Order of Reaction and rate constant

Question

The total pressure in atmospheres was recorded during the decomposition of acetaldehyde (CH3CHO(g)) at 701K for certain time intervals shown in the chart. Use the information to answer the questions below.
General equation for the decomposition of acetaldehyde:

CH3CHO(g) -> CH4(g) + CO(g)
Time (sec) 0 64 133 241 385
Total Pressure (atm) 0.600 0.717 0.801 0.886 0.956

a.) Determine the order of the reaction.
b.) Determine the rate constant using units of atm and sec.

Answer 1

You may want to re-write the table of values and include also the partial pressure of CH3CHO at each time point. For example, at zero time, partial pressure of CH3CHO is 0.600 atm.  At 64 seconds, some (x) has been converted to x CH4 and x CO, and the total pressure = 0.717 atm.  Thus, 0.600 - x + x + x = 0.717 atm and x = 0.117 atm CH4 and 0.117 atm CO and 0.600 - 0.117 = 0.483 atm CH3CHO.  So, partial pressure of CH3CHO at 64 sec = 0.483 atm.  At 133 sec, you have 0.600 - x + x + x = 0.801 atm and x = 0.201 atm = partial pressure of CH4 = partial pressure of CO and thus the partial pressure of CH3CHO at 133 sec = 0.399 atm.  In summary, you would have something like the following:

time(sec)......P(atm)....Partial Pressure CH3CHO(atm)

0...............0.6........................0.6.........

64............0.717......................0.483.....

133...........0.801......................0.399......

241..........0.886........................0.286...

You can graph the data as ln CH3CHO vs. time and if a straight line, it is first order.  If you plot 1/CH3CHO vs. time and it is a straight line it is second order.  It is not zero order, because the rate changes as CH3CHO changes.

If first order, you can calculate k from ln (pp CH3CHO) = ln (pp CH3CHO)_o - kt

Using 133 seconds as the time for example:  ln 0.399 = ln 0.6 - kt and k = 0.0030 atm/sec

Using 241 seconds as the time for example:  ln 0.286 = ln 0.6 - kt and k = 0.0030 atm/sec

 

If you happen to find after graphical analysis that it is 2nd order (doubtful), then k can be calculated from 

1/pp CH3CHO = 1/pp (CH3CHO)_o + kt