# Homework Help: Reacting Gases

1. May 13, 2006

### danago

Hi.

Heres the question:
A 6L flask containing carbon monoxide at 200kPa is connected by a closed stopcock to a 3L flask containing oxygen gas at 800kPa. If the stopcock is opened, calculate the pressure of the system, assuming constant temperature.

a) if no chemical reaction occurs
b) if a small amount of an appropriate catalyst is present

The balanced equation for this reaction is:

2 CO + O2 --> 2 CO2

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ok, i can do the first question. I found the partial pressures of each, and obtained an answer of 400kPa. Im stuck on the second question. Heres my working:

V(CO, 200kPa)=6L
$$\therefore$$ V(CO2, 200kPa)=6L

That working is according to gay-lussacs law. But that pressure of 200kPa is for 6L, and i need 9, so i did $$frac{200\times 6}{9}=133.33kPa[\tex], but that is incorrect. According to the book, the answer is 333.3kPa. Any help? Thanks, Dan. 2. May 13, 2006 ### Gokul43201 Staff Emeritus Your first part is correct. For the second part, work with the number of moles. Calculate the moles of CO and O2 present using the ideal gas law. You know that it takes 2 moles of CO to react with 1 mole of O2. From this you can determine which (if any) of the two reacting gases is in excess and find the excess amount in moles. Also use the fact that the reaction has a net loss of 1 mole (moles of product - moles of reactants) per mole of oxygen used up. From here you can find the final number of moles after the reaction goes to completion. From the number of moles, and assuming the temperature is at the same value, you can find the final pressure. 3. May 13, 2006 ### danago How do i find a number of moles if i am not given temperature? Can i just use any value of temperature, as long as i keep it constant? Heres my working from keeping it constant. [tex]PV=nRT$$
$$\therefore n=\frac{PV}{RT}$$
$$=\frac{200\times 6}{8.3145\times 273}=0.529$$

So there are 0.529 moles of 2 CO2.

$$=\frac{800\times 3}{8.3145\times 273}=1.057$$

So there are 1.057 moles of oxygen gas, which is present in excess, since only half the number of moles of carbon monoxide was required, so only 0.264 moles of oxygen will be used, leaving 0.793 moles of oxygen unreacted. So i then take the partial pressures of the oxygen gas, and the carbon dioxide gas:

$$PV=nRT$$
$$\therefore P_{CO_2}=\frac{nRT}{V}$$
$$=\frac{0.527\times 8.315\times 273}{9}=133.33$$

So the partial pressure of the carbon dioxide in the two flasks is 133.33kPa.

$$P_{O_2}=\frac{0.793\times 8.315\times 273}{9}=200$$

So i add them to get 333.33kPa?

Thanks alot for the help. But have i done the right thing in using any value for temperature?

Dan.

Last edited: May 13, 2006
4. May 13, 2006

### Gokul43201

Staff Emeritus
What you've done is essentially right, but there's a couple of problems with the way you've gone about it.

You can leave R and T as a variables, and they will cancel off in the end. This will also save some trouble with getting the units right.

The way you've calculated it right now, you've got various different units that you need to sort out. For instance, you are using P in kPa, V in L and R in J/K-mol. Those units are not compatible. Really, the units shouldn't matter because they will eventually cancel off, but the numbers you've written right now will not match the units used.

If you do the following, you don't have to worry too much about the units

n(CO) = P1V1/RT = 1200/RT moles (by leaving the units of R unspecified, I'm allowed to use any units I want for P and V)

n(O2) = P2V2/RT = 2400/RT moles - this is clearly in excess, as you've found

1200/RT moles of CO will react with 600/RT moles of O2. There's an excess 1800/RT moles of O2 which remains unreacted.

Also, each mole of CO produces a mole of CO2, so you end up with 1200/RT moles of CO2.

So, the total number of moles after reaction is 1200/RT + 1800/RT = 3000/RT moles

n(tot) = 3000/RT = P(final)V(tot)/RT, and V(tot) = 9L

So, P(final)*9L = 3000 kPa-L (RT cancels off in both denominators)

That will give you the same answer you got above.

As to the question of whether or not you can use some specific temperature, the answer in this particular case is "yes". The problem is independent of temperature (or else, the temperature would be specified), so by assuming a value of the temperature, you are not going to affect the result (but you need to state this reason before using a specific value in the calculation). However, in general, it is much better to solve it using a general temperature.

Last edited: May 13, 2006
5. May 14, 2006

### danago

Ohhhh i see now. Thanks very much for that.