Suppose a 250.mL flask is filled with 0.80mol of CO, 0.60mol of NO and 0.20mol of CO2. The following reaction becomes possible:NO2(g)+CO(g) <->NO(g)+CO2(g)The equilibrium constant K for this reaction is 9.16 at the temperature of the flask.Calculate the equilibrium molarity of NO. Round your answer to two decimal places.please check your work. most people get this wrong!

Question
Asked Nov 1, 2019
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Suppose a 250.mL flask is filled with 0.80mol of CO, 0.60mol of NO and 0.20mol of CO2. The following reaction becomes possible:

NO2(g)+CO(g) <->NO(g)+CO2(g)

The equilibrium constant K for this reaction is 9.16 at the temperature of the flask.

Calculate the equilibrium molarity of NO. Round your answer to two decimal places.

please check your work. most people get this wrong!

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Expert Answer

Step 1

The initial concentrations of CO, NO and CO2 are calculated as shown below.

250 mL = 0.250 L

No. of moles of solute
Concentration
Volume(mL
0.8
- 3.2M
0.25
Initial[CO
0.6
= 2.4M
0.25
Initi al [NO]
0.8
0.8M
Initial[CO,1=
0.25
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No. of moles of solute Concentration Volume(mL 0.8 - 3.2M 0.25 Initial[CO 0.6 = 2.4M 0.25 Initi al [NO] 0.8 0.8M Initial[CO,1= 0.25

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Step 2

Equilibrium molarity of NO3 can be determined:

NO
Со
NO
CO.
+
+
Initial
0
3.2
2.4
0.8
Change
+x
+x
X
-X
Equilibrium
3.2-x
2.4+x
0.8+x
х
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Image Transcriptionclose

NO Со NO CO. + + Initial 0 3.2 2.4 0.8 Change +x +x X -X Equilibrium 3.2-x 2.4+x 0.8+x х

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Step 3

K can be give...

[(2.4+x)(0.8+x
[(x)(3.2-x)
K=
[(2.4+x0.8)
9.16=
x)(3.2-x
x 2.49 or 0.07
[NO, 2.49 or 0.07
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[(2.4+x)(0.8+x [(x)(3.2-x) K= [(2.4+x0.8) 9.16= x)(3.2-x x 2.49 or 0.07 [NO, 2.49 or 0.07

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