Equilibrio. NOTE: [HI] at equilibrium is larger than the [H 2 ] or [I 2 ]; this means that the position of equilibrium favors products. If the reactant.

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Presentation transcript:

Equilibrio

NOTE: [HI] at equilibrium is larger than the [H 2 ] or [I 2 ]; this means that the position of equilibrium favors products. If the reactant concentrations decrease, then the forward reaction rate slows down.

Express the equilibrium constant for the chemical equation. Example 14.1 Expressing Equilibrium Constants for Chemical Equations SOLUTION For Practice 14.1 The equilibrium constant is the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients. Express the equilibrium constant for the combustion of propane in the balanced chemical equation.

Problem: Determine the K c value given the following concentrations at equilibrium for the chemical reaction at room temperature: Given:2 NOCl(g)  2 NO(g) + Cl 2 (g) [NOCl] eq = 1.34 [NO] eq = 0.66 [Cl 2 ] eq = 0.33 Solution:K c = [NO] 2 [Cl 2 ] / [NOCl] 2 K c = {[0.66] 2 [0.33]} / [1.34] 2 K c = (0.144)/(1.80) K c = Calculating K c When Given Equilibrium Concentrations

Example 14.8 Finding Equilibrium Concentrations When You Are Given the Equilibrium Constant and All but One of the Equilibrium Concentrations of the Reactants and Products SOLUTION Consider the reaction. In an equilibrium mixture, the concentration of COF 2 is M and the concentration of CF 4 is M. What is the equilibrium concentration of CO 2 ? SORT You are given the equilibrium constant of a chemical reaction, together with the equilibrium concentrations of the reactant and one product. You are asked to find the equilibrium concentration of the other product. STRATEGIZE Calculate the concentration of the product by using the given quantities and the expression for K c.

Example 14.8 Finding Equilibrium Concentrations When You Are Given the Equilibrium Constant and All but One of the Equilibrium Concentrations of the Reactants and Products Continued For Practice 14.8 Molecular iodine (I 2 ) decomposes at high temperature to form I atoms according to the reaction: In an equilibrium mixture, the concentration of I 2 is 0.10 M. What is the equilibrium concentration of I? SOLUTION CHECK Check your answer by substituting the given values of 3COF2 4 and 3CF4 4 as well as the calculated value for 3CO2 4 back into the equilibrium expression. SOLVE the equilibrium for [CO 2 ] and then substitute in the appropriate values to calculate it. [CO2] was found to be roughly equal to one. [COF 2 ] 2 ≈ 0.06 and [CF 4 ] ≈ Therefore K c is approximately 2, as given in the problem.

Example 14.4 Writing Equilibrium Expressions for Reactions Involving a Solid or a Liquid Write an expression for the equilibrium constant (K c ) for the chemical equation. SOLUTION Since CaCO 3 (s) and CaO(s) are both solids, you omit them from the equilibrium expression. For Practice 14.4 Write an expression for the equilibrium constant (K c ) for the chemical equation.

K Value and Equilibrium K > 1: Product Favored K < 1: Reactant Favored

Q problem: Determining if a reaction is at equilibrium Problem: Is the following reaction at equilibrium, and if not, in what direction must the reaction proceed to obtain equilibrium? Given: 2 H 2 S(g)   2 H 2 (g) + S 2 (g) K p = 2.4 x at 1073 K atm = (H 2 ) atm = (S 2 ) atm = (H 2 S) Strategy:1. Write the equilibrium expression for the reaction. 2. Determine Q value. 3. Compare calculated Q value to reaction’s K c value.

: Is the following reaction at equilibrium, and if not, in what direction must the reaction proceed to obtain equilibrium? Answer : Is the following reaction at equilibrium, and if not, in what direction must the reaction proceed to obtain equilibrium? Given: 2 H 2 S(g)   2 H 2 (g) + S 2 (g) K p = 2.4 x at 1073 K atm = (H 2 ) atm = (S 2 ) atm = (H 2 S) Solution:Q = {(H 2 ) 2 (S 2 )} / (H 2 S) 2 Q = {(0.112) 2 (0.055)} /( 0.455) 2 Q = (6.90 x )/(0.455) 2 Q = 3.33 x Q value is GREATER than K p value (Q > K c ). Reaction value is NOT at equilibrium. Reaction must shift to the reactant side to achieve equilibrium.

Example 14.2 Manipulating the Equilibrium Constant to Reflect Changes in the Chemical Equation SOLUTION You want to manipulate the given reaction and value of K to obtain the desired reaction and value of K. You can see that the given reaction is the reverse of the desired reaction, and its coefficients are twice those of the desired reaction. Begin by reversing the given reaction and taking the inverse of the value of K. Calculate the value of K'. Consider the chemical equation and equilibrium constant for the synthesis of ammonia at 25°C: Calculate the equilibrium constant for the following reaction at 25°C : Next, multiply the reaction bye ½ and raise the equilibrium constant to the ½ power.

SORT You are given K p for the reaction and asked to find K c. STRATEGIZE Use Equation 14.2 to relate K p and K c. SOLVE Solve the equation for K c. Calculate Δn. Substitute the required quantities to calculate K c. The temperature must be in kelvins. The units are dropped when reporting K c as described previously. Example 14.3 Relating K p and K c Nitrogen monoxide, a pollutant in automobile exhaust, is oxidized to nitrogen dioxide in the atmosphere according to the equation: SOLUTION Find K c for this reaction.

Continued Example 14.3 Relating K p and K c For Practice 14.3 CHECK The easiest way to check this answer is to substitute it back into Equation 14.2 and confirm that you get the original value for K p. Consider the following reaction and corresponding value of K c : What is the value of K p at this temperature? SOLUTION

Example 14.5 Finding Equilibrium Constants from Experimental Concentration Measurements 1. Using the balanced equation as a guide, prepare an ICE table showing the known initial concentrations and equilibrium concentrations of the reactants and products. Leave space in the middle of the table for determining the changes in concentration that occur during the reaction. 2. Calculate the change in concentration that must have occurred for the reactant or product whose concentration is known both initially and at equilibrium. A reaction mixture at 780°C initially contains [CO] = 0.500M and [H 2 ] = 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of the equilibrium constant?

Example 14.5 Finding Equilibrium Constants from Experimental Concentration Measurements Continued 3. Use the change calculated in step 2 and the stoichiometric relationships from the balanced chemical equation to determine the changes in concentration of all other reactants and products. Since reactants are consumed during the reaction, the changes in their concentrations are negative. Since products are formed, the changes in their concentrations are positive. 4. Sum each column for each reactant and product to determine the equilibrium concentrations. 5. Use the balanced equation to write an expression for the equilibrium constant and substitute the equilibrium concentrations to calculate K.

Example 14.5 Finding Equilibrium Constants from Experimental Concentration Measurements 1. Using the balanced equation as a guide, prepare an ICE table showing the known initial concentrations and equilibrium concentrations of the reactants and products. Leave space in the middle of the table for determining the changes in concentration that occur during the reaction. 2. Calculate the change in concentration that must have occurred for the reactant or product whose concentration is known both initially and at equilibrium. A reaction mixture at 780°C initially contains [CO] = 0.500M and [H 2 ] = 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of the equilibrium constant?

Example 14.5 Finding Equilibrium Constants from Experimental Concentration Measurements Continued 3. Use the change calculated in step 2 and the stoichiometric relationships from the balanced chemical equation to determine the changes in concentration of all other reactants and products. Since reactants are consumed during the reaction, the changes in their concentrations are negative. Since products are formed, the changes in their concentrations are positive. 4. Sum each column for each reactant and product to determine the equilibrium concentrations. 5. Use the balanced equation to write an expression for the equilibrium constant and substitute the equilibrium concentrations to calculate K.

Example 14.9 Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant 1. Using the balanced equation as a guide, prepare a table showing the known initial concentrations of the reactants and products. Leave room in the table for the changes in concentrations and for the equilibrium concentrations. 2. Use the initial concentrations to calculate the reaction quotient (Q) for the initial concentrations. Compare Q to K and predict the direction in which the reaction will proceed. A reaction mixture (at 2000 °C) initially contains [N 2 ] = M and [O 2 ] = M. Find the equilibrium concentrations of the reactants and products at this temperature.

Continued Example 14.9 Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant 3. Represent the change in the concentration of one of the reactants or products with the variable x. Define the changes in the concentrations of the other reactants or products in terms of x. It is usually most convenient to let x represent the change in concentration of the reactant or product with the smallest stoichiometric coefficient. 4. Sum each column for each reactant and product to determine the equilibrium concentrations in terms of the initial concentrations and the variable x.

Continued Example 14.9 Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant 5. Substitute the expressions for the equilibrium concentrations (from step 4) into the expression for the equilibrium constant. Using the given value of the equilibrium constant, solve the expression for the variable x. In some cases, such as Example 14.9 here, you can take the square root of both sides of the expression to solve for x. In other cases, such as the Example 14.10, you must solve a quadratic equation to find x. Remember the quadratic formula:

Example 14.9 Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant Continued 6. Substitute x into the expressions for the equilibrium concentrations of the reactants and products (from step 4) and calculate the concentrations. In cases where you solved a quadratic and have two values for x, choose the value for x that gives a physically realistic answer. For example, reject the value of x that results in any negative concentrations. 7. Check your answer by substituting the calculated equilibrium values into the equilibrium expression. The calculated value of K should match the given value of K. Note that rounding errors could cause a difference in the least significant digit when comparing values of the equilibrium constant. For Practice 14.9 Since the calculated value of K c matches the given value (to within one digit in the least significant figure), the answer is valid. The reaction in Example 14.9 is carried out at a different temperature at which K c = This time, however, the reaction mixture started with only the product, [NO] = M, and no reactants. Find the equilibrium concentrations of N 2, O 2 and NO at equilibrium.