Finding change in concentration over change in time

Click For Summary
SUMMARY

The discussion centers on calculating the change in concentration of NO2 over time for the reaction 2NO2(g) --> 2NO(g) + O2(g) in a 4.00 liter vessel, with a given rate of 0.88 mol · L−1· s−1. The correct approach involves recognizing the stoichiometric relationships in the reaction, specifically that the rate of change of NO2 is half the rate of formation of products due to the 2:2:1 mole ratio. Therefore, the change in concentration of NO2 over time, Δ[NO2]/Δt, is -0.44 mol · L−1· s−1.

PREREQUISITES
  • Understanding of chemical reaction rates
  • Knowledge of stoichiometry and mole ratios
  • Familiarity with concentration calculations (C = n/v)
  • Basic grasp of units in chemical kinetics
NEXT STEPS
  • Study stoichiometric calculations in chemical reactions
  • Learn about reaction rate laws and their applications
  • Explore the concept of molarity and its significance in kinetics
  • Investigate the effects of temperature on reaction rates
USEFUL FOR

Chemistry students, educators, and anyone involved in chemical kinetics or reaction rate analysis will benefit from this discussion.

Greywolfe1982
Messages
60
Reaction score
0

Homework Statement



At a certain temperature, the reaction 2NO2(g) --> 2NO(g) + O2(g) progresses in a 4.00 liter vessel at a rate of 0.88 mol · L−1· s−1. What is

\Delta[NO2]
\Deltat

under these conditions?
Answer in units of mol · L−1· s−1.

Homework Equations



C = n/v
...?

The Attempt at a Solution



I'm not at all sure where to start with this one. My only work with mol/Ls was finding the rate that a product was made compared to the reactant. Could I get a starting point?
 
Physics news on Phys.org
Honestly, I have no idea what the question is about. The way you worded it you are asked about speed and you are told what the speed is. Seems to me like the answer is just 0.88 mol L-1 s-1.

--
methods
 
Bleh, that was too simple. All you had to do was double it due to the mole ratio.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
4K
Replies
4
Views
3K
Replies
1
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K