CHEMICAL KINETICS Goal of kinetics experiment is to
- Slides: 30
CHEMICAL KINETICS Goal of kinetics experiment is to measure concentration of a species at particular time during a rxn so a rate law can be determined rates are obtained from concen. vs fct of time Rate of Rxn: describes how fast reactants used up & pdts formed Chem Kinetics: 1)study of rates, 2) factors that affect rxn rates, & 3) mechanisms (steps) by which rxns occur From a chem eqn, rate can be determined by following the Dconcen of any subst that is quantitatively detected
Order of rxn cannot be deduced from chemical eqn. of rxn Rate law expressions - calculate rate of rxn from rate constant & reactant concen - convert into eqn to determine concen of reactants @ any time 4 factors that affect chem rxns 1) nature of reactants 2) concen of reactants 3) temp 4) catalyst present Write rate law for rxn to describe how rate depends on concen. Rate Law is deduced experimentally from how its rate varies w/ concen
For rxn: A + B -----> pdts general form: rate = k[A]x[B]y k: rate constant exponents x & y - usually integers - value of x is the order of rxn w/ respect to A - y? ? Values for k, x, & y have no relation to coeff of balanced chem eqn. , remember, must be determined experimentally Exponent Define rate not depend on [reacts] 0 rate is directly proportional to [reacts] 1 rate is directly proportional to square of concen; [reacts]2 2 overall order of rxn = x + y
Examples of observed rate laws for following rxns 3 NO(g) ------> N 2 O(g) + NO 2 (g) rate = k[NO]2 order in NO: 2 nd 2 NO 2(g) + F 2(g) ------> 2 NO 2 F (g) rate = k[NO 2][F 2] order in NO 2: 1 st overall order: 2 nd order in F 21: st quick summary - order in rate law may not match coeff. in balanced eqn - no way to predict rxn orders overall from balanced eqn - orders must be determined experimentally
rate law can be determined by 2 methods: 1) Method of Initial Rates (if time) 2) using Integrated Rate Eqn
ZERO ORDER Has a rate which is independent of concentration of reactant(s), therefore, increasing concen. of rxning species not speed up rate A -----> pdts Rate is: Rate is a CONSTANT integration gives eqn called integrated zero-order rate law [A] = -kt + [A]o concentr of chemical @ particular time Initial concentr
eqn line: [A] = -kt + [A]o y = mx + b calculate k from plot of graph; straight line plot of [A] vs time, t; slope = -k [A] time, t
integration gives eqn called integrated zero-order rate law [A] = -kt + [A]o concentr of chemical @ particular time Initial concentr calculate k from plot of graph; straight line plot of [A] vs time, t; slope = -k Determine units:
If zero-order occurs, 1) closed system 2) no net build of intermediates 3) no competing rxns half-life describes time needed for half of reactant to be depleted
FIRST ORDER Depends on concentration of only 1 reactant, if other reactants present but each will be zero-order 1 st order rate constant, units of 1/time eqn for first-order reaction A -----> pdts rate is: know: integration gives eqn called integrated first-order rate law ln[A] = -kt + ln[A]o eqn line: y= mx + b
calculate k from plot of graph; plot of ln[A] vs time, t; gives straight line slope = -k Determine units: ln[A] time, t
half-life describes time needed for half of reactant to be depleted
second-order rate law SECOND ORDER A. A ------ > pdts depends on concentration of 2 nd-order reactant rate is: integrated in the form: [A]O @ t = 0 & [A] @ t: B. or, A + B = pdts two 1 st-order reactants: Another way to represents rate laws, take ln of both sides: ln r = ln k + 2 ln[A]
eqn line: Plot 1/[A] vs time, t; y= mx + b slope = 2 nd-order rate constant; +k 1/[A] time, t
Plot 1/[A] vs time, t; slope = 2 nd-order rate constant Determine units: half-life for 2 nd order dependent on one 2 nd order reactant:
FIRST ORDER REACTION 2 N 2 O 5 (aq) ----> 4 NO 2 (aq) + O 2 (g) DATA Time, s [N 2 O 5], M 0 600 1200 1800 2400 3000 3600 0. 0365 0. 0274 0. 0206 0. 0157 0. 0117 0. 00860 0. 00640 ln[N 2 O 5] - 3. 310 - 3. 597 - 3. 882 - 4. 154 - 4. 448 - 4. 756 - 5. 051 1/[N 2 O 5], M-1 27. 4 36. 5 48. 5 63. 7 85. 5 116 156
ln[N 2 O 5] time, s 1/[N 2 O 5] time, s rate = k[N 2 O 5] time, s
ln[N 2 O 5] time, s slope =
SECOND ORDER REACTION 2 NO 2 (g) ----> 2 NO (g) + O 2 (g) DATA Time, s 0 60 120 180 240 300 360 [N 2 O 5], M 0. 0100 0. 00683 0. 00518 0. 00418 0. 0350 0. 00301 0. 00264 ln[N 2 O 5] - 4. 605 - 4. 986 - 5. 263 - 5. 477 - 5. 655 - 5. 806 - 5. 937 1/[N 2 O 5], M-1 100 146 193 239 286 332 379
ln[NO 2] time, s 1/[NO 2] time, s rate = k[NO 2]2 time, s
1/[NO 2] time, s slope =
ZERO ORDER REACTION Can occur if: 1) rate limited by [catalyst] 2) photochemical rxn if rate determined by light intensity 3) most often occur when subst as a metal surface or enzyme required for rxn to occur 2 N 2 O (g) ----> 2 N 2 (g) + O 2 (g) N 2 O N 2 O N 2 O N 2 O N 2 O N 2 O Pt metal surface N 2 O N 2 O
N 2 O N 2 O N 2 O N 2 O N 2 O N 2 O N 2 O N 2 O Describe what is happening Rxn occurs on a hot Pt surface, when surface completely covered w/ N 2 O molecules, an increase of [N 2 O] has no effect on rate, since only N 2 O molecules on the surface are reacting. Therefore, the rate is constant because rsn is controlled by what happens on Pt surface rather than total [N 2 O].
[N 2 O] time, s rate = k[N 2 O]0
Determine: 0 ORDER 1 ST ORDER 2 ND ORDER UNITS HALF-LIFE
Summary for reaction orders 0, 1, 2, & n Zero-Order First-Order Rate Law Integrated Rate Law [A] = [A]Oe-kt [A] = [A]O - kt ln[A] = ln[A] - kt O Units of Rate Constant (k) Linear Plot to determine k y-intercept Half-life [A] vs t -k [A]O ln[A] vs t -k ln[A]O Second-Order nth-Order
NOTES Rate Rxn - describe rate rxn must determine concen of react/pdt at various times as rxn proceeds - devising methods is challenge for chemists -spectroscopic method: if 1 subst colored measure inc/dec in intensity of color 4 Factors: help control rates
METHOD OF INITIAL RATES Experiment 1 2 3 [A]O 1. 0 * 10 -2 M 2. 0 * 10 -2 M deduce rate law from experimental rate data [B]O 1. 0 * 10 -2 M 2. 0 * 10 -2 M 1. 0 * 10 -2 M initial rate 1. 5 * 10 -6 M. s-1 3. 0 * 10 -6 M. s-1 6. 0 * 10 -6 M. s-1 describing same rxn in each experiment, same rate law, form: rate = k[A]x[B]y Notice, [A]O same in #1 & #2, what would affect the rxn rate? Des in rxn rate due to diff initial concen of B Comparing the 2 experiments, [B] is Ded by factor of:
rate Des by factor of: Exponent y deduced from: rate ratio = ([B])y 2. 0 = (2. 0)y solving, y = 1 What order is rxn order in [B]? rate = k[A]x[B]1 Experiments 1 & 3 show [B]O same but [A]O different [A] is Ded by factor of: rate Des by factor of:
Exponent x deduced from: rate ratio = ([A])x 4. 0 = (2. 0)x solving, x = 2 What order is rxn order in [A]? rate = k[A]2[B]1 Rate constant, k, substitute data from any set of 3 sets into rate-law expression rate 1 = k[A]12[B]11 or, rate = 1. 5 M-2. s-1[A]2[B]
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