Question

4. Consider the following reaction:                       NO₂(g) à NO(g) + O(g)

The concentration of NO₂ was monitored at a fixed temperature as a function of time during the   decomposition reaction and the data is tabulated below:

Time (s)	[NO2] (M)	Time (s)	[NO2] (M)	Time (s)	[NO2] (M)	Time (s)	[NO2] (M)
0	0.0100	250	0.00611	550	0.00416	800	0.00329
50	0.00887	300	0.00567	600	0.00395	850	0.00316
100	0.00797	350	0.00528	650	0.00376	900	0.00303
150	0.00723	400	0.00495	700	0.00359	950	0.00292
200	0.00662	450	0.00466	750	0.00343	1000	0.00282

1. Using graphical analysis (using EXCEL) determine the overall reaction order. Be sure to attach a copy of the EXCEL graph with your assignment and show appropriate work!
2. Determine the rate constant.
3. Determine the [NO₂] at 2000 s (Show work!)
4. Determine the t_1/2 at t = 0 s (Show work!)

Answer 1

Part a.

The plot of ln[NO₂] versus time can be drawn as follows.

0.010 0.009 0.000 -0.007 time (s) 0.000 0.005 0.004 0.003 0. 002 100 1000 2 00400 600 [NO2] (M)

Part b.

Equation of the line: ln[NO₂] = -0.00121*t + (-4.75302)

Compare the above equation with ln[NO₂] = -kt + ln[NO₂]₀

i.e. The rate constant (k) = 0.00121 s^-1 = 1.21*10^-3 s^-1

Part c.

If t = 2000 s, ln[NO₂] = -0.00121*2000 + (-4.75302)

ln[NO₂] = -2.42 + (-4.75302) = -7.17302

i.e. [NO₂] = 0.000767 M

Part d.

At t = 0, [NO₂]₀ remains as such.

For a first-order reaction, the half-life period (t_1/2) is independent of the initial concentration.

i.e. t_1/2 = ln(2)/k = 0.693/0.00121 = 572.85 s ~ 573 s