Question

1)

A proposed mechanism for the reaction 2NO(g) +2H2(g) → N2(g) + 2H2O(g): Step 1: 2NO(g) + N2O2(g) (very fast, reversible) Step 2: N2O2(g) + H2(g) N20(g) + H20(g) (slow) Step 3: N2O(g) + H2(g) →N2(g) +H2O(g) (fast) What is the rate law for the overall reaction? Ok[NO2 Ok[NO]2[H212 O k[NO]2[H2] Ok[N20][H2] k[NO]1/2[H2]

2)

3)

Answer 1

Solution:

Given Reaction? 2 Nig+ 2H₂(g) N₂ Cg] + 2t (g) Step-1 2 No Cg) Ri Ng O₂ (g) every fast) reversible, Step-2 N, O, (g) by Hocg) + NO(g) (slow) F + H₂ Cg) Step 3 N₂O(g) + H₂ CG) R3 N₂ CgJ + H₂Ocg). (fast.) Let's take Rate Constant ki, Rz, Rz for step 1, 2 and 3.

ETI # when a reaction occurs in Steps, Rate Low is Pounded for slowest step. As slowest step, N₂O₂ (g) + H₂(g) N₂ Ocg) + H₂O(g) (slow] So, Rate = k [NO] [H] - where, h= Rate Constant [ J = Concent ration Now, as this Rate is not given in options. So, we need to apply steady state approximation on NO₂ Cq) in given steps ob reaction.d[N ] = o. =) d[N20₂] = R, [NOJ? (for step is det al = - d [N 2] = K₂ [N, Og] [Hz]. (for dt - step 2) . Now, applying steady state approximation k [NO] - K₂ [NO] [hy] =0 k [NO] ² = [NO] Putting in ① R₂ [H₂] = So, Rate = kx Ri [NO] [H] = kj [NO]2 18 [HT :CR = R rate Constants

Rate constant for overall reaction has found to be (a) all other options are not possible as we can see in the given picture.

So, (a) Rate= k[NO]2

As per Chegg's policy I've solved only first question.

Hope, it helps. :)