April 26, 2024, 08:19:35 PM
Forum Rules: Read This Before Posting


Topic: Partial pressure problem  (Read 6913 times)

0 Members and 1 Guest are viewing this topic.

Offline elliotyang

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
Partial pressure problem
« on: February 03, 2009, 07:35:56 PM »

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27664
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Partial pressure problem
« Reply #1 on: February 04, 2009, 03:29:53 AM »
Think, how number of molecules changes during decomposition. Do you remember Avogadro's hypothesis?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline aldoxime_amine

  • Full Member
  • ****
  • Posts: 191
  • Mole Snacks: +7/-4
Re: Partial pressure problem
« Reply #2 on: February 05, 2009, 03:28:06 AM »
Your first aim should be to determine order of reaction, and that has to be done by successive approximation.
To get you started off, assume, rate =k [C2H5NH2]x

Offline Donaldson Tan

  • Editor, New Asia Republic
  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3177
  • Mole Snacks: +261/-13
  • Gender: Male
    • New Asia Republic
Re: Partial pressure problem
« Reply #3 on: February 06, 2009, 07:57:58 PM »
Here are some pointers:

1. Assume the temperature and volume of the System is fixed. This allows you to work out the number of moles of gas in the System.

2. From the chemical equation, you can tell that every mole of ethylamine reacted produces 2 moles of products, so there is a net gain of one mole per mole consumed.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Offline elliotyang

  • Regular Member
  • ***
  • Posts: 18
  • Mole Snacks: +0/-0
Re: Partial pressure problem
« Reply #4 on: February 07, 2009, 12:04:42 AM »
since the rest are assumed constant, so n is directly proportional to p.
Let say initially there are p moles of [C2H5NH2 which gives rise to 55mmHg,
p-1 moles [C2H5NH2 will produce 2 moles of product.
Then how should i proceed?
It is not necessary that the at time t=2, the total number of moles are p+1
it suppose to be p+q because i do not know how many moles of product will be used and formed.
then at t=4,
the total number of moles would be p-q+r because the concentration of [C2H5NH2 would decrease as the experiment is going.


Offline aldoxime_amine

  • Full Member
  • ****
  • Posts: 191
  • Mole Snacks: +7/-4
Re: Partial pressure problem
« Reply #5 on: February 07, 2009, 09:44:08 AM »
Ok, so since you seem to be very new to this kind of problem solving, lets see how we can overcome this hurdle. Instead of assuming at that at time t=4, number of of moles is p+q-r+s, etc.., what we can do is generalize,

Consider A--->C+D
Say, i wanted to find at time t, how much reactant,A i have. Suppose after this time, a number of moles x of the reactant has dissociated. We don't know what this x is, but we know that after this time, reactant amount remaining is
(initial amount)-x=ao-x, [D]=(its initial amount)+x=do+x
Similarly, [C]=co+x

Note, the molar ratio is 1:1:1, hence we can say x moles of A will give x moles of D and x moles of C.

Total number of moles=ao+do+co+x

If at all all components are gases, then
partial pressure can be calculated using the formula,
(partial pressure of component)=(its mole fraction)*(total pressure)

Note again that when calculating partial pressure, consider the stoichiometry of only those components which are gaseous.

I am sure you can solve the stoichiometry part now.

Next, are you aware of kinetic rate laws? If you are, to solve this type of problem, a general approach that i use:

(1) Check if it is zero order
(2) Check if it is first order.
If both of them fail,
(3) Check for nth order.

Sponsored Links