Question

NOCl decomposes by the following reaction: 2NOCl (g) --> 2 NO (g) + Cl2 (g). THe table below is collected at the same temperature.

Time (s)	0	10	20	30	40
[NOCl]	0.0800 M	0.0606 M	0.0488 M	0.0408 M	0.0350 M

how would you find the rate law for this reaction and what is the rate law?

also what is the concentration of NOCl after 60 seconds?

Answer 1

Answer - We are given the concentration and time for reaction and using this data we need to draw graph. From the graph we can determine the order of reaction and rate law.

So first we need to draw the graph ln[NOCl] Vrs time and from that we can get whether it is a zero, first, or second order reaction.

The data ln [NOCl] Vrs time

Time(s)	ln[NOCl]
0	-2.5257
10	-2.8035
20	-3.0200
30	-3.1991
40	-3.3524

The graph ln[A] Vrs time as follow –

From the above graph we got the straight line with negative slope, so it is first order reaction. We know when the graph ln[A] Vrs time gives straight line with negative slope then it is first order of reaction.

Rate law = k [NOCl]

The straight line equation with negative slope

y = -0.0205x - 2.5703

slope = -k

so, k = 0.0205 s^-1

We know the integrated rate law for the first order –

ln [NOCl] /[NOCl]o = -k*t

[NOCl]o = 0.080 M , t = 60 s

ln [NOCl] / 0.0800 M = -0.0205 s^-1 * 60 s

                                    = -1.23

Taking antiln from both side

[NOCl] / 0.0800 M = 0.292

So, [NOCl] = 0.292 * 0.0800 M

                   = 0.0234 M

So, the concentration at time 60 seconds is 0.0234 M