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Chemical Kinetics Study of the TIME Vs RATE of Chemical Change CHEMICAL KINETICS DEALS WITH

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Chemical Kinetics Study of the TIME Vs RATE of Chemical Change CHEMICAL KINETICS DEALS WITH Chemical Kinetics Study of the TIME vs RATE of Chemical Change CHEMICAL KINETICS DEALS WITH 1. How FAST {Speed like miles per hour} and 2. By what MECHANISM does a reaction happen? Part I How Fast (RATE) Does a Chemical Reaction “Go” ? What does the speed (RATE) depend upon? REACTION RATES • The change in the concentration of a reactant or product with time (M/s) for A → B ∆[A] ∆[B] Rate = − Rate = ∆t ∆t Rate disappearance = Rate of formation AFFECTS THE RATE 2A → B A disappears at twice the rate B forms Rate = - ½ ∆[A] = ∆[B] Rate disappearance ≠ Rate of formation For 2 N2O5 4 NO 2 + O 2 -7 If rate of decomposition of N 2O5 = 4.2 x 10 M/s What is the rate of APPEARANCE of (a) NO 2 ? Twice rate of decomposition = ________ M/s (b) O 2 ? ½ the rate of decomposition = _________ M/s How is the rate of the disappearance of the reactants related to the appearance of the products for CH 4(g) + 2O 2(g) →→→ CO 2(g) + 2H 2O( g) 1 ∆[CH ] 1 ∆[O ] 1 ∆[CO ] 1 ∆[H ] rate = − 4 = − 2 = 2 = 2 1 ∆t 2 ∆t 1 ∆t 2 ∆t Consider the reaction 3A --> 2B The average rate of appearance of B is given by [B]/ t. How is the average rate of appearance of B related to the average rate of disappear- ance of A? (a) -2[A]/3 t (b) 2[A]/3 t (c) -3[A]/2 t (d) [A]/ t (e) -[A]/ t Answer on Text Book’s Web Site 1. Nature of Reactants {fixed 2. Concentration of Reactants {variable 3. Temperature {variable 4. Etc. {variable • Catalysts • Particle Size • Photochemical The “things” that affect the rate of reaction are Considered one at a time I. NATURE OF REACTING SUBSTANCES Zn(s) + HCl(aq) → H2 + ZnCl 2(aq) Fast Reaction Iron “Rusts” but not very fast Fe(s) + O 2 → Slow Reaction Whereas Silver does not react at all Ag(s) + O 2 → No Reaction THE NATURE OF REACTANTS IS FIXED REACTANTS Reactant → Products R → P rate ∝ [R] A + B→ P + Q rate ∝ [A] [B] A + B + C → P + Q + Z rate ∝ ___________ RATE CONSTANT: k A constant of proportionality between the reaction rate and the concentration of reactants. R → P rate ∝ [R] rate = k [R] REACTION ORDER R → P To complete the equation Concentrations are raised to a power of x rate = k [R] x [ x is determined experimentally] Reaction order is determined experimentally Reaction order = to the sum of the powers to which all reactant concentrations in the rate law are raised REACTION ORDER • Zero Order Reactions • First Order Reactions • Second Order Reactions • Third Order Reactions RATE LAW • Rate Law: Shows the relationship of the rate of a reaction to the rate constant & the concentration of the reactants raised to some powers. • For the general reaction: aA + bB → cC + dD rate = k [A] x [B] y REACTION ORDER Order = x + y The sum of the powers to which all reactant concentrations appearing in the rate law are raised x & y MAY OR MAY NOT BE THE STOICHIOMETRIC COEFFICIENTS Reaction order is determined experimentally A reaction A + B C obeys the following rate law Rate = k[A] 2[B] (a) If [A] is doubled, how will the ______ rate change? (b) Will the rate constant change? ______ (c) What are the reaction orders for ______ A and B? (d) Overall order ? ______ (e) Units of rate constant ______ The reaction A + B C obeys the following rate law : Rate =k[A] 2[B] (a) What is the order of the reaction (i) With respect to A ? (a) 1 st (b) 2 nd (c) 3 rd (ii) With respect to B ? (a) 1 st (b) 2 nd (c) 3 rd (iii)Overall order (a) 1 st (b) 2 nd (c) 3 rd Rate =k[A] 2[B] (b) How does the rate change if (i) [A] is doubled ? (a) doubles (b) triples (c) quadruples (d) no change (ii) [B] is doubled ? (a) doubles (b) triples (c) quadruples (d) no change (iii) [C] is doubled ? (a) doubles (b) triples (c) quadruples (d) no change Method of Initial Rates Uses Initial Rates to Determine Rate Law Nitrogen monoxide reacts with hydrogen to form nitrogen gas and water 1st Write reaction NO( g) + H 2(g) → N2(g) + H 2O( g) 2nd balance reaction 2 NO( g) + 2 H 2(g) → N2(g) + 2 H 2O( g) 2 NO( g) + 2 H 2(g) → N2(g) + 2 H 2O( g) x y Write Rate Law: Rate = k [NO] [H 2] Determine x and y using method of Initial rates What can you measure INITIALLY ? ______ How many UNKNOWNS in Rate Law? _____ How many experiments are needed ? _____ 2 NO( g) + 2 H 2(g) → N2(g) + 2 H 2O( g) x y Rate = k [NO] [H 2] METHOD OF INITIAL RATES Experiment [NO] initial [H ] 2 initial Initial Rate (M/s) 1 5.0 x 10–3 2.0 x10–3 1.3 x 10–5 2 10.0 x 10–3 2.0 x 10–3 5.0 x 10–5 3 10.0 x 10–3 4.0 x 10–3 10.0 x 10–5 2 NO( g) + 2 H 2(g) → N2(g) + 2 H 2O( g x y Rate = k [NO] [H 2] Exp 1: 1.3 x 10 -5 = k (5.0x10 -3)x (2.0x10 -5)y Exp 2: 5.0 x 10 -5 = k (10x10 -3)x (2.0x10 -5)y Exp 3: 10.0 x 10 -5 = k (10x10 -3)x (4.0x10 -5)y Experiment [NO] initial [H ] 2 initial Initial Rate (M/s) 1 5.0 x 10–3 2.0 x10–3 1.3 x 10–5 2 10.0 x 10–3 2.0 x 10–3 5.0 x 10–5 3 10.0 x 10–3 4.0 x 10–3 10.0 x 10–5 2 NO( g) + 2 H 2(g) → N2(g) + 2 H 2O( g) x y Rate = k [NO] [H 2] Exp 1: 1.3 x 10 -5 = k (5.0x10 -3)x (2.0x10 -5)y Exp 2: 5.0 x 10 -5 = k (10x10 -3)x (2.0x10 -5)y Exp 3: 10.0 x 10 -5 = k (10x10 -3)x (4.0x10 -5)y Take RATIO of Exp 2 to Exp 1 exp 2 5.0 x 10 -5 k (10x10 -3 )x (2.0x10 -5 )y : : = exp 1 1.3 x 10 5- k (5.0x10 3- )x (2.0x10 5- )y 2 NO( g) + 2 H 2(g) → N2(g) + 2 H 2O( g) Exp 1: 1.3 x 10 -5 = k (5.0x10 -3)x (2.0x10 -5)y Exp 2: 5.0 x 10 -5 = k (10x10 -3)x (2.0x10 -5)y exp 2 5.0 x 10 -5 k (10x10 -3 )x (2.0x10 -5 )y : : = exp 1 1.3 x 10 5- k (5.0x10 3- )x (2.0x10 5- )y 5.0 x 10 -5 (10x10 -3 )x 3.9 = = = (2) x 1.3 x 10 5- (5.0x10 3- )x What is x ? (a) 0 (b) 1 (c) 2 (d) 3 2 NO( g) + 2 H 2(g) → N2(g) + 2 H 2O( g) Exp 1: 1.3 x 10 -5 = k (5.0x10 -3)x (2.0x10 -5)y Exp 2: 5.0 x 10 -5 = k (10x10 -3)x (2.0x10 -5)y Exp 3: 10.0 x 10 -5 = k (10x10 -3)x (4.0x10 -5)y Take RATIO of Exp 3 to Exp 2 to find y exp 3 10 x 10 -5 k (10x10 -3 )x (4.0x10 -5 )y : : = exp 2 5 x 10 5- k (10x10 3- )x (2.0x10 5- )y 2 NO( g) + 2 H 2(g) → N2(g) + 2 H 2O( g) Exp 2: 5.0 x 10 -5 = k (10x10 -3)x (2.0x10 -5)y Exp 3: 10.0 x 10 -5 = k (10x10 -3)x (4.0x10 -5)y exp 3 10 x 10 -5 k (10x10 -3 )x (4.0x10 -5 )y : : = exp 2 5 x 10 5- k (10x10 3- )x (2.0x10 5- )y 10 x 10 -5 (4.0x10 -3 )y 2 = = = (2) y 5 x 10 5- (2.0x10 3- )y What is y = ? (a) 0 (b) 1 (c) 2 (d) 3 METHOD OF INITIAL RATES 2 NO( g) + 2 H 2(g) → N2(g) + 2 H 2O( g) x y Rate = k [NO] [H 2] The order with respect to NO is 2nd order st The order with respect to H 2 is 1 order 2 1 Rate = k [NO] [H 2] The overall order of the reaction is 2 + 1 = 3 2 NO( g) + 2 H 2(g) → N2(g) + 2 H 2O( g) 2 1 Rate = k [NO] [H 2] What is the rate constant ? Exp 1: 1.3 x 10 -5 = k (5.0x10 -3)2 (2.0x10 -5)1 Exp 2: 5.0 x 10 -5 = k (10x10 -3)2 (2.0x10 -5)1 Exp 3: 10.0 x 10 -5 = k (10x10 -3)2 (4.0x10 -5)1 Solve for k from any of the equations k = 2.5 x 10 4 Units???? EXAMPLE 2 2– – 2– – S2O8 (aq ) + 3 I (aq )→→→ 2SO 4 (aq ) + I 3 (aq ) 2– X – Y Rate = k [S 2O8 ] [I ] Determine – the rate law – the order – and rate constant 2– – 2– – S2O8 (aq ) + 3I (aq ) →→→ 2SO 4 (aq ) + I 3 (aq ) 2– Experiment [S 2O8 ] [I–] Initial Rate (M/ s) 1 0.080 0.034 2.2 x 10–4 2 0.080 0.017 1.1 x 10–4 3 0.16 0.017 2.2 x 10–4 Rate 1 = __________________________ Rate 2 = _________________________ Rate 3 = __________________________ 2– Experiment [S 2O8 ] [I–] Initial Rate (M/ s) 1 0.080 0.034 2.2 x 10–4 2 0.080 0.017 1.1 x 10–4 3 0.16 0.017 2.2 x 10–4 Divide Rate 1 by Rate 2 to find Y Y R k [ 08.
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