Problem 1

0.0689 mol of NO2 (g) is added to an evaculated 1L container.   When equilibrium is reached at 25 șC, the concentration of N2O4 (g) = 2.80 x 10-2 mol/L.  What is the value of K?

1. To solve this problem you should use an IRE table something like the following.  Which of the columns should you complete first?

2NO2 (g)

N2O4 (g)
I   0.0689     0
R
E

  2.80x10-2

2. Which of the following is a correctly set up IRE table before completing the equilibrium row?


 a)
2NO2 (g)

N2O4 (g)
I   0.0689     0
R -1.40x10-2   +2.80x10-2
E    

  2.80x10-2

b)
2NO2 (g)

N2O4 (g)
I   0.0689     0
R -5.60x10-2   +2.80x10-2
E    

  2.80x10-2

c)
2NO2 (g)

N2O4 (g)
I   0.0689     0
R -2.80x10-2   -2.80x10-2
E    

  2.80x10-2

d)
2NO2 (g)

N2O4 (g)
I   0.0689     0
R +2.80x10-2   -2.80x10-2
E    

  2.80x10-2

3. Which of the following is the value for the equilibrium [NO2] (g)?
a) 5.49x10-2 mol/L    b) 5.60x10-2 mol/L     c) 4.09x10-2 mol/L
d) 9.69x10-2 mol/L    e) none of these are correct

4. What is the calculated value of K for this reaction?

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