# physical chemistry

posted by
**ajones** on
.

I tried working out this problem but am unsure of a few steps. Please help me if you can.

Consider O3 --->1.5O2

At 210degrees Celsius and an initial gas pressure of 0.23atm, it is found that the rate of reaction is 1.8E-2 mol/(Lxs).

a) What is the change in the total gas pressure in 1.79E-2s?

b)If the reaction has an order of 1.5, what is the rate constant?

a) This is how I thought the approach to this problem should be:

Using the formula Rate = V/(uVRT) x (dP/dt) where u is stoichiometric coefficient, the two V's are the volume of the gas and the volume of the container, R is the universal gas constant and T is temperature. Dp/dt refers to the change in pressure of the gas versus the change in time.

Since both Oxygen and ozone are gases we need to use this equation twice, find the change in pressure for each and then add them together. THat's what i think anyway.

Do people out there agree with me that dPressure (change in pressure) would be written as final-initial (where final is what we are solving for)?

Solving for O3 I got: Given that the two Volume's are the same, I cancelled those out. We have Rate, u, R and T as well as dt so I put all those numbers in and wrote dP as (final-.23atm) and solved for the final, which I got 0.217atm. Would the stoichiometric coefficient be +1 or -1?

Using the similar technique, I got 0.249atm for 02, using +1.5 as the stoichiometric coefficient. I then took the (0.217atm-0.23atm) and added it to (0.249atm-0.23atm) and got 0.006 atm and said this was the total change in gas pressure.

Do people agree?

for b) using Rate = k[O3]^1.5, and solving for [O3] using Pressure of O3 x Volume = molesO3xRxT and isolated moles/V since that is concentration. I used the Pressure of O3 as 0.23atm, or should i use the change in pressure? I then solved for the [O3] and plugged in the reaction rate from before and got 40.74 for k, the rate constant. To me, this just seems to be big of an answer? Anyone have any suggestions?

Thanks a lot.