Equilibrium Examples. Ch. 15. Variations in Concentration problems.
The reaction of hydrogen and fluorine to form HF gas has an equilibrium constant of 1.15 x 102 at a given temp. 3.000 mol of each component was added to 1.500-L flask. Calculate the equilibrium concentrations of each.
Q < K so system shifts to products to reach equilibrium (to the right).
Kc = 1.15 x 102 = (2.000 + 2x)2
How convenient! 2 squared terms, so take square root of each side.
√1.15 x 102 = 2.000 + 2x
2.000 – x
19.4476 = 12.7238x x = 1.528
[H2] = [F2] = 2.000M – 1.528 = 0.472 M
[HF] = 2.000M + 2(1.528) = 5.056 M
The reaction of hydrogen and fluorine to form HF gas has an equilibrium constant of 1.15 x 102 at a given temp. 3.000 mol H2 and 6.000 mol F2 are added to a 3.000-L flask. Calculate the equilibrium concentrations of each.
3.000 L 3.000 L
Kc = 1.15 x 102 = (2x)2 _____________
(1.000 – x)(2.000-x)
115(2-3x+x2) = 4x2
111x2 – 345x + 230 = 0
X = -b ± √(b2 – 4ac)= 345 ± √ (3452-4(111)(230)
X = 2.14 and x = .968
[H2] = 1.000M-2.14 = not possible, so x = .968
[H2] = 1.000M-.968 = 0.032 M
[F2] = 2.000M – .968 = 1.032 M
[HF] = 2(0.968) = 1.936 M
Gaseous NOCl decomposes to form the gases NO and Cl2. At 35º C, the equilibrium constant is 1.6 x 10-5. In an experiment in which 1.0 mol of NOCl is placed in a 2.0-L flask, what are the equilibrium concentrations?
Kc = 1.6 x 10-5 = (2x)2 (x)___ Yikes!
(0.50 – 2x)2
But--- the system does not have to proceed far to reach equilibrium since K is so small. X must be very small compared to 0.50, so ignore x in the subtraction term (but not in the cubed term.)
0.50 – 2x ~ 0.50
(0.50)2(1.6 x 10-5) = 4x3
x3 = 1x 10-6 x = 0.01
[NOCl] = 0.50 -2x ~ 0.50 M (recall we approximated)
[NO] = 2x = 2(0.01) = 0.02M
[Cl2] = x = 0.01M
The reaction between nitrogen and oxygen to form nitric oxide has an equilibrium constant value of 4.1 x 10-4 at 2000 K. If 0.50 moles of nitrogen and 0.86 moles of oxygen are put into a 2.0 L container at 2000 K, what are the equilibrium concentrations of each species?
[N2]0 [O2]0 (.25)(.43)
H2SO3 (aq) H+ (aq) + HSO3- (aq)
If [H2SO3]0 = 1.5 M and the other concentrations are 0 M, find the equilibrium concentrations of all species at 25º C if K = 1.20 x 10-2.
[H2SO3] 1.50 - x
N2(g) + O2(g) 2NO(g) ΔH = +1.81 kJ
Kp = 3.3 x 1030
What happens to the direction of equilibrium if the following changes were made?
This makes Q smaller. When Q<K, more product must be made. Shift to right.
Helium is inert and does not affect any partial pressures of these gases. No change to the equilibrium.
This favors the side with more gas molecules because they can exist with fewer collisions per unit time. Both sides have same number of molecules at equilibrium, so no change occurs in equilibrium.
Predict the effect of each of the following changes to the system on the direction of the equilibrium.
Moves to right (Q < K if P removed).
Moves to right (Q < K if H2 added).