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.
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?
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