Chemistry Partial pressure and moles problem?

AI Thread Summary
The discussion revolves around a combustion reaction between carbon monoxide (CO) and oxygen (O2) at 300 K, with a balanced equation of 2CO + O2 → 2CO2. Participants seek to determine the initial moles of reactants, the moles of products at completion, and the partial pressures of each gas post-reaction. It is established that 2 moles of CO react with 1 mole of O2, leading to 2 moles of CO2 remaining after the reaction. The limiting reagent is identified as O2, resulting in its complete consumption and a calculated partial pressure of CO2, while O2's pressure becomes zero. The conversation highlights the need for accurate data and clarifications regarding the initial conditions to solve the problem effectively.
gh_pluvilias
Messages
8
Reaction score
0

Homework Statement


Here is the problem: Carbon monoxide and oxygen are kept at exactly 300 K in separate chambers of the apparatus shown below. When the stopcock is opened, the combustion reaction begins and is allowed to proceed to completion, while being maintained at exactly 300 K.


Homework Equations


PV=nRT
I believe the balanced equation is: 2CO + O2 → 2CO2

The Attempt at a Solution


I have 3 questions to answer regarding it. I have typed up what I think are the answer or what I have so far. If you can help me answer these it would be greatly appreciated.

1. How many moles of each reactant are present before the reaction begins?
2 moles of CO2 react with 1 mole of O2 in this reaction.

2. How many moles of each gaseous reactant and product are present when the reaction goes to completion? Assume the connecting tube has negligible volume.
After the reaction is complete, 2 moles of CO2 remain (equal amount of starting CO).

3. What is the partial pressure of each gas when the reaction goes to completion? Assume the connecting tube has negligible volume.
CO2:O2 ratio = 2:1 For NO, 0.5*2/(0.082*300) = 0.041 mol of NO
For O2, 1*1/(0.082*300) = 0.041 mol of O2, 0.02 corrected for ratio.
O2 is the limiting reagent (there won't be any left), so pressure due to O2=0, mass of O2=0.
Not sure what to do from here. =/
 
Physics news on Phys.org
gh_pluvilias said:
1. How many moles of each reactant are present before the reaction begins?
2 moles of CO2 react with 1 mole of O2 in this reaction.

The answer is not related to the question. And the data you have listed so far is not enough to answer the question, could be there was additional data on the image.

For NO, 0.5*2/(0.082*300) = 0.041 mol of NO

Where did you got nitrogen from? You started with CO and O2 only.
 
Oh sorry... it was late. Must've have just typed NO instead of CO. lol Guess I get my oxides mixed up.

There is an image attached, but I thought I had included it in my attempts. Sorry. There's 1L of O2 at 1atm and 2L of CO at 0.5atm. And that is all the information I'm given.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...

Similar threads

Replies
16
Views
3K
Replies
1
Views
4K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
5
Views
4K
Back
Top