AutoTutored Practice Problem 15.3.4 cOUNTS TOWARDS GRADEClose ProblemUse K and initial concentrations to calculate equilibrium concentrations.Consider the equilibrium system involving the decomposition of nitrogen monoxide.N2] [02N2(g) + 02(g)2NO(g)K =6.18x10at 284 K[NO2A flask originally contains 0.226 M nitrogen monoxide. Calculate the equilibrium concentrations of the three gases.MNO]=[N2]M[02]Show ApproachCheck & Submit Answer

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Asked Sep 23, 2019
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Tutored Practice Problem 15.3.4 cOUNTS TOWARDS GRADE
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Use K and initial concentrations to calculate equilibrium concentrations.
Consider the equilibrium system involving the decomposition of nitrogen monoxide.
N2] [02
N2(g) + 02(g)
2NO(g)
K =
6.18x10
at 284 K
[NO2
A flask originally contains 0.226 M nitrogen monoxide. Calculate the equilibrium concentrations of the three gases.
M
NO]=
[N2]
M
[02]
Show Approach
Check & Submit Answer
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Auto Tutored Practice Problem 15.3.4 cOUNTS TOWARDS GRADE Close Problem Use K and initial concentrations to calculate equilibrium concentrations. Consider the equilibrium system involving the decomposition of nitrogen monoxide. N2] [02 N2(g) + 02(g) 2NO(g) K = 6.18x10 at 284 K [NO2 A flask originally contains 0.226 M nitrogen monoxide. Calculate the equilibrium concentrations of the three gases. M NO]= [N2] M [02] Show Approach Check & Submit Answer

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Expert Answer

Step 1

Given:

T = 284 K

K = 6.18 X 10-2

Conc. Of NO = 0.226 M

We need to calculate the equilibrium concentrations of the reactants and products.

Step 2

The balanced chemical equation:

2NO2 <-----> N2 + O2

 

0.226 M                 0        0


0.226 M -2x           x       x

 

Step 3

Formula :

[N2][O2]/[NO] 2 =  k

Substitute the known values,

6.18 X 10-2 = [x][x]/[ 0.226 M -2x] 2

 6.18 X 10-2 = x 2 / [ 0.226 M -2x] 2

0.248  =  x / 0.226  -2x

0.248 (0.226 -2x)  =  x  

0.056 – 0.496x = x

0.056 = x +  0.496x

0.056 = x(1 +  0.496)

0.056 = x (1.496)

0.056/...

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