# rate = k [NO] 2 [H 2 ] CHEMICAL KINETICS Review Exam 3 1. How FAST {Speed like miles per hour and 2. By what MECHANISM Does a Reaction Take Place?

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1 Review Chap 14: CHEMICAL KINETICS Review Exam Chapters CHEMICAL KINETICS DEALS WITH 1. How FAST {Speed like miles per hour and. By what MECHANISM Does a Reaction Take Place? Given the following data, determine the rate law for the all gas reaction of nitrogen monoxide reacting with hydrogen forming nitrogen and water Experiment moles NO moles H Initial Rate (M/s) x x x 10 - Write, balance, and interpret reaction NO (g) + H (g) N (g) + H O (g) Rate [NO] X [H ] Y NO (g) + H (g) N (g) + H O (g) Rate [NO] X [H ] Y Experiment moles NO moles H Initial Rate (M/s) x x x 10 - From Exp 1 & When [H ] doubles, Rate doubles Rate [H ] 1 NO (g) + H (g) N (g) + H O (g) Rate [NO] X [H ] Y Experiment moles NO moles H Initial Rate (M/s) x x x 10 - From Exp & When [NO] doubles, Rate quadruples Rate [NO] rate k [NO] [H ] 1 st order w.r.t H nd order w.r.t. NO rd order overall k is the rate constant. 1

2 TEMPERATURE and RATE As temperature increases The rate constant (k) The rule of thumb is: so does the reaction rate is temperature dependent rate doubles for each 10 degree rise in temp Half-life {t 1/ } is defined as the time required for one-half of a reactant to react For A B at t 1/, the concentration of A is one-half the initial concentration of A HALF-LIFE time required for one-half of a reactant to react Pressure of CH NC is ½ of the original Pressure at t 1/ In this example at 1,000 sec P CH NC 0.5 P 0 75 torr All radioactive decay is 1 st order [ A ] 0 ln k t [ A ] At t 1/ concentration is ½ initial [ A] 0 ln k t [ A] 1 0 ln k t 1/ 1/ In a chemical reaction bonds are broken & bonds are formed The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. The mechanism MUST agree with experiment Reaction mechanisms Reactions occur through several discrete steps Each of these is known as an elementary reaction or elementary process The molecularity of a process tells how many molecules are involved in the process

3 (a) (b) (c) It has been proposed that the mechanism of O O proceeds by the following mechanism step 1 O O + O step O + O O Describe the molecularity of each elementary reaction in this mechanism. Does mechanism agree with balanced reaction? Identify the intermediate(s). (a) 1. 1 st order. nd order (b) Yes (c) O Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen Write and balance reaction O (g) + NO (g) N O 5 (g) + O (g) The experimental rate law is rate k[o ][NO ] Interpret results 1 st order with respect to Ozone & nitrogent dioxide The proposed mechanism for the reaction of Ozone with Nitrogen dioxide is Step 1 O (g) + NO (g) NO (g) + O (g) step NO (g) + NO (g) N O 5 (g) Does the proposed mechanism agree with experiment? Yes, steps add to give O + NO N O 5 + O What is the intermediate? NO (g) What can be said about the relative rates of the two steps of the mechanism? O (g) + NO (g) N O 5 (g) + O (g) Mechanism step 1 O (g) + NO (g) NO (g) + O (g) step NO (g) + NO (g) N O 5 (g) The experimental rate law is rate k[o ][NO ] Because the rate law conforms to the molecularity of the first step, that must be the rate-determining step. The second step must be much faster than the first one. The decomposition of Hydrogen Peroxide H O (aq) H O (liq) + O (g) is catalyzed by iodide ion By experiment the rate law is found to be Rate k[h O ][I - ] The proposed mechanism Step 1 : H O + I - H O + IO - Step : H O + IO - H O + I - Is this an acceptable mechanism? yes THE COLLISION THEORY (Three parts) 1. Molecules can only react if they collide. collide with sufficient energy*. *Activation Energy (E a ): The energy barrier that must be surmounted before reactants can be converted to products.. correctly oriented molecules

4 Figure Not all collisions lead to products Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and E a : k A e -E a RT Taking the natural logarithm of both sides, the equation becomes E ln k - a 1 ( ) + ln A R T Arrhenius Equation E a ln k - ( ) + ln A R 1 T y m x + b If k is determined experimentally at several temperatures, E a can be calculated from the 1 slope of a plot of ln k vs.. T Chapter 15 Chemical Equilibrium The reaction for the production of ammonia can be written in a number of ways (a) N (g) + H (g) NH (g) (b) ½ N (g) + / H (g) NH (g) (c) 1/ N (g) + H (g) / NH (g) Write the equilibrium constant expressions and how they are related K (a) N (g) + H (g) NH (g) (b) ½ N (g) + / H (g) NH (g) (c) 1/ N (g) + H (g) / NH (g) [ NH ] [ N ][ H ] [ NH ] 1/ [ N ] [ H ] a K b K / c / [ NH ] 1/ [ N ] [ H ] K a K b K a K c K b K c The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq k f [NO ] k r [N O 4 ] 4

5 The Initial Change Equilibrium method is used to calculate equilibrium constant SO (g) + O (g) SO (g) Initial Moles Moles Moles Change x -x + x Equilibrium Moles Moles Moles What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium. If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium. Change to Concentrations at Equilibrium H (gas) + I (gas) HI(gas) H, Moles I, Moles HI, Moles I nitially x x 10-0 C hange -9.5 x x x 10 - E quilibrium 6.5 x x x 10 - K c the equilibrium constant 51 [HI] [H ] [I ] (1.87 x 10 - ) (6.5 x 10-5 )(1.065 x 10 - ) UNITS? Relationship between K c and K p Where Kp Kc (RT) n n (moles of gaseous product) (moles of gaseous reactant) Kp Kc For H (gas) + I (gas) HI(gas) WHY? For N O 5 (g) 4 NO (g) + O (g) The K p and K c expressions : K p (P NO ) 4 (P O ) / (P NO5 ) K C [NO ] 4 [ O ] / [N O 5 ] Is K p K C? No! 5

6 HETEROGENEOUS EQUILIBRIUM Pure Solids and Liquids do not have a concentration therefore they do not appear in the equilibrium expression Ag + (aq) + Cl - (aq) AgCl (solid) or AgCl (solid) Ag + (aq) + Cl - (aq) K c 1 / [Ag + (aq) ] [Cl - (aq) ] or K sp [Ag + (aq) ] [Cl - (aq) ] 1/ K c Le Châtelier s Principle Le Châtelier s Principle The Effect of Changes in Pressure The Effect of Changes in Temperature Catalysts increase the rate of both the forward and reverse reactions. Equilibrium is achieved faster, but the equilibrium composition remains unaltered. O (g) O (g) H 84 kj What would be the effect of (a) Decreasing the volume of the system (b) Increasing the pressure by adding O (c) Decreasing the temperature (d) Adding a catalyst Chapter 16 Acids and Bases Arrhenius Acid: Hydronium ion (H O + ) Base: Hydroxide ions (OH - ) Brønsted Lowry Acid: Proton donor Base: Proton acceptor Lewis Acid electron-pair acceptor Base is an electron-pair donor 6

7 No Equilibrium for the 7 Strong Acids or 7 Strong Bases HCl (aq) HBr (aq) HI (aq) HNO (aq) HClO (aq) HClO 4 (aq) H SO 4 (aq) LiOH(aq) NaOH(aq) KOH(aq) RbOH(aq) CsOH(aq) Ba(OH) (aq) Sr(OH) (aq) Conjugate Acids and Bases: An Acid and a Base that differ only in the presence {or absence} of a proton. What are the conjugate bases of the following Brønsted Lowry acids Which one of the conjugate bases of the following Brønsted-Lowry acids is incorrect? HClO 4 H SO 4 HSO 4 H O + ClO - 4 HSO - 4 SO 4 H O (a) ClO - for HClO (b) HS - for H S (c) NH for NH + 4 (d) SO - 4 for HSO - 4 (e) H SO 4 for HSO - 4 (e) Ion-Product Constant This special equilibrium constant is referred to as the ion-product constant for water, K w K W [H O + ] [OH ] At 5 C, K w ph is defined as ph log [H O + ] 7

8 Determine the ph of 0.50 M HF solution at 5 C. K a 7.1 x ST Write & Balance Reaction HF(aq) H + (aq) + F - (aq) nd Write Equilibrium Expression K a + [H ][ F ] [ HF] Initial Change Equilibrium Determine ph of 0.50 M HF solution K a 7.1 x 10 4 (aq) (aq) (aq) HF H + + F Initi al (M): Change(M): x +x +x Eqm (M): 0.50 x x x K a + [H ][ F ] [ HF] Ka 7.1 x 10-4 ; [HF] 0.50 x ; [H + ] [F - ] x 4 [x][ x] x x [0.50 x] x 0.50 x (7.1 x 10-4 )(0.50).55 x 10-4 x 1.9 x 10 - WAS THE APPROXIMATION VALID? [x][ ] x x x 10 [0.50 x ] x 0.50 No! Why not? Because x 1.9 x 10 - is not small compared to x 0.50 VALID APPROXIMATIONS? What was the difference between the two examples { HCN(aq) and HF(aq) }? Both are monoprotic acids Both has an initial concentration of 0.50 M Right the difference was in the size of K a K a 7.1 x 10-4 for HF K a 4.9 x for HCN What is the poh of a 0.001M monoprotic weak acid whose K a is 1.6 x 10 10? HA (aq) H + (aq) + A - (aq) Initial Change x + x + x Equil x x x K A + [H ][A ] ( x)( x) 1.6x10 [HA] (0.001) ph log [H + ] log x -log (4.0x10-7 ) 6.4 ph + poh 14 poh

9 % dissiciation of a 0.001M monoprotic weak acid whose K a is 1.6 x 10 10? + [H ] % Dissociation 100% [HA] 7 [4.0x10 ] Dissociation [1x10 ] % % 4 x 10 - What is the ph of a 0.001M monoprotic strong acid? HA(aq) H + (aq) + A (aq) Initial Change End ph log [H + ] log Find poh of a 0.10 M solution of Aniline (C 6 H 5 NH ), K b.8 x Write, Balance and interpret C 6 H 5 NH (aq) + H O C 6 H 5 NH + (aq) + OH (aq) I C - x x x E 0.10 x x K b + 6H5NH ][OH ] [x][x] [C x [C H NH ] [ x] 6 5 x [OH - ] 6. x 10-5 poh -log[oh - ] 4. ph of a 0.10 M solution of ammonia, K b 1.8 x 10-5 K b NH (aq) NH 4 + (aq) + OH - (aq) [NH4 ][OH ] K b [NH ] + [OH - ] 1.4 x 10 - poh -log [OH - ].9 ph [NH4 ][OH ] [x][x] 1.8x10 [NH ] [0.10- x] 5 Acid Base Properties of Salts Salts that produce neutral solutions are those formed from strong acids and strong bases. For example NaCl Salts that produce basic solutions are those formed from weak acids and strong bases. For example MgCl Salts that produce acidic solutions are those formed from strong acids and weak bases. For example NaClO 9

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