Consider the following data for the decomposition of nitrous oxide (N20) to N2 a

Question

Consider the following data for the decomposition of nitrous oxide (N20) to N2 and O2 gas at 273 K: [N20] Initial Rate Experiment 0.359 1.077 2.513 (M/min) 0.00213 0.0191 0.104 2 Mechanism 1 NO N2+O fast equilibrium slow Mechanism 2 2 N N402 fast equilibrium slow 4 2 Mechanism 3 2 N° N402 fast equilibrium slow Mechanism 4 N2O + N2O 2 N2 + O + O slow fast equilibrium Plausible Mechanism 2 Plausible Mechanism 3 Plausible' Mechanism 1 Not valid Mechanism 4 Submit Answer Incorrect. Tries 2/3 Previous Tries

Explanation / Answer

We know slowest step is always the rate determining step.

So now check only slowest step. Such that the slow step constains (N2O) beacause from table rate is changing with chenge in N2O concentration.

From this info, only Mechanism 1 and Mechanism-4 are possible since their slow step contains N2O

Now,

let consider rate = -k (N2O)m

0.00213 = k (0.359 )m ...(1)

0.0191 = k (1.077)m .... (2)

Divide equation (2) with (1)

0.0191/0.00213 = (1.077/0.359)m

9 = 3m

m=2

The power of N2O in mechanism -4 is equal to 2.

So correct answer is mechanism-4

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