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Equilibrium•Limiting reagent•Concentrations become constant
•Dynamic situation•Reversible reactions
Equilibrium Constant (K)
represents a ratio of the concentrations ofproducts to reactantsat equilibrium:aA +bB cC + dD
Equilibrium Constant (K)
[C]c[D]d
[A]a[B]bK=
Equilibrium, or K,expression
Equilibrium Constant (K)
• “[]” represents concentration in mol/L for (g) and (aq), only
• Each “[]” must be raised to the power of its coefficient
Equilibrium Constant (K)
• K < 1 indicates little product formation
• K > 100 indicates great amount of product formation
Equilibrium Constant (K)
• Write the K expression for the dimerization of nitrogen dioxide.
Equilibrium Constant (K)
[N2O4] [NO2]2
What will the units ofK be in this example?
K= L/mol
Equilibrium Constant (K)
• At 25°C, the equilibrium concentrations of NO2 and N2O4 are 0.0370M and 0.2315M. What is the value of K at this T?
Equilibrium Constant (K)
[0.2315] [0.037]2
K=
K=0.2315mol/L 0.001369mol2/L2
K=169 L/mol
N2 + 3H2 2NH3
• Write the K expression for the synthesis of ammonia.
Equilibrium Constant (K)
[NH3]2
[N2][H2]3
What will the units ofK be in this example?
K= L2/mol2
Equilibrium Constant (K)
• At 300°C, the equilibrium concentrations are:
[N2]eq= 2.59M[H2]eq=2.77M[NH3]eq=1.82M
What is the value of K at thistemperature?
Equilibrium Constant (K)
[1.82]2
[2.59][2.77]3K=
K=3.3124mol2/L2 55.05mol4/L4
K=0.0602 L2/mol2
Equilibrium Constant (K)
• Small K (<1) means…• Big K (>100) means…• Different manner of
solving problems
Equilibrium Constant (K)
• If a reaction is reversed, then the value of K for the reversed reaction is the reciprocal of K.
Equilibrium Constant (K)
• So, if the dimerization of NO2 is reversed to be the decomposition of N2O4…
Equilibrium Constant (K)
• K = (169 L/mol)-1 or
0.00592 mol/L
Equilibrium Constant (K)
• At 25°C, the initial concentration of N2O4 is 0.750M. What are the eq. conc. of both species at this temperature?
Equilibrium Constant (K)
• You will make an equilibrium chart to indicate the initial, change, and equilibrium concentrations.
Equilibrium Constant (K)
[N2O4] [NO2]
Initial 0.750 0
Change
Eq.
Equilibrium Constant (K)
[N2O4] [NO2]
Initial 0.750 0
Change -x +2x
Eq.
Equilibrium Constant (K)
[N2O4] [NO2]
Initial 0.750 0
Change -x +2x
Eq. 0.75 - x 2x
Equilibrium Constant (K)
[2x]2
[0.75 – x]0.00592=
[2x]2
[0.75]0.00592=
4.44 x 10-3=4x2
Equilibrium Constant (K)
1.11 x 10-3=x2
0.0333=x
Equilibrium Constant (K)
5% rule…is what youremoved less than 5%of the smaller initialvalue?
Equilibrium Constant (K)
If so, then yourassumption that whatyou removed was so small it is negligibleis correct
Equilibrium Constant (K)
5% rule test:1(0.0333)1x
0.75X 100 =
4.44% < 5%…assumption is good
Equilibrium Constant (K)
[N2O4] [NO2]
Initial 0.750 0
Change -0.0333 +2(0.0333)
Eq. 0.7167 0.0666
Equilibrium Constant (K)
[0.0666]2
[0.7167]K=
Check your answer:
K=0.00619
Equilibrium Constant (K)
• At 25°C, the initial concentration of NO2 is 0.500M. What are the eq. conc. Of both species at this temperature? Remember that K = 169L/mol.
Equilibrium Constant (K)
Since K is big, lots of productwill be made. Thus, almost allof the initial amount of reactantwill be used. You need to maketwo charts for a big Kproblem…Stoichiometry andEquilibrium
Equilibrium Constant (K)
Stoichiometry Chart[NO2] [N2O4]
Initial 0.500 0
Change
Final
Equilibrium Constant (K)
Stoichiometry Chart[NO2] [N2O4]
Initial 0.500 0
Change -0.5 +0.25
Final
Equilibrium Constant (K)
Stoichiometry Chart[NO2] [N2O4]
Initial 0.500 0
Change -0.5 +0.25
Final 0 0.25
Equilibrium Constant (K)
Equilibrium Chart[NO2] [N2O4]
Initial (=final)
0 0.25
Change
Eq.
Equilibrium Constant (K)
Equilibrium Chart[NO2] [N2O4]
Initial 0 0.25
Change +2x -x
Eq.
Equilibrium Constant (K)
Equilibrium Chart[NO2] [N2O4]
Initial 0 0.25
Change +2x -x
Eq. 2x 0.25-x
Equilibrium Constant (K)
[0.25 - x] [2 x]2
169=
[0.25]4x2
169=
676x2=0.25
Equilibrium Constant (K)
Since the assumption that xwas so small it was negligibleis bad, then you must re-insert xand solve the equation with xpresent.
Equilibrium Constant (K)
You will get two values of x. Ifboth are positive, then you willalways select the smaller one. If one is positive and the othernegative, you will select thepositive one.
Equilibrium Constant (K)
Equilibrium Chart[NO2] [N2O4]
Initial 0 0.25
Change +2(0.0185)
-0.0185
Eq. 0.037 0.2315