Calculating N2O4 Reaction Rate in 4L Container

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Discussion Overview

The discussion revolves around calculating the reaction rate of N2O4 converting to NO2 in a 4L container, specifically addressing a homework problem involving initial moles, concentration, and the rate constant.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents a calculation for the concentration of N2O4 and attempts to determine the reaction rate using the rate constant k = 0.050 s-1.
  • The calculation involves determining the concentration after converting 3/4 of the initial 0.20 moles of N2O4.
  • Another participant questions how much N2O4 remains after the conversion of 3/4.
  • Further inquiries are made regarding the methodology used to arrive at the calculated values.
  • There is a noted discrepancy between the calculated reaction rate and an expected answer, prompting further discussion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations or the expected answer, with multiple viewpoints and questions raised regarding the methodology and results.

Contextual Notes

Participants have not clarified certain assumptions regarding the reaction conditions or the definitions of terms used in the calculations, leaving some steps unresolved.

Nanu Nana
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Homework Statement


N2O4 --------------> NO2 N2O4 for k = 0.050 s-1
The process starts from 0.20 mole of N2O4 in a container of 4.0L. Calculate v for N2O4.
if 3/4 of the N2O4 is converted

Homework Equations


To calculate concentration
c= mole/L

The Attempt at a Solution


0.20 mole /4.0 L =0.05 M
(3/4)x0.05 =0.0375 M
V(reaction rate) = k .0.0375M
v= 0.050 s-1 x 0.0375 M
v=0.001875
But the answer should be 6.3x10^-4 [/B]
 
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How much is left

Nanu Nana said:
if 3/4 of the N2O4 is converted

?
 
Nanu Nana said:

if 3/4 of the N2O4 is converted

How did you arrive at this?
 
Suraj M said:
How did you arrive at this?

That's a given.
 

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