Booking.com: Chea..TripAdvisorFacebook[References]The reaction2 NO2(s) N20.(g)has an equilibrium constant, K. of 170 at 25 °C. If 4.5 x 10 mol of NO2 is present in 10.-L flask along with 3.5 x 10 mol of N204, is the system at equilibrium? If it is not atequilibrium, does the concentration of NO2 increase or decrease as the system proceeds to equilibrium?The system isV. The concentration of NO2 willVas the system proceeds to equilibrium.Submit AnswerTry Another Version2 Item attempts remalningPreviousEmail InstructorSave and ExitmentCeneare Learning CeneaeA

Given reaction for the process is 2NO2(g) → N2O4(g) Given data : Kc = 170 Number of moles of NO2 =…

Answered: [References] The reaction 2…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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The reaction
2 NO2(s) N20.(g)
has an equilibrium constant, K. of 170 at 25 °C. If 4.5 x 10 mol of NO2 is present in 10.-L flask along with 3.5 x 10 mol of N204, is the system at equilibrium? If it is not at
equilibrium, does the concentration of NO2 increase or decrease as the system proceeds to equilibrium?
The system is
V. The concentration of NO2 will
Vas the system proceeds to equilibrium.
Submit Answer
Try Another Version
2 Item attempts remalning
Previous
Email Instructor
Save and Exit
ment
Ceneare Learning Ceneae
A

Expert Solution

Step 1

Given reaction for the process is

$2 N O_{2} (g) \to N_{2} O_{4} (g)$

Given data :

K_c = 170

Number of moles of NO₂ = 4.5 x 10^-3 moles

Number of moles of N₂O₄ = 3.5 x 10^-3 moles

Volume of Solution = 10 L

So,

$C o n c e n t r a t i o n o f N O_{2} = \frac{N u b e r o f m o l e s}{V o l u m e o f S o l u t i o n (i n L)} = \frac{4.5 x 10^{- 3}}{10} = 4.5 \times 10^{- 2} M C o n c e n t r a t i o n o f N_{2} O_{4} = \frac{N u b e r o f m o l e s}{V o l u m e o f S o l u t i o n (i n L)} = \frac{3.5 x 10^{- 3}}{10} = 3.5 \times 10^{- 2} M$

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