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A+ 200C, Ke for the following reaction is 3.6 2 NO₂Cl = N₂04 + Cl₂ K₁₂ = 3,6 If 2.0 mol of NO₂Cl are placed in 2.0L container 8 permitted to react at this temp, what will the Cone of N₂O4 at equilibrium? contain

A+ 200C, Ke for the following reaction is 3.6 2 NO₂Cl = N₂04 + Cl₂ K₁₂ = 3,6 If 2.0 mol of NO₂Cl are placed in 2.0L container 8 permitted to react at this temp, what will the Cone of N₂O4 at equilibrium? contain

Chemistry
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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I am reviewing my old tests for my final .I got the answer correct, but I cant for the life of me figure out where I got the 3.79. can someone help?

H с E 2 NO₂Cl = N₂0₂ + (2 - 2x 1-2X K₂= 3.6: = = 1.90 = +x X. X = X T-2x X 1-2x X 1-2X X Clz + X X = 1.90- let X=A[№₂0}]} X = 1.90 -3,79 X 0= 1.90-4.79 X -1.90=4.79 X X = 6.39 = [NO]
Transcribed Image Text:H с E 2 NO₂Cl = N₂0₂ + (2 - 2x 1-2X K₂= 3.6: = = 1.90 = +x X. X = X T-2x X 1-2x X 1-2X X Clz + X X = 1.90- let X=A[№₂0}]} X = 1.90 -3,79 X 0= 1.90-4.79 X -1.90=4.79 X X = 6.39 = [NO]
At 200C, K₂ for the following reaction is 3.6 2 NO₂Cl = N₂04 + Cl₂ K₁₂ = 3,6 If 2.0 mol of NO₂Cl are placed in 2.0L container 8 permitted to react at this temp, what will the Cone of N₂O4 at equilibrium? contain
Transcribed Image Text:At 200C, K₂ for the following reaction is 3.6 2 NO₂Cl = N₂04 + Cl₂ K₁₂ = 3,6 If 2.0 mol of NO₂Cl are placed in 2.0L container 8 permitted to react at this temp, what will the Cone of N₂O4 at equilibrium? contain
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