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
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.
post yourwork, we will take a look.
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.
To calculate the value of Kc for this reaction at 200 degrees C, you need to use the given information and apply the ideal gas law and the concept of equilibrium.
First, let's establish the initial concentration of NO2. We're given that the initial amount of NO2 is 2.00 mol, and it is placed in an 80.0 L vessel. Thus, the initial concentration of NO2 ([NO2]initial) is:
[NO2]initial = amount of NO2 / volume of vessel
[NO2]initial = 2.00 mol / 80.0 L
[NO2]initial = 0.025 mol/L
Next, let's determine the equilibrium concentration of NO2. We know that 6.0% of the NO2 decomposes, which means that 94.0% remains. Thus, the equilibrium concentration of NO2 ([NO2]eq) is:
[NO2]eq = [NO2]initial * (1 - % decomposition)
[NO2]eq = 0.025 mol/L * (1 - 0.06)
[NO2]eq = 0.025 mol/L * 0.94
[NO2]eq = 0.0235 mol/L
Now that we have the equilibrium concentration of NO2, we can determine the equilibrium concentrations of the other species (NO and O2). According to the balanced equation, the stoichiometry is 2:2:1 for NO2:NO:O2. This means that the equilibrium concentrations of NO ([NO]eq) and O2 ([O2]eq) will also be 0.0235 mol/L.
Finally, we can calculate the value of Kc using the equation:
Kc = ([NO]^2 * [O2]) / [NO2]^2
Substituting the equilibrium concentrations we found:
Kc = (0.0235 mol/L)^2 * (0.0235 mol/L) / (0.0235 mol/L)^2
Kc = 0.000013176 / 0.000013176
Kc = 1.000
Therefore, the value of Kc for the reaction at 200 degrees C is 1.000.
If you have worked through the problem multiple times and are still getting the wrong answer, double-check your calculations and ensure that you are using the correct units. Pay attention to decimal places and rounding errors, as they can affect the final answer.
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.