Question

In: Chemistry

1. Consider the following reaction: 2NO2 + F2  2NO2F Calculate the average rate of formation of NO2F

 

1. Consider the following reaction: 2NO2 + F2  2NO2F Calculate the average rate of formation of NO2F (Δ[NO2F]/Δt) between 410.5 seconds and 679.2 seconds after mixing of the reactants given the following information. (Sample Exercise 13.2) (3 pts) Time [F2] 410.5 s 1.79 x 10-2 M 679.2 s 9.73 x 10-3 M

2. Consider the following generic reaction: 2A + B + C  2D + 3E From the following initial rate determinations, write the rate law for the reaction, determine the overall order for the reaction and calculate the rate constant, k. (Sample exercise 13.3) (7 pts) Initial rate [A] [B] [C] 1.27 x 10-4 M s-1 0.0125 M 0.0125 M 0.0125 M 2.56 x 10-4 M s-1 0.0250 M 0.0125 M 0.0125 M 1.27 x 10-4 M s-1 0.0125 M 0.0250 M 0.0125 M 5.06 x 10-4 M s-1 0.0125 M 0.0125 M 0.0250 M

3. Consider the following reaction and its rate law: 2N2O5  4NO2 + O2; rate = k[N2O5] At 25oC, the half-life for this reaction is 96.3 minutes; calculate the initial N2O5 concentration, [N2O5]o, if [N2O5] = 1.80 x 10-2 M after 60.0 minutes when the reaction proceeds at 25oC.

Solutions

Expert Solution

Question 1.

First, let us anlayse the first reaction:

2NO2 + F2 = 2NO2F

1 mol of F2 will react to form 2 mol of NO2F,

Our analysis must be perfored via F2 concentration FIRST

then, let us first find rate of reaction for F2, then relate via stocihometry to the product, NO2F

the definition of average rate or reaction ( AVG)

AVG Rate = dC/dt

where, dC is the change in concentration and dt the change in time

AVG RAte = (Cfinal - Cinitial)/(tfinal - Tinitial)

Tfinal - Tinitial = 679.2 - 410.5 = 268.7 s

Cfinal - Cinitial = (9.73*10^-3 - 1.79*10^-2) = -0.00817 M of F2 ( Very important to note that this is reacting!)

note that this is only for F2, we must relate via stoichiometry

relate via:

1 mol of F2 reacts to form 2 mol of NO2F (product)

Then, the rate can be related as follows:

dNO2F/dt = -2*dF2/dt

substitute known data

dNO2F/dt = -2*(-0.00817)/(268.7) = 0.0000608113 M/s of NO2F

this is positive, since it is the rate of FORMATIO/PRODUCTION of NO2F, a product


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