Answer Key, Problem Set 1, Written

Transcription

1 Cheistry 1 Mies, Sprig, 018 Aswer Key, Proble Set 1, Writte ; (add part (e): Estiate / calculate the iitial rate of the reactio); 3. NT1; 4. NT; ; ; ; 8. NT3; ; 10. NT Rates of Foratio / Loss (ad Relatio to Coefficiets), Average & Istataeous Rates, ad Rates of Reactio Cosider the reactio (represeted by): NO (g) NO(g) + The followig data were collected for the cocetratio of NO as a fuctio of tie: 1 O (g) Tie (s) [NO (M) (a) What is the average rate of the reactio betwee 10 ad 0 s? Betwee 50 ad 60 s? (b) What is the rate of foratio of O betwee 50 ad 60 s? Aswers: (a) M s ad M s ; (b) M s Reasoig: Avg.Rateofeactio1R[NO [NO (a) here because the coefficiet of NO is 1. Thus, 1 t t [NO f [NO i M 0.951M M Fro 10 s to 0 s, Avg. Rate M t t 0 s 10 s 10. s Fro 50 s to 60 s, Avg.Rate [NO t s f i [NO t M M M 60 s 50 s 10. s f i M s f i (b) R(for, of O ) = ½ R(loss of NO ) because of the coefficiets i the equatio (1 : ½ for NO : O). Thus, the (average) rate of foratio of O betwee 50 s ad 60 s ½ x M s (fro (a)) M s Cosider the reactio (represeted by): H O (aq) H O (l) + O (g) The graph below (to the right i this key) shows the cocetratio of H O as a fuctio of tie. Use the graph to calculate each quatity: (a) The average rate of the reactio betwee 10 ad 0 s. (b) The istataeous rate of the reactio at 30 s. (c) The istataeous rate of foratio of O at 50 s. (d) If the iitial volue of the H O is 1.5 L, what total aout of O (i oles) is fored i the first 50 s of reactio? (e) [added Estiate / calculate the iitial rate of the reactio. Aswers: (y results): (a) M s ; (b) M s ; (c) M s ; (d) 0.59 ol O; (e) M s (a) Strategy/Explaatio: [The average rate of the reactio betwee 10 ad 0 s. Avg. rate of reactio equals the rate of loss of a reactat (or rate of foratio of a product) divided by its coefficiet i the balaced cheical equatio (see Eq i Tro, or hadout). Sice we have data for HO, the appropriate expressio for the rate of reactio is: Rate of rx 1 [H O t because the coefficiet of HO is a. So, PS1

2 0.01) Fid the values of [HO at t = 10 ad 0 s. Aswer Key, Proble Set 1 1 [H O ) Calculate t Executio of Strategy: Note: I scaed the plot ad elarged it i this key for better precisio. At 10. s, [HO 0.75 M; at 0. s, [HO 0.56 M (see plot above; poits added by e). Thus, 5Ms1 [HO M M Rate (of rx) t s 10. s (b) Strategy/Explaatio: [The istataeous rate of the reactio at 30 s. Istataeous rate eas the rate at a sigle poit i tie, rather tha over a tie iterval. The istataeous rate of loss of HO ca be estiated by fidig the slope of the taget to the curve for [HO at a poit i tie ad the takig its opposite (to ake the value positive). Sice this questio asks for istataeous rate of reactio, oe ust divide the rate of loss of HO by the coefficiet of HO i the balaced equatio (see (a)). So i this proble oe ust: 1) Draw a taget to the curve at the poit represetig t = 30 s; ad ) Fid the slope of the taget lie by pickig ay two poits o the lie ad usig the defiitio of y [HO slope which here eas. x t 1 [HO 3) Multiply the result i () by -½ to get rate of reactio i.e., i the ed, deterie t Executio of Strategy: 1) I created a taget lie o the plot below at t = 30 s by adjustig a lie to ake it go through the poit o the curve at t = 30 s, ad also have a slope equal to the slope of the curve at 30 s (see short-dashed lie o plot): 09PS1-

3 0.0Aswer Key, Proble Set 1 ) ad 3): Lookig at the dotted lie ad pickig as y two poits, the itercepts (just for coveiece you could use ay two poits), I get the poits (0 s, 0.78 M) ad (6 s, 0 M). Thus: 3Ms1 [HO M M Rate (of rx at 30. s) M s t s 6. s 06Ucertaity probably about M s due to slope estiatio (c) Strategy/Explaatio: [The istataeous rate of foratio of O at 50 s. This is siilar to (b) except that the rate of foratio of O is asked for (ot rate of reactio ). However, sice the coefficiet of O is 1, the rate of foratio of O equals the rate of reactio i this proble! Thus, you ca actually do exactly the sae thig you did i (b) [i.e., fid the slope of the plot for HO ad take ½ of that except do it at the poit correspodig to t = 50 s. Executio of Strategy: 1) I created what I believe to be a reasoably good taget lie to the curve at t = 50 s (log-dashed lie o plot below). PS1-3

4 0.0s-Aswer Key, Proble Set 1 ) ad 3): Lookig at the log-dashed lie, the y-itercept is (0 s, 0.55 M), ad aother poit is (60. s, 0.15 M) [I did t pick the x-itercept here because it was slightly off y scale. Thus: 3M11 [O 1 [HO M M M Rate (of O foratio) 1 t (d) Strategy/Explaatio: s t s 60. s [If the iitial volue of the H O is 1.5 L, what total aout of O (i oles) is fored i the first 50 s of reactio? First of all, the proble s wordig is a bit tricky, sice it says the iitial volue is 1.5 L, perhaps iplyig that the volue ight chage with tie. Hopefully you realize(d) that the volue of the cotaier is just 1.5 L ad this does ot chage durig the reactio. As such, if you step back ad really aalyze this proble, hopefully you will recogize that it is actually just a stoichioetry proble (it has othig to do with rate really)! You are asked for oles of O fored i the first 50 s, ad you are effectively give the cocetratio of HO lost durig this sae tie iterval (because you ca get this fro the graph by deteriig [HO50 s [HO0 s). So, how do you relate the oles of O fored to the olar of HO lost (whe you kow the volue is 1.5 L)? There are two basic ways: Oe way: olarity of HO ([H O ) oles of HO oles of O I this approach, the first step ivolves ultiplyig M by V (i L) to get oles. The secod step ivolves 1ol O ultiplyig by a ole ratio fro the balaced equatio. ol HO d way: olarity of HO ([H O ) olarity of O ([O ) oles of O I this approach, the first step ivolves ultiplyig by a ole ratio fro the balaced equatio 1ol O [You could do this i a ICF type of table, or ot. The secod step would ivolve ol HO ultiplyig the resultig M (i.e., [O fored) by V (i L) to get oles (of O fored). 03Executio of Strategy: [HO50 s 0. M; [HO0 s 1.00 M (see plot below [poits added by e). 1 st Approach: PS1-4

5 0.0Aswer Key, Proble Set 1 Thus, [HO (i 1 st 50 s) 0. M 1.00 M = M 0.78 M HO was lost (i 1 st 50 s) 0.78 ol H O L 1ol O fored x 1.5 L x ol O (fored i 1 st 50 s) ol H O lost d Approach (with ICF table): H O (aq) O (g) + H O (l) [I swapped the order o the right for coveiece oly [HO (M) [O (M) [HO (M) Iitial 1.00 (ot kow/eeded) (ot kow/eeded) Chage Fial (t = 50 s) = [HO (1 st 50 s) 0. M 1.00 M = M HO 0.9ol0.78 ol H O 1ol O fored O fored x Lx L ol H O lost - ½ (-0.78) = 0.39 M (ot asked for) 0. (ot asked for) (ot kow/eeded) L ol 0.59 ol O (fored i 1 st 50 s) (e) Strategy/Explaatio: [Estiate / calculate the iitial rate of the reactio. Iitial rate eas the istataeous rate at t = 0. Thus, do the sae thig as i (b) except at t = 0 s. Executio of Strategy: I drew a approxiate taget lie to the curve at t = 0 s (dot-dash lie; see below). Lookig at the dot-dash lie, the y-itercept is (0 s, 1.00 M), ad the x-itercept is (36. s, 0. M) Thus: 14Ms1 [HO M M Iitial Rate of Rx M s t s 36. s Ucertaity probably about M s due to slope estiatio PS1-5

6 Aswer Key, Proble Set 1 3. NT1. Cosider a reactio represeted by the eq aa + bb cc ad the followig average rate data over soe tie period t: A M s M s t B C M s t t Deterie the coefficiets (a, b, ad c) i the stadard balaced cheical equatio (i.e., lowest whole # s) Aswer: A + 3 B 4 C (i.e., a = ; b = 3; c = 4) Explaatio: The ratio of coefficiets i a balaced cheical equatio ust be equal to the ratio of chages i ole values (i absolute value) of reactats ad/or products durig a cheical reactio. This is essetially the eaig of the coefficiets i a balaced equatio. Assuig a fixed volue ad a fixed tie iterval (t), the rate of chage of a species is directly proportioal to chage i cocetratio ad thus chage i oles: Rate of chage of X (M/s) [X (M) oles of X (ol) This eas that the ratio of coefficiets (= ratio of oles) also equals the ratio of rates of chage. For this proble, the, sice : : equals 8 : 1 : 16 (divide all values by ) which equals : 3 : 4, the set of lowest whole uber coefficiets would be, 3, ad 4. (Differetial) Rate Laws / Meaig of Orders (i additio to ideas fro the previous sectio) 4. NT; The reactio represeted by the followig equatio was foud to have the rate law show. N O 5 (g) 4 NO (g) + O (g); Rate Law: Rate = k[n O 5 (g) (a) What is the order of the reactio with respect to N O 5? Aswer: First order (because the expoet of [NO5(g) i the rate law is 1). (b) How does the rate of foratio of O copare with the rate of loss of N O 5? (State the correct aswer fro aog the stateets below ad the explai your aswer). (i) O fors half as fast as N O 5 is lost. (ii) O fors twice as fast as N O 5 is lost. (iii) O fors four ties as fast as N O 5 is lost. (iv) It depeds o how uch N O 5 you start with. [O i.e., t 1 [NO t 5 Explaatio: Sice the aout of oles of O produced is always half the aout of oles NO5 used up (based o the : 1 ratio of NO5 : O i the balaced equatio), the rate of foratio of O ust be half the value of the rate of loss of NO5. (c) A specific trial of the above reactio is carried out, ad the cocetratio vs. tie plot of oe of the species ivolved is plotted below. Idetify which species is depicted i the plot below, ad the draw curves represetig the other two species ivolved i the reactio (copy the plot oto your paper, ad be careful to be reasoably quatitatively accurate!) Give brief reasoig NO Cocetratio Cocetratio N O 5 O Tie Tie (oe possible) Aswer Explaatio: The balaced cheical equatio is: NO5(g) 4 NO(g) + O(g) PS1-6

7 Aswer Key, Proble Set 1 What does this ea? It eas that for every two oles of NO5(g) that udergo reactio, four oles (i.e., twice as ay) of NO ad oe ole (i.e., half as ay) of O will be fored. Thus, the liitig value for the aout of NO produced (i.e., the value whe essetially all of the NO5 has reacted) is twice the iitial value of NO5, whereas the liitig value for the aout of O produced is half the iitial value of NO5. Note that the products cocetratios should INCREASE with tie sice they are beig produced; the reactat s cocetratio should DECREASE with tie sice it is beig used up. If you assue that you started with o products at t 0, the plots for NO ad O would start at the origi, ad the graph would look soethig like what is show i the aswer box (see above, right). Note that the slope at ay poit i tie (o a plot of cocetratio vs. tie) represets the rate of chage of that species cocetratio at that tie. So this is yet aother way to see that at ay poit i tie durig this reactio, the rate of icrease of NO is 4 ties the value of the rate of icrease of O, ad the rate of foratio of O is half the value of the rate of loss of NO5 (see part (b) above). NOTE #: There is really o eed to assue that you start without ay products preset (although that is what is ofte doe, for coveiece, whe we begi the discussio of kietics). I the ext chapter we will see ay probles i which soe products are preset at the start of soe reactio. So IF you were to assue that you started with [NO 0 = ½ [N O 5 0 ad [O 0 = 1.5 x [N O 5 0, the plot would look as show below (all curves have the sae exact shape ad size as above, but those for NO ad O are just oved up relative to where they were i the prior plot): Aother possible correct aswer NO O Obviously, you could have started with ay iitial cocetratios of NO or O, so I ca t write all of the possible correct aswers here! Cocetratio N O 5 Tie What are the uits of k for each type of reactio? (a) first-order reactio s (b) secod-order reactio M s Aswers (c) zero-order reactio M s Reasoig / Proof Write a exaple rate law for each case ad the solve for k with uits oly (o ubers): 1 st order: R k[a R M s k i ters uits of [A M 1 s 1 d order: R k[a R k [A i ters of uits M s M 1 s M 1 M 1 s 1 0 th order: R k[a 0 k k R i ters of uits M s A reactio i which A, B, ad C react to for products is first order i A, secod order i B, ad zero order i C. (a) Write a rate law for the reactio. (b) What is the overall order of the reactio? (c) By what factor does the reactio rate chage if [A is doubled (ad the other reactat cocetratios are held costat)? PS1-7

8 Aswer Key, Proble Set 1 (d) By what factor does the reactio rate chage if [B is doubled (ad the other reactat cocetratios are held costat)? (e) By what factor does the reactio rate chage if [C is doubled (ad the other reactat cocetratios are held costat)? (f) By what factor does the reactio rate chage if the cocetratios of all three reactats are doubled? Aswers: (a) R = k[a[b [C 0 (or just R = k[a[b ) (b) 3 (c) factor of (d) factor of 4 (e) factor of 1 (o chage) (f) factor of 8 Explaatios: (a) A order is a expoet of a cocetratio ter. If the reactio is first order i A, that eas that the expoet of [A i the rate law is a 1. If it is secod order i B, that eas that the expoet of [B i the rate law is a. If it is zero order i C, that eas that the expoet of [C is zero (but sice aythig to the zero power is oe, that ter ca be left off etirely). (b) Overall order su of idividual orders (c) Whatever factor of chage there is i the cocetratio, that value raised to the order is the factor of chage i the rate. Whe the order is 1, the factor of chages are equal, so if cocetratio doubles, rate doubles. Proof : R k[a 1. Let the first [A be [A(1); the after the cocetratio is doubled, [A() x [A(1) R(1) k[a(1), ad after doublig the cocetratio, R() k ( x [A(1)) 1 x k[a(1) x R(1). The last atheatical equatio eas Rate after doublig [A = ties the origial Rate (d) Whatever factor of chage there is i the cocetratio, that value raised to the order is the factor of chage i the rate. Whe the order is, the factor of chage of the rate equals the square of the factor of chage of the cocetratio, so if cocetratio doubles, rate quadruples. Proof : R k[b. Let the first [B be [B(1); the after the cocetratio is doubled, [B() x [B(1) R(1) k[b(1), ad after doublig the cocetratio, R() k ( x [B(1)) x k[b(1) x R(1). The last atheatical equatio eas Rate after doublig [B = 4 ties the origial Rate (e) If a reactio is zeroth order i C, that eas that the rate is idepedet of [C. I.e., if you double [C, the rate reais the sae. (f) If ore tha oe chage i cocetratio occurs, the factor chage i the rate is the product of the factors of chages that would have occurred if each chage had occurred o its ow. Proof : R k[a[b. Let the first [A be [A(1); the after the cocetratio is doubled, [A() x [A(1). Let the first [B be [B(1); the after the cocetratio is doubled, [B() x [B(1) R(1) k[a(1) [B(1), ad after doublig both cocetratios, R() k( x [A(1)) ( x [B(1)) x k[a(1)[b(1) x R(1) 8 x R(1) The last atheatical equatio eas Rate after doublig [A ad [B = 8 ties the origial Rate (Note: I left off [C here for siplicity. We kow that the rate is uaffected by a chage i [C because it is zero order i C.) Deteriig (Differetial) Rate Laws / Method of Iitial Rates Cosider the data below showig the iitial rate of a reactio (A products) at several differet cocetratios of A. (a) What is the order of the reactio? (b) Write a rate law for the reactio icludig the value of the rate costat, k. Experiet (trial) [A0 (M) Iitial Rate (M/s) PS1-8

9 Aswer Key, Proble Set Aswers: order is ; R = k[a, where k = 5.3 M s or 5.5 M s Reasoig: To fid the order: Verbal: Goig fro Trial 1 to Trial, the (iitial) [A doubles (0.100 M to 0.00 M, ad the (iitial) rate quadruples (0.053 to 0.10 M/s; 0.10 / ). That eas the order of A ust be ( 4 = ). This ca be checked / cofired by otig that i goig fro Trial 1 to Trial 3, the [A triples, ad the rate goes up by / = ties, which equals 3. Thus (agai), the order is. Brute Force Approach: Set up the rate law: R = k[a (1) Substitute values fro Trials 1 ad ito the rate law (1) above: Trial 1: M/s k(0.100 M) Trial : 0.10 M/s k(0.00 M) Now look at the ratio of rates ad the substitute ito the equatio usig the right-had side expressios for the rates (fro above): (Note: Although ot ecessary, I prefer to divide the equatio havig the larger rate i it by the oe with the saller rate sice I fid it easier to work with values greater tha oe rather tha fractios.) (ote the cacellatio of the k s, ad uits i the uerator ad deoiator of both sides) So, dividig Trial s equatio by Trial 1 s, you get: R() 0.10 M/s k(0.00 M) R(1) M/s k(0.100 M) To fid k: Use ay trial ad just substitute i the values for iitial rate ad [A. Now that the order (, here) is kow, k will be the oly ukow! Trial 1: M/s k(0.100 M) k M/s (0.100 M) M s M s M M 1 s 1 As a check: Trial : : 0.10 M/s k(0.00 M) k 0.10 M/s (0.00 M) 0.10 M s M s M M 1 s 1 (NOTE: These two values are cosistet with oe aother. The best thig to do i priciple [e.g., if this were a actual set of experiets would be to calculate all three k s ad the average the [this is what you will do for Experiets 0 ad 1 i lab, but I will ot do that here. O a exa, for a siilar proble, you ay choose ay ONE trial ad leave it at that.) PS1-9

10 Aswer Key, Proble Set 1 8. NT3. The followig data were deteried for the reactio represeted by the equatio: ClO (aq) + OH - - (aq) ClO 3 (aq) + - ClO (aq) + H O(l) The teperature was the sae i all trials. Trial [ClO0 (M) [OH - 0 (M) Iitial Rate (M/s) x x x 10 - (a) Deterie the rate law (usig k to represet the rate costat) (b) Deterie the (value ad uits of the) rate costat (at the teperature at which these data were collected). (c) What would be the iitial rate for a trial with [ClO 0 = M ad [OH - 0 = M (at the sae T)? Aswers: (a) R k[clo [OH - ; (b) k 115 M - s ; (c) 0.98 M/s Reasoig: The rate law will be of the for: Rate = k[clo [OH - (1) To fid the orders ( ad ): Verbal: Goig fro Trial 1 to Trial, the (iitial) [ClO doubles (0.05 M to 0.10 M) while the (iitial) [OH - is kept the sae, ad the (iitial) rate goes up by (1.15 x 10 M/s)/(.88 x 10 - M/s) ties. That eas the order of ClO ust be. I goig fro Trial 3 to Trial, the [OH - doubles while the [ClO is kept the sae, ad the rate also doubles (1.15 x 10 M/s divided by 5.75 x ). That eas the order of OH - ust be 1. Brute Force Approach: R() 1.15 x 10 R(1).88 x 10 - M/s k(0.100 M) (0.100 M) M/s k( M) (0.100 M) ~ 4.00 R() 1.15 x 10 R(3) 5.75 x 10 - M/s k(0.100 M) (0.100 M) M/s k(0.100 M) ( M) Substitutig ito (1) with the two (ow kow) orders yields: (a) R k[clo [OH - (b) To get the value of the rate costat, k, pick a trial ad substitute i the values of R ad cocetratios (alog with the values of the orders). Here, I ll use Trial 1: -.88 x 10 - Ms = k( M) (0.100 M) 1.88 x 10 M s k ( M) (0.100 M) M s M ( M) ( M) (c) To get the iitial rate for a experiet with [ClO0 = M ad [OH - 0 = M, just substitute ito the rate law (with k ow kow): R = (115. M - s )[ClO [OH - = (115. M - s ) (0.175 M) ( M) - = = 0.98 M/s s 1 PS10

11 Aswer Key, Proble Set The data below were collected for a reactio represeted by this equatio: CH 3 Cl(g) + 3 Cl (g) CCl 4 (g) + 3 HCl(g) Trial [CH3Cl0 (M) [Cl0 (M) Iitial Rate (M/s) Brute Force Approach: Aswers: (a) R = k[ch3cl[cl 1/ ; (a) Write a expressio for the reactio rate law ad calculate the value of the rate costat, k. (b) What is the overall order of the reactio? k = 1.9 M/ s The rate law will be of the for: (b) 1.5 Rate = k[ch3cl [Cl (1) (a) To fid the orders ( ad ): Verbal: Goig fro Trial 1 to Trial, the (iitial) [CH3Cl doubles (0.050 M to M) while the (iitial) [Cl is kept the sae, ad the (iitial) rate goes up by (0.09 M/s)/(0.014 M/s).07 ties. That eas the order of CH3Cl ust be 1. I goig fro Trial to Trial 3, the [Cl doubles while the [CH3Cl is kept the sae, ad the rate goes up by (0.041 M/s divided by ties. If oe 1 otices that 1.41 is very close to it is clear that the order of Cl ust be ½. R() R(1) 0.09 M/s M/s k(0.100 M) (0.050 M) k( M) (0.050 M) ~.00 R(3) R() 0.041M/s 0.09 M/s k(0.100 M) (0.100 M) k(0.100 M) ( M) *As will be show i lab, oe ca also solve for as follows: (see below*) l(1.41) l(.00) l(1.41) l(.00) l(1.41) x l(.00) Substitutig ito (1) with the two (ow kow) orders yields: (a) R = k[ch3cl[cl 1/ To get the value of the rate costat, k, pick a trial ad substitute i the values of R ad cocetratios (alog with the values of the orders). Here, I ll use Trial 4 (just to get 3 SF s): Ms = k(0.00 M)(0.00 M) 1/ M s k 1/ (0.00 M)(0.00 M) M s ( M)( M) 1/ 1.9 M / s 1 (b) The overall order is 1 + ½ = 1.5 PS11

12 Aswer Key, Proble Set 1 (Differetial) Rate Laws / Coceptual Ideas 10. NT4. State whether the stateet is true or false AND IF FALSE, CHANGE A FEW WORDS TO MAKE IT CORRECT. Note: I will ask a few questios like this o each of y exas! YOU MUST CORRECT ANY FALSE STATEMENT CORRECTLY TO RECEIVE CREDIT ON AN EXAM FOR THESE! (a) It is possible to chage the rate costat for a reactio by chagig the teperature. TRUE (ore discussio about this durig Week ). (b) The reactio rate reais costat as a first-order reactio proceeds at a costat teperature. FALSE. The reactio rate will cotiuously decrease as a reactio proceeds because the cocetratios of reactats will cotiuously decrease. It is the rate costat that reais costat durig ay cheical reactio as log as the T is kept costat. Thus, two good ways to correct this stateet are: decreases The reactio rate reais costat as a first-order reactio proceeds at a costat teperature. OR (c) The rate costat for a reactio is idepedet of reactat cocetratios. OR here s yet a third way: chage first to zero! costat The reactio rate reais costat as a first-order reactio proceeds at a costat teperature. TRUE. Notice that ost of these questios are forcig you to distiguish betwee rate ad rate costat. It is also iportat for you to distiguish both of these fro rate law. The rate of reactio varies with cocetratios, but the rate costat does ot vary with cocetratios! ================================ END OF SET ============================== Additioal probles which ay or ay ot have bee assiged i Masterig Cosider the reactio (represeted by): N O(g) N (g) + O (g) (a) Express the rate of the reactio i ters of the chage i cocetratio of each of the reactats ad products. (b) I the first 15.0 s of the reactio, ol of O is produced i a reactio vessel with a volue of L. What is the average rate of the reactio over this tie iterval? (c) Predict the rate of chage i the cocetratio of N O over this tie iterval. I other words, what is Δ[N O? tδ1 [NO 1 [N [O Aswers: (a) Rate of reactio t t t (b) M s (over the 1 st 15 s) (c) M s Explaatio/Reasoig: NOTE: The aswers here are those for the values give i the textbook proble. I Masterig, the give values ay be differet, which would obviously affect the fial aswers. Be careful! (a) Sice the rates of loss ad foratio of the reactats ad products are proportioal to the coefficiets i the balaced equatio (i.e., these rates are ot all equal), a sigle rate of reactio will be obtaied if PS1

13 Aswer Key, Proble Set 1 each rate of loss or foratio is divided by the species coefficiet i the balaced equatio. That is why 1 [X 1 [ X rate of reactio is defied as if X is a reactat, ad if X is a product (with coefficiet x). x t x t (b) Average rate of reactio (expressed i ters of O ) (c) ol O [O (over st 15 ) L [O t 1 s MO [b/c M oles / L Thus, 1 [O M M t 15.0 s [N O t s [O t [N O t [O t s (0.000 M s ) M I.e., the rate of chage of NO is twice the rate of foratio of O, which ca be see fro the :1 ratio of NO lost : O fored i the balaced equatio. The egative sig is eeded here because rate of chage of NO was asked for rather tha rate of loss (ad rates of loss are always the opposite of rates of foratio). PS13