NO2 + H2 <-> NO + H2O
[H2] = .30 mol/L
[NO2] = .20 mol/L
[NO] = [H20] = .70 mol/L
@ 1000K
A. What is the mole fraction of NO in the equilibrium mixture?
B. Calculate Kc
C. Determine Kp in terms of Kc
D. If system is cooled to a lower temperature, 35% of NO is converted to NO2. What is Kc then?
E. In a different experiment, 0,86 mol of H2 is mixed with 0.86 mole of NO2 in a 2.0 L vessel at 1000K. What is the equilibrium concentration, in mol/L, of NO and NO2 at this temperature?
I assume those concns listed are at equilibrium.
Take 1 L solution so you have the mols present in the 1L
XNO = mols NO/total mols
b. Substitute the equilibrium conditions into the Kc expression and solve.
c. Kp = Kc(RT)^dN
d. I would simply take 0.35x0.70 for the products and convert them to reactants. Then plug the new cocns in and solve for the new Kc. I think the problem does not want you to rewrite the equation.
d.
(H2) = 0.86/2 = ? = 0.43
(NO2) = 0.86/2 = ? = 0.43
...............NO2 + H2 ==> NO + H2O
I............0.43...0.43.....0.....0
C..............-x....-x......x.....x
E............0.43-x.0.43-x...x.....x
Substitute the E line into kc and solve for x and 0.43-x.
Post your work if you get stuck.
To answer these questions, we need to apply the principles of chemical equilibrium and use the given information.
A. Mole fraction of NO:
The mole fraction (X) of a component in a mixture is calculated by dividing the moles of that component by the total moles in the mixture.
Total moles in the mixture = moles of NO + moles of NO2 + moles of H2 + moles of H2O
Total moles in the mixture = 0.70 mol/L + 0.20 mol/L + 0.30 mol/L + 0.70 mol/L = 1.9 mol/L
Mole fraction of NO = moles of NO / total moles in the mixture
Mole fraction of NO = 0.70 mol/L / 1.9 mol/L ≈ 0.368
B. Kc (equilibrium constant in terms of concentration):
For the given balanced chemical equation:
NO2 + H2 ↔ NO + H2O
Kc = [NO][H2O] / [NO2][H2]
Kc = (0.70 mol/L) / (0.20 mol/L)(0.30 mol/L)
Kc ≈ 11.67
C. Kp (equilibrium constant in terms of partial pressure):
To determine Kp in terms of Kc, we will use the ideal gas law:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
For the balanced equation:
NO2 + H2 ↔ NO + H2O
Kp = (PNO)(PH2O) / (PNO2)(PH2)
Kp = Kc(RT)Δn
Δn = (moles of product gas) - (moles of reactant gas)
Δn = (1 mol + 1 mol) - (1 mol + 1 mol)
Δn = 0
Since Δn is zero, Kp = Kc.
D. If the system is cooled and 35% of NO is converted to NO2:
Let x be the moles of NO converted to NO2.
The new equilibrium concentrations will be:
[NO] = 0.70 mol/L - x
[NO2] = 0.20 mol/L + x
[H2] will remain the same.
Using the new equilibrium concentrations, we can calculate Kc:
Kc = [NO][H2O] / [NO2][H2]
Kc = ([NO] - x)(0.70 mol/L) / ([NO2] + x)(0.30 mol/L)
Kc = (0.70 mol/L - x)(0.70 mol/L) / (0.20 mol/L + x)(0.30 mol/L)
E. In a different experiment, 0.86 mol of H2 is mixed with 0.86 mol of NO2 in a 2.0 L vessel at 1000K. To determine the equilibrium concentrations of NO and NO2 at this temperature, we need to use an ICE (initial, change, equilibrium) table.
Initial:
[NO2] = 0.86 mol / 2.0 L = 0.43 mol/L
[H2] = 0.86 mol / 2.0 L = 0.43 mol/L
[NO] = 0 mol/L (since none is initially present)
[H2O] = 0 mol/L (since none is initially present)
Change:
Let x be the amount of NO and NO2 formed.
[NO2] decreases by x
[H2] decreases by x
[NO] increases by x
[H2O] increases by x
Equilibrium:
[NO2] = 0.43 mol/L - x
[H2] = 0.43 mol/L - x
[NO] = x
[H2O] = x
Now, calculate the equilibrium concentration in terms of x by substituting these values into the Kc expression:
Kc = [NO][H2O] / [NO2][H2]
Kc = x * x / (0.43 mol/L - x)(0.43 mol/L - x)
Solving this quadratic equation will give the equilibrium concentration of NO and NO2 in mol/L at 1000K.