Question

The tabulated data show the concentrations of N2O5 versus time for this reaction: N2O5 (g) --> NO3 (g) + NO2 (g).

Time(s)	[N2O5] (M)
0	1.000
25	0.822
50	0.677
75	0.557
100	0.458
125	0.377
150	0.310
175	0.255
200	0.210

1a. Determine the order of the reaction by graphing:

Zero Order: Time vs [N2O5]

First Order: Time vs ln[N2O5]

Second Order: Time vs 1/[N2O5]

1b. Determine the rate constant

1c. Predict the concentration of N2O5 at 250 seconds.

Answer 1

SOLUTION:

Step - 1:

Following is the given Data in Table format... in image format......

Step - 2:

Following is the plot of [N₂O₅ ] (M) vs Time ....obtained using the given Data (above)...in..image format....

Step - 3:

Following is the plot of ln[N₂O₅ ] vs Time ....obtained using the given Data (above)...in..image format....

Step - 4:

Following is the plot of 1/[N₂O₅ ]  vs Time ....obtained using the given Data (above)...in..image format....

--  Solution to Question - 1(a):

Since, as we can see the ln[N2O5] vs. Time (sec.) plot is a streight line, which indicates that the reaction is of first order in [N2O5]............... (Answer)

--  Solution to Question - 1(b):

The Rate constant (k) of the given reaction will be equal to ...the Slope of the obtained graph of ln[N2O5] vs Time.

Therefore...

Slope: k =  y/x =   ( ln[N₂O₅ ] ) /   Time = [ - 1.56 - 0.0 ] / [ 200.0 - 0.0 ] = - 7.8 x 10 ^{- 3} sec^-1

Reaction rate constant (k) =   7.8 x 10 ^{- 3} sec^-1 (Answer - Q 1(b) )

--  Solution to Question - 1(c):

We know for a first order recation, following is the Integrated rate law...

  ln[N₂O₅ ] = - kt + ln[N₂O₅ ]_o ................................. (Equation - 1 )

....and we have the following:

k = 7.8 x 10 ^{- 3} sec^-1

  t = 250 sec.

ln[N₂O₅ ]_o = 0 (i.e. 'zero')

Therefore ....substituting the above values into Equation - 1 , we get ....

ln[N₂O₅ ] _{t = 250 sec.} =  - 1.95

Or........... [N₂O₅ ] _{t = 250 sec.} = 0.1422 M

Answer to Question - 1(c):

.........answer is ............. [N₂O₅ ] _{t = 250 sec.} = 0.1422 M