Question

2. Consider the reaction below between nitric oxide and hydrogen gas: 2 NO(g) + 2 H2(g) - N2 + 2 H2009) Upon being studied experimentally, it has been determined that the rate law expression for this reaction is first order in Hz and second order in NO(g). With this information in mind, answer the following: [Tip: You can assume that the steady-state approximation is valid for any intermediate here.) a. As written, is the reaction above a valid mechanism? In other words, is it consistent with the experimentally determined rate law? b. Is the proposed mechanism below consistent with the experimentally determined rate law? Explain H2(g) + 2 NO(g) - N20(g) + H20() H2(g) + N2009) N2(g) + H2009) c. Prove that the proposed mechanism below is consistent with the experimentally determine rate law. Note: For the first reaction in the mechanism, consider the forward reaction as having a rate constant k, and the reverse as having k-1. Also, the notation saying that this reaction is "fast" simply means that the equilibrium is achieved quickly, therefore the equilibrium is maintained throughout the mechanism and you can use the preequilibrium approximation. 2 NO(g) = N202(g) (fast) H2(g) + N2026) - N20(g) + H2009) H2(9) + N2069) N2(9) + H2O(g) Page 2 of 2

Answer 1

Given data, The Deaction between hitoric Oxide and hydoogen gas. 2 No C) + 2Ha (9) > pi(a) targo (3) a) Elementary Deactions are those Deactions in which there is Corses pondence between Stoichiometry of the reaction and Date Expression. Accordingly, the above Deaction. . hehe The rate of Deaction: which is different from Elence the reaction experimental result. is not elementary and is not bi The proposed mechanism below consistent with the experimentally determined Bate law H2(g) + 2NO (9) K Nao (g) + H20 (g) Ha (a) + Ngo (9) ka, na(3) + Hyo (3) The first reaction with state Constant ki and The Second reaction with rate Constant g..

Rate of foomation of intermediate is 020 (3) Det Date = 0 It is formed from 1st reaction and decomposed in the Second reaction, Hence. K. (Ha)(roja- Kalwa][ta] = 0 [@go] =(k, /K2) [0032 The date of reaction is -k,[moja (Ha] + K3[Hz] [20] - [Ha][k, wo]z +Ky xk (20) K2] =[Halak, (voda c) The proposed mechanism below is consistent with the Experimentally determine date law. a oo (3)= 3 (3) fast H2 (8) +230g (9). K2, pao(3)-4 Hao (3) 4 (3)-oạo () Đ (3)+1go (3) The the first reaction mechanism is 2 No wao It is fast.

and K, [no] 2 = K.; [@202] ki is forward rate Constant kias backward date Constant [dava] - CK, /k_)(oo)a = K[oo], 0) Where Hoo is intermediate Hence net date=0. It is formed from reaction - and decomposed from geaction -3. wel date = ka[ha] (na0a] - Ks [ha wao]=0, Kalta] [2202] = k3ſt6[220] -4 Rate of decomposition of the is K2 [Hz][2009] + K3 (Ha][220] - The rate [Ha] [K> [map] + K3 (220) = a*a[2008] [4] Rate akak[w0Ja C#a]. offence. The proposed Date law . mechanism Explain the