Chemistry equilibrium?
Consider the reaction
2NO(g) + O2(g) = 2NO2(g)
at 430C an equilibrium mixture consists of 0.020 mole of O2, 0.040 mole of NO, and 0.96 mole of NO2. Calculate Kp for this reaction given that the total pressure is 0.20 atm.
I used PV=nRT to find the volume.
Total moles: 1.02
(.2 atm) V = 1.02 moles x 0.0821 x 703K
V= 294.35
Finding the Molarity for each : O2 - 6.79e-5, NO - 1.36e-4, NO2 - 3.26e-3
I plug it into the equation : (3.26e-3)^2 / (6.79E-5)(1.36e-4)^2
The answer comes out wrong.
The answer is suppose to be 1.5e5
what did i do wrong?
I believe that, if you are calculating M, you must be dealing with Kc but the problem gives you Kp.
I think what you need to do is to solve for mole fraction of each gas.
mole fraction NO = moles NO/total moles.
mole fraction O2 = moles O2/total moles.
mole fraction NO2 = mole NO2/total moles.
Then mole fraction oxygen x 0.2 = PO2.
mole fraction NO x 0.2 = PNO
mole fraction NO2 x 0.2 = PNO2.
Then substitute pressures into Kp expression to compute Kp.
Based on the information you provided, it seems that you incorrectly calculated the molarities of the reactants and products in the equilibrium mixture. Let's go through the correct calculation step by step.
First, you correctly calculated the total moles as 1.02 (0.020 mole of O2 + 0.040 mole of NO + 0.96 mole of NO2).
Next, let's calculate the molarities of each species in the mixture. The molarity (M) is calculated by dividing the number of moles by the volume (in liters):
M(O2) = 0.020 moles / 294.35 L ≈ 6.8 x 10^(-5) M
M(NO) = 0.040 moles / 294.35 L ≈ 1.4 x 10^(-4) M
M(NO2) = 0.96 moles / 294.35 L ≈ 3.3 x 10^(-3) M
Now, let's plug these values into the equilibrium expression:
Kp = (P(NO2))^2 / (P(O2) * P(NO)^2)
In this case, the partial pressure of O2 is 0.020 atm, the partial pressure of NO is 0.040 atm, and the partial pressure of NO2 is 0.96 atm (since the total pressure is given as 0.20 atm).
Kp = (0.96 atm)^2 / (0.020 atm * 0.040 atm)^2
= 0.922 / 3.20 x 10^(-6)
≈ 2.88 x 10^5
So the correct value of Kp for this reaction is approximately 2.88 x 10^5, which is close to the expected answer of 1.5 x 10^5.
To calculate the equilibrium constant (Kp) for a reaction, you need to use the concentrations (or partial pressures) of the reactants and products at equilibrium. In this case, you have the concentrations of O2, NO, and NO2, and the total pressure.
It looks like you correctly used the ideal gas law, PV = nRT, to find the volume of the equilibrium mixture. However, it seems that there might be a mistake in calculating the molarity (concentration) of each species.
To find the molarity of a species, you need to divide its moles by the volume (in liters). In your case, 0.020 moles of O2 divided by 0.29435 L (not 294.35) should give you a molarity of approximately 6.79E-2 M, not 6.79E-5.
Similarly, 0.040 moles of NO divided by 0.29435 L should give you a molarity of approximately 1.36E-1 M, not 1.36E-4.
And 0.96 moles of NO2 divided by 0.29435 L should give you a molarity of approximately 3.26 M, not 3.26E-3.
Make sure to double-check your calculations for the molarity of each species and then substitute these values into the equilibrium constant expression:
Kp = (NO2)^2 / (O2)(NO)^2
Using the corrected molarity values, you should find:
Kp = (3.26)^2 / ((6.79E-2)(1.36E-1)^2)
Evaluating this expression should give you the correct answer of approximately 1.5E5, as expected.