Half-Life and Order of Reaction Problems

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[kelly86]

Hey Everyone, I'm completely new at this and I am glad a board like this exists!

I have two problems, I keep trying to find a way but I'm just not getting it, ESPECIALLY the first problem.

#1. At 400K, the half-life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340s. When the pressure was 28.9 kPa, the half life was 178s. Determine the order of the reaction. (A: Reaction Order = 0)

#2. The rate constant for the first order decomposition of a compound A in the reaction 2A --> P is k = 2.78 x 10^-7 /s at 25C. What is the half-life of A? What will be the pressure, initially 32.1 kPa, at a) 10s, b) 10 min after initiation of the reaction? (A: t(1/2) = 1.80 x 10^6 sec; a) P = 31.5 kPa; b) 29.0 kPa)

*There is another problem identical to this one on the book, and I got it right. When I attempted this one, my answer was incorrect. The problem started right off the bat when calculating half-life using the t(1/2) = ln 2/ 2*k (multiplying the constant by 2 because of the coefficients in front of A...

Rate of Consumption (A) = Rate * Coefficient of A
Therefore, R.O.C = 2*k (P)                                   P = pressure)

Any help would be appreciated! Thanks =)
- Kelly

[kyle1990]

the definition of a zero order reaction is one in which the rate of the reaction is independent of the concentration of the reactant(s). Consider the reaction A---->C
if the concentration [A] changes and the rate does not change, the reaction is zero order. Now let's figure out how they obtained a zero order reaction from the problem:

By definition:
Rate= change in reactant concentration/change in time

so,  55.5kPa  = Rate= .16 kPa/s
          340s

and also 28.9kPa   = Rate =  .16 kPa/s
                178s

Even though the concentration in the first calculation is higher than the concentration in the second calculation, the rate is unaffected. therefore, the rate does not depend on the concentration of the reactant and is thus zero order.
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Question 2

the half life of a first order reaction is given by the equation :

t_1/2 =.693
           k

the half life of first order reactions is independent of the initial concentration. So plug in k to the equation and you have your half life for A. If the 'correct' answer they gave you was 1.80X10^6 sec, then that is incorrect. The value i obtained is 2.50X10^6 sec. 

a) what is the pressure (initially at 32.1kPa) after 10s into the reaction? How about afer 10 min?

Solution:
the following equations will prove to be very useful when calculating half-lifes and such for ONLY first order reactions:

Time elapsed=(t_1/2)(n)

where t_1/2 is the half life and n is the number of half lives.

we want to know the pressure of the vessel after 10 seconds. So we plug in 10 seconds=time elapsed, and we solved our half life earlier as 2.49X10^6 sec:

10.sec=(2.49X10^6 sec)n
n=4.0X10^-4 half lives

Now we use the equation:

[A]_t=[A]₀(.5)ⁿ

where [A]_t is the concentration of A at some time t, [A]₀ is the initial concentration, and n is the number of half lives. plug in the values
[A]₀=32.1kPa
n=4.0X10^-4
[A]_10s= unknown

[A]_t=(32.1kPa)(.5)^{4.0X10^-4}
=32.099 kPa


use the same equations for solving for the concentration of A at 10 min, but don't forget to multiply 10min by 60 so that you have your units in seconds.
after solving for [A]_10min=32.095

Essentially the  concentrations at both times are the same, even though the second one ran almost for ten minutes. This is because the half-life is so large that it requires a LOT of time for the concentration to go down even by a little bit.


as for the last bit of your question, the correct rate of consumption is written as

Rate= -1 change in[A]
           2  change in time