Question

The reaction:   N2O5 (g)      NO3 (g)   +   NO2 (g)

a. Determine the rate order by plotting the data.

(Hint: Which plot gives a straight line? Conc vs time, ln(conc) vs time, or 1/conc vs time)

b. Predict the concentration of N2O5 at 250 seconds.

Time (s) [N2O5]

0 1.00

25 0.822

50 0.677

100 0.557

125 0.458

150 0.377

Answer 1

Solution to Part-(a) of the given Question....

Following is the given data in the problem....as a Table ...in ....image format.... (obtained using MS Excel ):

1) Data:

• Plot of [N₂O₅ ] vs. Time (sec.);----- (i.e. in ...image format... obtained using MS Excel....)

• Plot of ln [N₂O₅ ] vs. Time (sec.):..................................(in image format) obtained using MS Excel....

• Plot of 1/[N2O5] vs. time (sec.)....obained using MS Excel....in...image format.....

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Since from the above Plots....as it seems...we have the plot of... ln([ N₂O₅ ])...vs. Time (sec.)...which is almost Linear...(i.e. Streight Line...)

Answer: (a)

As suggested by the plotted data...the Rate Order...of the given reaction, would be.... of first order.... (reaction)

Solution to Part-(b):

From ln( [ N₂O₅ ]) vs. Time (sec.) data.... we get the Slope (k) = {0 - ( - 0.196 )} / (25 - 0 ) = 7.84 x 10 ^{- 3}

Therefore at Time = 250 sec.... we get corresponding

ln ([N₂O₅]) = Slope x Time = (7.84 x 10 ^{- 3} ) x 250 sec. = - 1.96

Step- 2:

        Therefore.... [N₂O₅ ] = 0.14 ...i.e. at 250 ^oC...........(Answer...(b))