Question

e obtained for the de PROBLEMS The data in the following table (molar concentration of N,Os versus time) were obtained composition of dinitrogen pentoxide: 2 N,O3(g) → 4NO2(g) + O2(g) 10 1000 2000 8000 9000 15000 16000 20000 conc in M 0.75 0.65 0.57 0.24 0.21 0.092 0.080 0.048 00 0.092 16000 0.080 20000 0.046 time in s 21000 0.040 1. Determine--for example by making the appropriate graphs-whether this decomposition is zero, first, or second order with respect to N,05

Answer 1

Part 1 To Calculate the order of Heretion we have to fout the given values of concentration time in the rate equations of 0, 1st 2nd ouder reactions. Thow there we calculate of rate constant k. 1h the value of K remains fairly constant, then that is the order of Heaction > Cutting the values i tate equation of lost ouden. Henction Well know, you first order K = L In a Det t -2 Dese, the values of decomposition ie a-x are given. and 0.75 zs are initial concentration ki= I en a conc. of N205 left after ty Z a-X is giveng ki = ln 0.15 1000 I 0. 65 a ln 1.15 1000 I X0.13 1000 = 1.0.00013 |

K₂ = I ln 0.75 0.57 = 000 en 1.31 1 x0.27 fairly conštant This proves that the given reaction is a Frest Ouder Meaction The graph will lue as followus goro K2 = 0.00013 Kz K3 ´8000 m 1 in 0.75 0-24 kg = 0-0001.61 Ku = 1000 m 0.15 Bicone 9000 en 5:57 Ky < 0.00014] time K5 = 7 en 0.750 15000 0.092 5 I en 815 15000 2.09 15000 k5= 0.00013/ Similarly, when you calculate for kog kg etc Dan will see that the topizdité of K remains kart 2 The stoihiometuie coefficient of Marthe neaction is 2. 2N205 Eg] 4N02(g) +0209) laut sale = K[N205 This is because, the reaction o takes place in the following steps :-

+ NO N20 ki, Naz + NO3 (slow) NO2 + Naz K2, No 10 2 + No (part) NO + N20 K3, 3N02 (fast) N205 4NO2 + 0₂ (duerall reaction) Kince, Rate of Heaction is dēlermined by thi Kaviest vteß which here only depenas on the concentration of N2O5, Therefore r = K[N 2057 ie Fúst onder heition. Rate constant has been calculated in part 1 wang equation ks I en a su wasan t av It came out to be 0.00013 = 1.3810-4 Mints - The formula to calcutate units of k for any order of reaction is (moll) 1-ng- no 090104 of the Heatlion O CO here n=1 OT ON POCO F1

(mel (-) 'st (mol 2-1) 05-1 Bettina / x9 = 1 1xst = S-' Therefore k = 13 x 10-45-1/ Part 4 5N05 4Now to at time t=0 conc= 120 a 2 ť a-x at time to x i at t= 1000 a £ Britial cane ) - 0.75 | a-x [come of N205 lift after decomposition) = 0.65 X (Imount of Nails decampored) n = a- (a-x) 0.75 - 0.65 0.10 blimilarly, calculating x sou time ginen and plotting the values of (N205] [NO] [On? СО Reactant Procluct TON POCO F1

0.75 o 70 0.65 0.55 a Rate of formation of Products NO and New o so 0.45 - ( with time) 0-40 0.35 of 0:30 Rate of decomposition reactant Nos (bo with time) 0.25 e 0-20 0:15 0.10 + 2000 youo 6000 8000 10,070 12010 19000 16070 18000 2000 22 Time 09.03> Rate of dreamboortion of lN2067 o Rade of formation of (NO2) and [027