UNIT 7: CHEMICAL KINETICS AND EQUILIBRIUM Chapter 19: Reaction Rates and Equilibrium 19.1: Rates of Reaction Reaction Rates: - the speed of which the concentration of a reactant or product changes over time. Collision Model: - a model that state for a reaction to occur, molecules must collide with each other. Factors Affecting the Collision Model: 1. Activation Energy (E a ): - the threshold energy molecules needed to overcome to cause a chemical reaction that was first proposed by Svante Arrhenius. - E a is the highest energy (top of the hill - E max ) minus the sum of energy of the reactants (ΣH reactants ) on the potential energy diagram. - in general, the Lower the Activation Energy, the Faster is the Rate of Reaction. E a Ea Activation Energy of an Endothermic Reaction Activation Energy of an Exothermic Reaction. Temperature (T): - the effective number of collisions increases exponentially with temperature. Because T > T 1, there are more molecules colliding with enough Kinetic Energy causing a reaction to occur. In general, the Higher the Temperature, the Faster the Rate of Reaction. age 150.
Unit 7: Chemical Kinetics and Equilibrium. article Size: - the Smaller the article Size (the Larger the Surface Area exposed), the Faster the Reaction Rate. Example 1: Grain sugar dissolves faster than ual mass of sugar cubes because of smaller particle size and therefore increased surface area of the grain sugar. 4. Concentration: - the higher the Concentration (the more molecules in an available space, the higher the chance of collision), the Faster the Reaction Rate. 5. Catalyst: - a substance that speeds up the reaction without being consumed in the reaction. - unlike intermediates, catalyst is used and recycled in the reaction. - lowers activation energy by providing an alternate reaction pathway. ( E a is lowered but H rxn remains the same.) - in general, the Addition of a Catalyst INCREASES the Rate of Reaction. Example : Enzyme is a catalyst in the body that speeds up certain bodily reaction. E a without Catalyst E a with Catalyst H rxn 6. Inhibitor: - a substance that inhibit the function of a catalyst to speed up the reaction. - in general, the Addition of an Inhibitor DECREASES the Reaction Rate. Example : Ammonia is formed from its elements using heterogeneous catalyst such as t (s): N (g) + H (g) t NH (g) (Check out Video at http://www.dac.neu.edu/physics/b.maheswaran/phy111/data/ch11/anim/anim11-5b.mov) Example 4: The catalytic converter converts NO (g) (result of burning nitrogen at high temperature) to N (g) and O (g). The O (g) along with the catalytic converter is used to produce CO (g) from CO (g). catalytic NO (g) converter catalytic CO (g) + O (g) converter N (g) + O (g) CO (g) (Left): A catalytic converter for most modern vehicle. Leaded gasoline, an inhibitor, deactivates catalytic converter. Hence, it is not legally used in vehicles age 151.
Example 5: The destruction of ozone in upper atmosphere can be attributed to NO (g) and CFCs acting as catalysts. NO (g) is produced from the combustion of N (g) at high temperature commonly found in internal combustion engine (high-altitude aircraft produces lots of NO gas). CFCs (Chloro-Fluoro-Carbon compounds) are found in aerosol can propellants, refrigerators, and air conditioners. They break down to form Cl (g) with the presence of light. NO (g) + O (g) NO (g) + O (g) O (g) + NO (g) NO (g) + O (g) O (g) + O (g) O (g) Cl (g) + O (g) ClO (g) + O (g) O (g) + ClO (g) Cl (g) + O (g) O (g) + O (g) O (g) Factors Affecting Reaction Rate 1. Activation Energy (E a ) Reaction Rate. Temperature (T) Reaction Rate. article Size (Surface Area ) Reaction Rate 4. Concentration Reaction Rate 5. Catalyst Reaction Rate 6. Inhibitor Reaction Rate Assignment 19.1: pg. 58 #1 to 5; pg. 57 #9 to 4; pg. 57 #74 (Above) The Ozone Hole over the South ole (Sept 000). A similar hole is present over the Arctic. (Left) rocess of Ozone Depletion. Ozone blocks harmful UV rays that can otherwise cause skin cancer. age 15.
Unit 7: Chemical Kinetics and Equilibrium 19.A: Reversible Reactions Reversible Reactions: - reactions that can go from the right hand side of the uation (products) to the left hand side of the uation (reactants). Chemical Equilibrium: - the state at which the concentrations of all reactants and products remain constant with time (the Forward Reaction Rate = Reverse Reaction Rate). - the uilibrium state is dynamic (not static). Chemical species are continuously converting from reactants to products and vice versa. It appears that the reaction has stopped only because the rate of consumption = rate of production. - if an uilibrium state is disturbed (changing concentrations of species, pressure, volume and temperature change), the reaction will shift towards one side in order to re-establish the new uilibrium state. - like reaction rate, Equilibrium is affected by Temperature. aa + bb cc + dd a and b: Coefficients of reactant species A and B c and d: Coefficients of product species C and D uilibrium symbol (indicating forward and reverse reactions) Equilibrium State [D] Concentration [C] [B] [A] Time to reach Equilibrium Time (Check out Video at http://www.carlton.paschools.pa.sk.ca/chemical/mtom/contents/chapter/images/hosimulation.mpg) The N O 4 (g) NO (guilibrium. Initially (icture A), there were very little NO (g). As time proceeded forward, more NO (g) (brown color) is produced (ictures B & C). However, there are still N O 4 (g) present at the uilibrium state (icture D). age 15.
Le Châtelier s rinciple: - a qualitative method to predict the shift on an uilibrium system if it is disturbed by means of changing concentration, pressure and temperature. - the uilibrium will shift in the direction that minimizes the change imposed on the system. 1. Effects of a Change in Concentration: a. An ADDITION of a species on one side of the uilibrium will Drive the System TOWARDS the Opposite Side. (There is more concentration of the species being added. Hence, the system will shift towards the opposite side to reduce the increased amount of that particular species.) b. A REMOVAL of a species on one side of the uilibrium will Drive the system TOWARDS the Same Side. (There is less concentration of the species being removed. Hence, the system will shift towards the removal side to compensate.) Concentration [D] [C] [B] [A] Changes in Concentrations on Equilibrium System aa + bb cc + dd Time indicates an Increase in [D]. As [D], uilibrium shifts to the left (aa + bb cc + dd). Hence, [A], [B], and [C]. indicates a Decrease in [A]. As [A], uilibrium shifts to the left (aa + bb cc + dd). Hence, [B], [C], and [D]. indicates an Increase in [B]. As [B], uilibrium shifts to the right (aa + bb cc + dd). Hence, [A], [C], and [D]. indicates a Decrease in [C]. As [C], uilibrium shifts to the right (aa + bb cc + dd). Hence, [A], [B], and [D]. age 154.
Unit 7: Chemical Kinetics and Equilibrium. Effects of a Change in ressure: a. Adding an Inert Gas has NO CHANGE on the uilibrium system. This is because an inert gas does not participate in the forward or reverse reaction. b. Reducing the Volume will Drive the System TOWARDS the Side With LESS Gaseous Molecules. Since there are less space for the number of molecules, the system will have to shift to the side with lesser gaseous molecules to compensate. c. Conversely, Expanding the Volume will Drive the System TOWARDS the Side With MORE Gaseous Molecules. Now that there is more room for the molecules to move about, the system will shift to the side that has more gaseous molecules to adjust to the new condition. d. When there are Equal Number of Gaseous Molecules on Both Side of the Equilibrium, any Change in Volume will NOT Affect System. Changes in ressures on a Gaseous Equilibrium System aa + bb cc + dd when (c + d) > (a + b) [D] [C] Concentration [B] [Inert Gas] [A] indicates an Increase in Volume. As V, uilibrium shifts to the right (aa + bb cc + dd) since there are more gaseous molecules on the product side [(c + d) > (a + b)]. Hence, [A], [B], [C], and [D]. indicates a Decrease in Volume. As V, uilibrium shifts to the left (aa + bb cc + dd) since there are LESS gaseous molecules on the reactant side [(c + d) > (a + b)]. Hence, [A], [B], [C], and [D]. indicates an Addition of an Inert Gas. There is no shifting of the uilibrium. (aa + bb cc + dd) as inert gas does not affect the system. Hence, [A], [B], [C], and [D] remain unchanged. Time (See the Video at http://www.sasked.gov.sk.ca/docs/chemistry/missionmars/contents/chapter/images/pno4.mpg) age 155.
. Effects of a Change in Temperature: - look at the energy (written in the reactant or product side) as a chemical species. Then, the predictions will be the same as those found with changing the concentrations. a. For an Exothermic Equilibrium System: aa + bb cc + dd + Energy - an Increase in Temperature will drive the system to the left (aa + bb cc + dd + Energy). There is more heat added and because energy is written on the product side, the system will shift to the opposite side to re-establish uilibrium. Hence, [A], [B], [C], and [D]. - a Decrease in Temperature will drive the system to the right (aa + bb cc + dd + Energy). There is less heat overall and because energy is written on the product side, the system will shift to the products to compensate. Hence, [A], [B], [C], and [D]. Changes in Temperature on an Exothermic Equilibrium System aa + bb cc + dd + Energy [D] Concentration [C] [B] [A] Time indicates a Decrease in Temperature. As T, uilibrium shifts to the right of an exothermic system (aa + bb cc + dd + Energy). Hence, [A], [B], [C], and [D]. indicates an Increase in Temperature. As T, uilibrium shifts to the left of an exothermic system (aa + bb cc + dd + Energy). Hence, [A], [B], [C], and [D]. b. For Endothermic Equilibrium System: aa + bb + Energy cc + dd - a Decrease in Temperature will drive the system to the left (aa + bb + Energy cc + dd). There is less heat overall and because energy is written on the reactant side, the system will shift to the reactants to compensate. Hence, [A], [B], [C], and [D]. - an Increase in Temperature will drive the system to the right (aa + bb + Energy cc + dd). There is more heat added and because energy is written on the reactant side, the system will shift to the opposite side to re-establish uilibrium. Hence, [A], [B], [C], and [D]. age 156.
Unit 7: Chemical Kinetics and Equilibrium Changes in Temperature on an Endothermic Equilibrium System aa + bb + Energy cc + dd [D] Concentration [C] [B] [A] Time indicates an Increase in Temperature. As T, uilibrium shifts to the right of an endothermic reaction (aa + bb + Energy cc + dd). Hence, [A], [B], [C], and [D]. indicates a Decrease in Temperature. As T, uilibrium shifts to the left of an endothermic reaction (aa + bb + Energy cc + dd). Hence, [A], [B], [C], and [D]. Example 1: The uilibrium system, NF (g) + 14.9 kj N (g) + F (g), is put under the following changes. redict the shift of the system and the resulting concentrations of all species for each case. a. an increase in the concentration of F (g). b. a decrease in the concentration of N (g) The system will shift to the LEFT. NF (g) + 14.9 kj N (g) + F (g) Effects: [NF ] (increase); [N ] (decrease) The system will shift to the RIGHT. NF (g) + 14.9 kj N (g) + F (g) Effects: [NF ] (decrease); [F ] (increase) c. a decrease in the concentration of NF (g). d. a decrease in Temperature. The system will shift to the LEFT. NF (g) + 14.9 kj N (g) + F (g) Effect: [N ] (decrease); [F ] (decrease) The system will shift to the LEFT. NF (g) + 14.9 kj N (g) + F (g) Effect: [NF ] (increase); [N ] (decrease); [F ] (decrease) age 157.
e. an addition of He (g). f. an increase in volume. There will be NO SHIFT on the system. NF (g) + 14.9 kj N (g) + F (g) (He is an inert gas and does not involve with the uilibrium system) Effect: [F ], [N ] and [F ] remain the same. The system will shift to the RIGHT. NF (g) + 14.9 kj N (g) + F (g) (There are more gaseous molecules on the product side 4 moles of N (g) and F (g) versus moles of NF (g) ) Effect: [NF ] (decrease); [N ] (increase); [F ] (increase) Example : The Haber-Bosch process, N (g) + H (g) NH (g) + 9 kj is essentially an uilibrium system. A chemical engineer would like the highest yield of ammonia. List all the possible method of production that will ensure maximum amount of NH (g) produced. N (g) + H (g) NH (g) + 9 kj (Desire Effect: [NH ], which means driving the system forward.) 1. Increase the concentrations of N (g) or H (g) or both will drive the system forward.. Decrease the concentration of NH (g) as it is produced will shift the system forward.. Lower the Temperature will drive the system to the product side. 4. Decrease the Volume of the system will shift the system to the right due to smaller number of gaseous molecules on the product side. Assignment 19.A: pg. 544 #6 and 7; pg. 548 #14 and 16; pg. 57 #44 to 47; pg. 57 #67, 70 and 76; pg. 575 # age 158.
Unit 7: Chemical Kinetics and Equilibrium 19.B: Equilibrium Constants, Concentrations and ressures Equilibrium Expression: - an expression relating the concentrations or pressures of the reactants and products when they are at the state of uilibrium. - it takes the form of the individual products raised to the power of their respective coefficients divided by the individual reactants raised to the power of their respective coefficients. - the uilibrium expression is unique for each reaction, but it is the same for that particular reaction regardless of temperature. Equilibrium Constant (K): - the unitless numerical value of the uilibrium expression. - the uilibrium constant is the same for a particular reaction if it remains at the same temperature. - the symbol for uilibrium constant when the expression deals with concentrations is simply K or K. When the expression deals with pressures, it is symbolized as K. - reversing uilibrium reaction will cause the reciprocate the uilibrium constant (1/K) Equilibrium Expression and Constant of a Reaction aa + bb cc + dd c d C D K = a A b B K = c C, a A, d D, b B, a and b: Coefficients of reactant species A and B c and d: Coefficients of product species C and D [A], [B], [C] & [D] = [A], [B], [C] & [D] = Equilibrium Concentrations of Chemicals (mol/l) A, B, C & D = A,, B,, C, & D, = Equilibrium ressures of Chemicals (atm) K or K = Equilibrium Constant (Concentrations) Equilibrium Expressions Both K and K are Unitless K = Equilibrium Constant (ressures) Reversing Equilibrium Reaction aa + bb cc + dd cc + dd aa + bb a b A B K = [ C] c [ D] d = c [ C] [ D] a [ A] [ B] d b 1 = K 1 = K 1 K = Reverse Equilibrium Constant age 159.
Example 1: Write the uilibrium expression of the following reactions. a. Br (g) + 5 F (g) BrF 5 (g) b. N H 4 (g) + 6 H O (g) NO (g) + 8 H O (g) [ BrF5 ] [ Br ][ F ] 5 BrF 5 K = 5 Br F 8 NO H O [ N H ][ H O ] 6 4 8 NO H K = 6 NH4 O HO Example : For the following reaction and the uilibrium concentrations at 00 K. C H 6 (g) + NH (g) + O (g) C H N (g) + 6 H O (g) [C H 6 ] = 0.500 M [NH ] = 0.50 M [O ] = 0.50 M [C H N] =.50 M [H O] =.00 M a. Write the uilibrium expression and determine the uilibrium constant. b. Write the uilibrium expression and calculate the uilibrium constant for the following reaction with the same uilibrium concentrations. C H N (g) + 6 H O (g) C H 6 (g) + NH (g) + O (g) a. Equilibrium Expression: Equilibrium Constant: 6 [ CH N] [ H O] [ C ] H 6 NH O 6 [ CHN] [ H O] [ C H ] [ NH ] [ O ] C H 6 NH O b. Equilibrium Expression: K = [ C H N] [ H O] 6 C H6 NH O Equilibrium Constant: K = [ C H N] [ H O] 6 Example : The German s Haber-Bosch process developed in 191 utilizes an iron surface that contains traces of aluminum and potassium oxide as a catalyst to manufacture ammonia from nitrogen and hydrogen. It is an important process as ammonia is commonly used in fertilizer and ammunition. In 1918, the scientist Fritz Haber won the Nobel rize in chemistry for his contribution. N (g) + H (g) NH (g) At 400 K, NH = 0.104 atm, N =.8084 atm and H expression in terms of pressure and calculate K. 6 = 6 (.50) (.00) ( 0.500) ( 0.50) ( 0.50) (Reverse Equilibrium) 6.80 10 6 6 CHN H O = [ CH 6 ] [ NH ] [ O ] = K 1 K = (6.80 10 6 ) 1 K = 1.47 10 7 NH Equilibrium Expression: K = Equilibrium Constant: K = N H ( 0.104) (.8084)( 0.010) 1 = 0.010 atm. Write the uilibrium K =.5 10 age 160.
Unit 7: Chemical Kinetics and Equilibrium Equilibrium osition: - the concentrations or pressures of all chemical species at uilibrium state. - depends strongly on the Initial Concentrations of the chemical species. (In contrast, K does NOT depend on initial concentrations, only on temperature and the specific reaction.) - since there all many possible initial concentrations for any one reaction, there are infinite number of uilibrium position for a particular reaction. Note: Do NOT confuse initial concentrations [A] 0 with uilibrium concentration [A]!! We only use Equilibrium Concentrations to calculate K by substituting them into the uilibrium expression. Example 4: The formation of HI (g) is an uilibrium reaction. Several experiments are performed at 710 K using different initial concentrations. H (g) + I (g) HI (g) Experiment 1 Experiment Initial Equilibrium Initial Equilibrium [H ] 0 = 0.100 M [H ] = 0.0 M [H ] 0 = 0 M [H ] = 0.050 M [I ] 0 = 0.100 M [I ] = 0.0 M [I ] 0 = 0.0100 M [I ] = 0.0450 M [HI] 0 = 0 M [HI] = 0.156 M [HI] 0 = 0.50 M [HI] = 0.80 M Experiment Experiment 4 Initial Equilibrium Initial Equilibrium [H ] 0 = 0.00150 M [H ] = 0.0150 M [H ] 0 = 0 M [H ] = 0.044 M [I ] 0 = 0 M [I ] = 0.015 M [I ] 0 = 0 M [I ] = 0.044 M [HI] 0 = 0.17 M [HI] = 0.100 M [HI] 0 = 0.400 M [HI] = 0.11 M a. Write the uilibrium expression for the formation of HI (g). b. Calculate the uilibrium constant for each experiment, and average them for an overall value. a. Equilibrium Expression: b. Equilibrium Constant: [ HI] Experiment 1: K 1 = H I K 1 = Experiment : K = K = [ ] [ ] ( 0.156) ( 0.0)( 0.0) [ HI] [ H ] [ I ] ( 0.100) ( 0.0150)( 0.015) [ HI] [ H ][ I ] Experiment : K = K 1 = 49.4 K = Experiment 4: K 4 = K = 49. K 4 = [ HI] [ H ] [ I ] ( 0.80) ( 0.050)( 0.0450) [ HI] [ H ] [ I ] ( 0.11) ( 0.044)( 0.044) K = 49.8 K 4 = 49.5 Average 49.4 + 49.8 + 49. + 49.5 4 K avg = 49.5 age 161.
Homogeneous Equilibria: - an uilibrium system where all chemical species are in the same phase. Heterogeneous Equilibria: - an uilibrium system where some chemical species are in different phase compare to the others. - chemical species that are ure Solid or ure Liquid are NOT Included in the Equilibrium Expression. This is due to the fact that pure solids and liquids do not have concentrations. Example 5: Write the uilibrium expression for the following systems. a. NaN (s) Na (s) + N (g) b. NH (g) + CO (g) H NCONH (s) + H O (g) [N ] K = N (NaN and Na are ure Solids) c. Ag SO 4 (s) Ag + (aq) + SO 4 (aq) d. HF (aq) + H O (l) H O + (aq) + F (aq) [Ag + ] [SO 4 ] (Ag SO 4 is a ure Solid) (No K because there are no gases.) [ H O] [ NH ] [ CO] K = NH (H NCONH is a ure Solid) [ O ][ F ] H + [ HF] H O CO (H O is a ure Liquid) (No K because there are no gases.) Important Notes Regarding the Size of the Equilibrium Constant (K): aa + bb cc + dd c d C D K c = a A b B K = c C, a A, d D, b B, 1. When K >> 1, the uilibrium system favours the products. There are more products than reactants at the state of uilibrium. ([C] and [D] or C, and D, >> [A] and [B] or A, and B, ). When K << 1, the uilibrium system favours the reactants. There are less products than reactants at the state of uilibrium. ([A] and [B] or A, and B, >> [C] and [D] or C, and D, ). When K 1, the uilibrium system favours neither the products nor the reactants. There are roughly the same amount of products and reactants at the state of uilibrium. ([C] and [D] or C, and D, [A] and [B] or A, and B, ) 4. The Size of K has NO Relationship with the Rate of Reaction to reach the state of uilibrium. Reaction Rate is dependent on Activation Energy and Temperature (T) (NOT K). age 16.
Unit 7: Chemical Kinetics and Equilibrium Example 6: The reaction, Cl (g) + F (g) ClF (g) at 50 K has 50., If the uilibrium concentrations of Cl (g) and ClF (g) are 0.149 M and 0.05 M respectively, what is the uilibrium concentration of F (g)? [Cl ] = 0.149 M [ClF ] = 0.05 M 50. [F ] =? 50. = [ ClF ] [ Cl ] [ F ] ( 0.05) ( 0.149)[ F ] [F ] = [F ] = ( 0.05) ( 50.)( 0.149) ( 0.05) ( 50.)( 0.149) [F ] = 0.178 M ICE Box: - stands for Initial, Change, and Equilibrium. It is a table that organizes information to calculate final uilibrium concentrations given the uilibrium constant and initial concentration. Example 7: The formation of HCl (g) from its elements, H (g) + Cl (g) HCl (g) has 0.404 at 50 K. A 5.00 L flask at 50 K contained an initial concentration of.00 mol of HCl (g) and.85 mol of H (g). When the system reached uilibrium, it was found that there were 0.860 mol of Cl (g). Determine the concentrations of H (g) and HCl (g) at uilibrium. 0.404.85 mol [H ] 0 = = 0.770 M 5.00 L.00 mol [HCl] 0 = = 0.600 M 5.00 L [Cl ] 0 = 0 M 0.860 mol [Cl ] = = 0.17 M 5.00 L [H ] =? [HCl] =? The system must shift to the left because initially, we are missing one reactant ([Cl ] 0 = 0 M). Hence, the change to the H is positive, and the change to the HCl would be negative. Since there is 0.17 M of Cl at uilibrium, it means 0.17 M of H is added (1:1 mol ratio between Cl and H ). It also means that there is (0.17 M) less HCl (:1 mol ratio between Cl and HCl). H (g) + Cl (g) HCl (g) Initial 0.770 M 0 M 0.600 M Change + 0.17 M + 0.17 M (0.17 M) Equilibrium 0.94 M 0.17 M 0.56 M Verify with K: [ HCl] [ H ] [ Cl ] = ( 0.56) ( 0.94)( 0.17) 0.404 (This matches with K given in the question.) Therefore, the uilibrium concentrations are: [H ] = 0.94 M, [Cl ] = 0.17 M and [HCl] = 0.56 M Assignment 19.B: pg. 545 #8; pg. 546 #9 and 10; pg. 547 #11 and 1; pg. 548 #1, 15, 17 to 19; pg. 57 #48 to 50; pg. 57 #71 and 7; pg. 575 #1 age 16.
19.C: Equilibrium Calculation with Initial Concentrations and ressures rocedure to calculate Equilibrium Concentrations/ressures from Initial Concentrations/ressures: 1. Balance the Equilibrium Chemical Equation (if needed).. Set up the ICE Box and the uilibrium expression.. Find the Initial Concentrations or ressures. 4. Decide on which way the system will shift. 5. Using mole ratios, fill out the Change row using x as 1 mole of reactant and/or product used/produced. 6. Write out the expressions for the Equilibrium row of the ICE Box. 7. Substitute these Final Concentrations into the Equilibrium Expressions. 8. Solve for x using the Solve Function of your calculator. 9. Calculate the actual uilibrium concentrations of all chemical species. Using the SOLVE function of the TI-8 lus Calculator: 1. Manipulate the Equation so one-side is ual to 0. x 0.4 x Example:.4 10 = ( ) becomes 0 = ( 0.4 x) x.4 10. Callout the solve function. solve (Equation Expression, X, guess, {lower boundary, upper boundary}) nd CATALOG 0, X,T,θ,n Enter 0 nd { ( Enter 0 Enter Smallest Initial Concentration / ressure or 1E99 Scroll Down from S e x S LN nd } ) ENTER Enter arameters ENTER Select solve ( Value of x age 164.
Unit 7: Chemical Kinetics and Equilibrium Example 1: The uilibrium between dinitrogen textraoxide and nitrogen dioxide, N O 4 (g) NO (g) has 4.66 10 at C. If 1.40 mol of N O 4 (g) is injected into a.00 L container at C, determine the concentrations of both gases at uilibrium. NO [ NO4] = 4.66 10 1.40 mol [N O 4 ] 0 = = 0.700 M.00 L [NO ] 0 = 0 M [N O 4 ] =? [NO ] =? The system must shift to the right because initially, we are missing the product ([NO ] 0 = 0 M). Hence, the change to the NO is positive, and the change to the HCl would be negative. Since the mole ratio is 1:1 mol ratio between N O 4 and NO, for every x mol of N O 4 used, x mol of NO is produced. Substitute uilibrium concentrations into the uilibrium expression. NO [ NO4] 4.66 10 = ( x) x or 0 = ( 0.700 x) ( 0.700 x) Using the SOLVE function of the TI-8 lus Calculator. (0 x 0.700) x = 0.0798 mol/l [N O 4 ] = (0.700 0.0798) [N O 4 ] = 0.67 mol/l [NO ] = (0.0798) [NO ] = 0.0560 mol/l Or expanding the original uation 4.66 10 (0.7 x) = 4x 0.006 (4.66 10 ) x = 4x 0 = 4x + (4.66 10 ) x 0.006 (Quadratic Equation: Apply the Quadratic Formula!) x = b ± b 4ac a a = 4 b = 4.66 10 c = 0.006 ( 4.66 10 ) ± ( 4.66 10 ) 4( 4)( 0.006) x = ( 4) x = 0.079804009 x = 0.091454009 (omit negative x) Verify with K: NO = ( 0.0560 ) [ NO4] ( 0.67) 0.00467 = 4.67 10 Therefore, the uilibrium concentrations are: N O 4 (g) NO (g) Initial 0.700 M 0 M Change x M + x M Equilibrium (0.700 x) M x M [N O 4 ] = 0.67 mol/l, and [NO ] = 0.0560 mol/l 4 4.66 10 (This matches with K given in the question.) or age 165.
Example : The uilibrium, N O (g) N (g) + O (g) has K = 84.1 at 5 C. If 1.50 atm of N (g) and 1.80 atm of O (g) are injected into a flask at 5 C, determine the uilibrium pressures of all gases. ( ) ( ) N O ( ) N O ( N O ) 0 = 0 atm ( N ) 0 = 1.50 atm ( O ) 0 = 1.80 atm ( N O =? ( N =? ( O =? = 84.1 Substitute uilibrium concentrations into the uilibrium expression. ( ) ( ) N O ( ) N O 84.1 = ( ) 1.50 x ( 1.80 x ) ( x) Using the SOLVE function of the TI-8 lus Calculator. (0 x 1.50) x = 0.09514405 atm ( O O N = (0.09514405) ( N ( N = 1.50 (0.09514405) ( ( O = 1.80 (0.09514405) ( = 0.187 atm N = 1.1 atm O = 1.71 atm Or expanding the original uation: 6.4x = (.5 6x + 4x )(1.8 x) 6.4x = 4.05.5x 10.8x + 6x + 7.x 4x 4x +.x + 1.05x 4.05 = 0 (Use the SOLVE function instead of solving a cubic function by hand!) Verify with K: ( ) ( ) N O ( ) N O = ( ) 1.1 ( 1.71 ) ( 0.187) Therefore, the uilibrium pressures are: The system must shift to the left because initially, we are missing the reactant ( ( N O ) 0 = 0 atm). Hence, the change to the N O is positive, and the change to the N and O would be negative. Since the mole ratio is 1::1 mol ratio between N O 4, N, and O, for every x mol of N O 4 produced; x mol of N and x mol of O is used. N O (g) N (g) + O (g) Initial 0 atm 1.50 atm 1.80 atm Change + x atm x atm x atm Equilibrium x atm (1.50 x) atm (1.80 x) atm or 0 = ( )( 1.5 x 1.8 x ) ( 4x ) 84.1 8.9 (This is a very close match with K given in the question.) ( N O = 0.187 atm, ( N = 1.1 atm, and ( O = 1.71 atm or Assignment 19.C: Equilibrium Calculations Worksheet age 166.
Unit 7: Chemical Kinetics and Equilibrium Equilibrium Calculations Worksheet 1. Given: A (g) + B (g) C (g) + D (g). One mole of A and one mole of B are placed in a 0.400 L container. After uilibrium has been established, 0.0 mole of C is present in the container. Calculate the uilibrium constant, K c, for the reaction.. NO (g) and O (g) are mixed in a container of fixed volume and kept at 1000 K. Their initial concentrations are 0.000 mol/l and 0.000 mol/ L, respectively. When the reaction, NO (g) + O (g) NO (g) has come to uilibrium, the concentration of NO (g) is.1 10 mol/l. Calculate a. the concentration of NO (g) at uilibrium, b. the concentration of O (g) at uilibrium, and c. the uilibrium constant, K c, for the reaction.. For the reaction, CO (g) + H O (g) CO (g) + H (g), the value of the uilibrium constant, K c, is 1.845 at a given temperature. If 0.500 mole CO (g) and 0.500 mole H O (g) are placed in a 1.00-L container at this temperature, and allow the reaction to reach uilibrium, what will be the uilibrium concentrations of all substances present? 4. The uilibrium constant for the reaction, SO (g) + O (g) SO (g) is K c = 79 at a given high temperature. a. What is the value of the uilibrium constant for the reaction, SO (g) SO (g) + O (g) at this temperature? b. 4.00 mol of SO (g) and.00 mol of O (g) are initially injected into a.00 L flask. What are the uilibrium concentrations for all species? 5. At 448 C, the reaction indicated by the uation, H (g) + I (g) HI (g) has an uilibrium constant, K, ual to 50.5. If.00 atm of H (g) and 1.00 atm of I (g) were present initially, what is the uilibrium pressures for all species? 6. The reaction NO (g) + Cl (g) NOCl (g), has an uilibrium constant, K, ual to 0.6 at 700 K. What are the uilibrium pressures for all species if initially 0.650 atm of NO (g) and 1.45 atm of NOCl (g) were placed in a container with fixed volume? 7. For the reaction, Cl (g) + F (g) ClF (g), K = 19.9. Determine the finial pressures of all gases if the pressures for chlorine gas and fluorine gas were 4.50 atm and.5 atm respectively. 8. For the reaction described by the uation, N (g) + C H (g) HCN (g), K c =.0 10 4 at 00 C. What is the uilibrium concentration of hydrogen cyanide if the initial concentrations of N and acetylene (C H ) were 5.5 mol/l and 7.00 mol/ L respectively? 9. At 7 K, K = 0.416 for the uilibrium, NOBr (g) NO (g) + Br (g). If the original pressures of NOBr (g) and Br (g) were 0.760 atm and 0.450 atm respectively, what is the uilibrium pressures for all species? 10. The uilibrium constant, K c, for BrF (g) Br (g) + F (g), is 0.0181. What are the uilibrium concentrations of all these gases if the initial concentration of Br (g) and F (g) was both 0.00 M? age 167.
11. At 18 C, K = 4.8 10 for the uilibrium, NH 4 HS (s) NH (g) + H S (g). Calculate the uilibrium pressures for all gases if an excess amount of NH 4 HS (s) and 1.5 atm of NH (g) were initially placed in a closed container? 1. K c = 96. at 400 K for the reaction, Cl (g) + Cl (g) Cl 5 (g). What are the uilibrium concentrations of all species if the initial concentrations were 0.50 mol/l for Cl 5 and 6.00 mol/l for Cl? 1. For the uilibrium, IBr (g) I (g) + Br (g), K = 8.5 10 at 150 C. If 0.685 atm of I (g) and 0.4 atm of IBr (g) were placed in a closed container, what are the uilibrium pressures of all gases? 14. The reaction of iron and water vapor results in an uilibrium, Fe (s) + 4 H O (g) Fe O 4 (s) + 4 H (g), and an uilibrium constant, K c, of 4.60 at 850 C. What are the uilibrium concentrations of all gaseous species if the reaction is initiated with a. 0.0 g of H O (g) and excess Fe in a 10.0 L container? b. 10.0 g of H (g) and excess iron oxide, Fe O 4, in a 16.0 L container? 15. The dissociation of calcium chromate, CaCrO 4 (s) Ca + (aq) + CrO 4 (aq), has a K c = 7. 10 4. What are the uilibrium concentrations of Ca + and CrO 4 in a saturated solution of CaCrO 4? 16. The dissociation of calcium fluoride, CaF (s) Ca + (aq) + F (aq), has a K c =.9 10 11. What are the uilibrium concentrations of Ca + and F in a saturated solution of CaF? Answers: 1. K c = 0.15 a. 0.0179 mol/l b. 0.090 mol/l c. K c = 8.8 10. [CO] = [H O] = 0.1 mol/l; [CO ] = [H ] = 0.88 mol/l 4a. K c =.58 10 b. [SO ] = 0.147 mol/l; [O ] = 0.57 mol/l; [SO ] = 1.85 mol/l 5. ( H = 1.065 atm; ( I = 0.0650 atm; ( HI = 1.87 atm 6. ( NO = 1.56 atm; ( Cl = 0.455 atm; ( NOCl = 0.59 atm 7. ( Cl = 1.9 atm; ( F = 0.684 atm; ( ClF = 5.1 atm 8. [N ] = 0.007 mol/l; [C H ] = 1.75 mol/l; [HCN] = 10.5 mol/l 9. ( NOBr = 0.418 atm; ( NO = 0.4 atm; ( Br = 0.61 atm 10. [BrF] = 0.47 mol/l; [Br ] = [F ] = 0.066 mol/l 11. ( NH = 1.5 atm; ( H S =.85 10 atm 1. [Cl ] = 4. 10 4 mol/l; [Cl ] = 6.00 mol/l; [Cl] = 0.50 mol/l 1. ( IBr = 0. atm; ( I = 0.686 atm; ( Br = 6.7 10 4 atm 14a. [H O] = 0.0676 mol/l; [H ] = 0.0989 mol/l b. [H O] = 0.16 mol/l; [H ] = 0.184 mol/l 15. [Ca + ] = [CrO 4 ] = 0.07 mol/l 16. [Ca + ] =.1 10 4 mol/l; [F ] = 4. 10 4 mol/l age 168.