A 2.00 mol sample of nitrogen dioxide was placed in an 80.0L vessel. At 200 degrees C, the gen dioxide was 6.0% decomposed according to the equation:
2NO2 in equilibrium 2NO + O2
Calculate the value of Kc for this reaction at 200 degrees C.
I have worked through this problem 3 times and am still wrong -- Help
4 answers
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This was my thinking:
For each mole of NO2 that reacts (1.00-0.060=0.94)mol remains. Starting with 2.00 mol NO2, 2 x 0.94=1.88 mol of NO2 remains. Because the volume is 80.0 L, the concentration of NO2 is 0.0235M.
2NO2 equil 2NO + O2
Init 2.00 O 0
chan -2x +2x x
final 2.00-2x
=0.0235
Because 2.00-2x=).0235, x = 0.94. Therefore the equilibrium concentrations are [NO2]=0.0235, [NO]= 1.88 and [O2} = 0.94.
Next I divided all the conc by 80.0L to give me M (why I don't know) and got NO2 = 2.9375 x 10 -4, NO = 2.35 x 10 -2, and O2 = 1.175 x 10 -2.
Then I substituted the values into the Kc formula of Kc=products/reactants. After squaring and multiplying the final that I got was
Kc = 6.49 x 10 -6/8.628 x 10 -8, my answer was 75.22. The answer was supposed to be 3.1 x 10 -6.
For each mole of NO2 that reacts (1.00-0.060=0.94)mol remains. Starting with 2.00 mol NO2, 2 x 0.94=1.88 mol of NO2 remains. Because the volume is 80.0 L, the concentration of NO2 is 0.0235M.
2NO2 equil 2NO + O2
Init 2.00 O 0
chan -2x +2x x
final 2.00-2x
=0.0235
Because 2.00-2x=).0235, x = 0.94. Therefore the equilibrium concentrations are [NO2]=0.0235, [NO]= 1.88 and [O2} = 0.94.
Next I divided all the conc by 80.0L to give me M (why I don't know) and got NO2 = 2.9375 x 10 -4, NO = 2.35 x 10 -2, and O2 = 1.175 x 10 -2.
Then I substituted the values into the Kc formula of Kc=products/reactants. After squaring and multiplying the final that I got was
Kc = 6.49 x 10 -6/8.628 x 10 -8, my answer was 75.22. The answer was supposed to be 3.1 x 10 -6.
2 mol/80 L = 0.025 M
equn
equn   I   C     E
2NO2   0.025   -0.0015  0.0235
|
v
2NO   0.0   +0.0015   0.0015
+
O2   0.0   +0.00075   0.00075
Substitute into the Kc expression. I get 3.06E-6 which rounds to 3.1E-6.
equn
equn   I   C     E
2NO2   0.025   -0.0015  0.0235
|
v
2NO   0.0   +0.0015   0.0015
+
O2   0.0   +0.00075   0.00075
Substitute into the Kc expression. I get 3.06E-6 which rounds to 3.1E-6.
This one may may a little more sense. I tried to write the equation DOWN to control spacing. The ICE stands for initial, change, equilibrium. There are three columns, one for I, one for C, and one for E.
2 mol/80 L = 0.025 M
equn   I   C     E
2NO2   0.025   -0.0015   +0.0235
|
v
2NO   0.0   +0.0015   0.0015
+
O2   0.0   +0.00075   0.00075
Substitute into the Kc expression. I get 3.06E-6 which rounds to 3.1E-6.
2 mol/80 L = 0.025 M
equn   I   C     E
2NO2   0.025   -0.0015   +0.0235
|
v
2NO   0.0   +0.0015   0.0015
+
O2   0.0   +0.00075   0.00075
Substitute into the Kc expression. I get 3.06E-6 which rounds to 3.1E-6.