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### Topic: decomposition of N2O4  (Read 17080 times)

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#### plu ##### decomposition of N2O4
« on: June 07, 2006, 01:06:57 PM »
The equilibrium constant for the reaction N2O4 <-> 2NO2 is 0.113.  If one were to start with 1 mole of N2O4 at 1 atm pressure and 298 K, what fraction of it would decompose if pressure and temperature are both maintained at 1 atm and 298 K?

By doing a simple equilibrium calculation:
0.113 = (PNO2)2 / (PN2O4)
= (2x)2 / (1 - x)
I get a value of 0.155 for x.

However, the solution my instructor gives:
0.113 = (PNO2)2 / (PN2O4)
= (2x / 1 + x)2 / ( (1 - x) / (1 + x))
Where (1 + x) represents the total pressure of the system at equilibirum, results in an x value of 0.166.  Which solution is correct and why?
« Last Edit: June 07, 2006, 07:50:53 PM by plu »

#### Borek ##### Re: decomposition of N2O4
« Reply #1 on: June 07, 2006, 02:16:45 PM »
Which solution is correct and why?

Both. One is for constant volume, one is for constant pressure. Unless I am wrong ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### plu ##### Re: decomposition of N2O4
« Reply #2 on: June 07, 2006, 05:55:39 PM »
Which solution is correct and why?

Both. One is for constant volume, one is for constant pressure. Unless I am wrong Assuming that you are right ( ), which is which?

#### Borek ##### Re: decomposition of N2O4
« Reply #3 on: June 07, 2006, 06:31:40 PM »
I was wrong.

You were wrong.

Use definition of partial pressure.
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#### plu ##### Re: decomposition of N2O4
« Reply #4 on: June 07, 2006, 07:50:39 PM »
Use definition of partial pressure.

I don't understand.  The way my instructor solved the problem, he used pressure/mole fractions instead of actual partial pressures (which are simply the individual pressures of the gases)...

#### Borek ##### Re: decomposition of N2O4
« Reply #5 on: June 08, 2006, 03:13:30 AM »
I don't understand.  The way my instructor solved the problem, he used pressure/mole fractions instead of actual partial pressures (which are simply the individual pressures of the gases)...

What's a difference then between partial pressure and pressure fraction?
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#### plu ##### Re: decomposition of N2O4
« Reply #6 on: June 08, 2006, 03:27:01 PM »
What's a difference then between partial pressure and pressure fraction?

Pressure fractions are equivalent to mole fractions (assuming ideal gases): the pressure of the individual gases divided by the total pressure of the system.  Partial pressures are simply the individual pressures of the gases.

#### Borek ##### Re: decomposition of N2O4
« Reply #7 on: June 08, 2006, 03:45:10 PM »
Pressure fractions are equivalent to mole fractions (assuming ideal gases): the pressure of the individual gases divided by the total pressure of the system.  Partial pressures are simply the individual pressures of the gases.

1. How are they related to each other?

2. Which should be used in the reaction quotient?
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#### plu ##### Re: decomposition of N2O4
« Reply #8 on: June 09, 2006, 09:37:45 AM »
1. How are they related to each other?

2. Which should be used in the reaction quotient?

The partial pressures should be used in the reaction quotient...

#### Borek ##### Re: decomposition of N2O4
« Reply #9 on: June 09, 2006, 09:53:34 AM »
OK, try to express partial pressure in terms of amounts of both substances (and in terms of x).

pNO2 = ...

pN2O4 = ...
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#### plu ##### Re: decomposition of N2O4
« Reply #10 on: June 09, 2006, 10:00:37 AM »
PNO2 = 2x

PN2O4 = 1 - x

Where x is the amount of N2O4 that has decomposed...

#### Borek ##### Re: decomposition of N2O4
« Reply #11 on: June 09, 2006, 10:11:08 AM »
Use definition of partial pressure for ideal gases:

pA = nA/(nA + nB + ...) * ptotal
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#### plu ##### Re: decomposition of N2O4
« Reply #12 on: June 09, 2006, 03:36:09 PM »
Use definition of partial pressure for ideal gases:

pA = nA/(nA + nB + ...) * ptotal

A-ha!  My understanding of the definition of partial pressure was flawed.  Many thanks, Borek