible reaction remains constant under constant conditions of pressure and temperature. • The numerical value of this ratio is called the equilibrium constant for the given reaction.**The Equilibrium Constant**H2 + I2< -- > 2HI Rate of forward rxn: Ratef = kf [H2][I2] Rate of reverse rxn: Rater = kr[HI]2 At equilibrium: kf [H2][I2] =kr[HI]2 Therefore: kf= [HI]2 kr[H][I] Kc= [HI]2 [H][I]**For the general equation above, the equilibrium-constant**expression would be: The Equilibrium Constant • The equilibrium-constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation.**The Equilibrium Constant**• The equilibrium-constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation. • The molar concentration of a substance is denoted by writing its formula in square brackets.**The Equilibrium Constant**• The equilibrium constant, Kc, is the value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted. • A large Kc indicates large concentrations of products at equilibrium. • A small Kc indicates large concentrations of unreacted reactants at equilibrium.**Consider the equilibrium established in the Haber process.**The Equilibrium Constant • The law of mass action states that the value of the equilibrium constant expression Kc is constant for a particular reaction at a given temperature, whatever equilibrium concentrations are substituted.**Note that the stoichiometric coefficients in the balanced**equation have become the powers to which the concentrations are raised. The Equilibrium Constant • The equilibrium-constant expression would be**Calculating Equilibrium Concentrations**• Once you have determined the equilibrium constant for a reaction, you can use it to calculate the concentrations of substances in the equilibrium mixture.**For example, consider the following equilibrium.**Calculating Equilibrium Concentrations • Suppose a gaseous mixture contained 0.30 mol CO, 0.10 mol H2, 0.020 mol H2O, and an unknown amount of CH4 per liter. • What is the concentration of CH4 in this mixture? The equilibrium constant Kc equals 3.92.**0.020 mol**0.30 mol 0.10 mol ?? 1.0 L 1.0 L 1.0 L Calculating Equilibrium Concentrations • First, calculate concentrations from moles of substances.**0.30 M**0.10 M ?? 0.020 M • The equilibrium-constant expression is: Calculating Equilibrium Concentrations • First, calculate concentrations from moles of substances.**0.30 M**0.10 M ?? 0.020 M Calculating Equilibrium Concentrations • First, calculate concentrations from moles of substances. • Substituting the known concentrations and the value of Kc gives:**0.30 M**0.10 M ?? 0.020 M Calculating Equilibrium Concentrations • First, calculate concentrations from moles of substances. • You can now solve for [CH4]. • The concentration of CH4 in the mixture is 0.059 mol/L.**Calculating Equilibrium Concentrations**• Suppose we begin a reaction with known amounts of starting materials and want to calculate the quantities at equilibrium.**Calculating Equilibrium Concentrations**• Consider the following equilibrium. • Suppose you start with 1.000 mol each of carbon monoxide and water in a 50.0 L container. Calculate the molarity of each substance in the equilibrium mixture at 1000 oC. • Kc for the reaction is 0.58 at 1000 oC.**1.000 mol**1.000 mol 50.0 L 50.0 L Calculating Equilibrium Concentrations • First, calculate the initial molarities of CO and H2O.**0.0200 M**0.0200 M 0 M 0 M Calculating Equilibrium Concentrations • First, calculate the initial molarities of CO and H2O. • The starting concentrations of the products are 0. • We must now set up a table of concentrations (starting, change, and equilibrium expressions in x).**The equilibrium-constant expression is:**Calculating Equilibrium Concentrations • Let x be the moles per liter of product formed.**Substituting the values for equilibrium concentrations, we**get: Calculating Equilibrium Concentrations • Solving for x.**Calculating Equilibrium Concentrations**• Solving for x. • Or:**Calculating Equilibrium Concentrations**• Solving for x. • Taking the square root of both sides we get:**Calculating Equilibrium Concentrations**• Solving for x. • Rearranging to solve for x gives:**Calculating Equilibrium Concentrations**• Solving for equilibrium concentrations. • If you substitute for x in the last line of the table you obtain the following equilibrium concentrations. 0.0114 M CO 0.0086 M CO2 0.0114 M H2O 0.0086 M H2**Calculating Equilibrium Concentrations**• The preceding example illustrates the three steps in solving for equilibrium concentrations. • Set up a table of concentrations (starting, change, and equilibrium expressions in x). • Substitute the expressions in x for the equilibrium concentrations into the equilibrium-constant equation. • Solve the equilibrium-constant equation for the values of the equilibrium concentrations.**Calculating Equilibrium Concentrations**• In some cases it is necessary to solve a quadratic equation to obtain equilibrium concentrations. • The next example illustrates how to solve such an equation.**Calculating Equilibrium Concentrations**• Consider the following equilibrium. • Suppose 1.00 mol H2 and 2.00 mol I2 are placed in a 1.00-L vessel. How many moles per liter of each substance are in the gaseous mixture when it comes to equilibrium at 458 oC? • Kc at this temperature is 49.7.**The equilibrium-constant expression is:**Calculating Equilibrium Concentrations • The concentrations of substances are as follows.**Substituting our equilibrium concentration expressions**gives: Calculating Equilibrium Concentrations • The concentrations of substances are as follows.**Calculating Equilibrium Concentrations**• Solving for x. • Because the right side of this equation is not a perfect square, you must solve the quadratic equation.**Calculating Equilibrium Concentrations**• Solving for x. • The equation rearranges to give:**Calculating Equilibrium Concentrations**• Solving for x. • The two possible solutions to the quadratic equation are:**Calculating Equilibrium Concentrations**• Solving for x. • However, x = 2.33 gives a negative value to 1.00 - x (the equilibrium concentration of H2), which is not possible.**Calculating Equilibrium Concentrations**• Solving for equilibrium concentrations. • If you substitute 0.93 for x in the last line of the table you obtain the following equilibrium concentrations. 0.07 M H2 1.07 M I2 1.86 M HI**Le Chatelier’s Principle**Obtaining the maximum amount of product from a reaction depends on the proper set of reaction conditions. • Le Chatelier’s Principle states that when a system in a chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the equilibrium will shift in a way that tends to counteract this change. • See LeChatelier’s Principle animation 44**Removing Products or Adding Reactants**Let’s refer an illustration of a U-tube. “reactants” “products” • It’s a simple concept to see that if we were to remove products (analogous to dipping water out of the right side of the tube) the reaction would shift to the right until equilibrium was reestablished. 45**Removing Products or Adding Reactants**Let’s refer back to the illustration of the U-tube in the first section of this chapter. “reactants” “products” • Likewise, if more reactant is added (analogous to pouring more water in the left side of the tube) the reaction would again shift to the right until equilibrium is reestablished. 46**Effects of Pressure Change**A pressure change caused by changing the volume of the reaction vessel can affect the yield of products in a gaseous reaction only if the reaction involves a change in the total moles of gas present (see Figure 15.12). CO + 3H2 CH4+H2O 3 mol 9 mol 3 mol 3 mol 47**Effects of Pressure Change**If the products in a gaseous reaction contain fewer moles of gas than the reactants, it is logical that they would require less space. • So, reducing the volume of the reaction vessel would favor the products. • If the reactants require less volume (that is, fewer moles of gaseous reactant) • decreasing the volume of the reaction vessel would shift the equilibrium to the left (toward reactants). 48**Effects of Pressure Change**Literally “squeezing” the reaction will cause a shift in the equilibrium toward the fewer moles of gas. • It’s a simple step to see that reducing the pressure in the reaction vessel by increasing its volume would have the opposite effect. • In the event that the number of moles of gaseous product equals the number of moles of gaseous reactant, vessel volume will have no effect on the position of the equilibrium. 49**Effect of Temperature Change**Temperature has a significant effect on most reactions (see Figure 15.13). • Reaction rates generally increase with an increase in temperature. Consequently, equilibrium is established sooner. • In addition, the numerical value of the equilibrium constant Kc varies with temperature. 50