homogeneous equilibria
DESCRIPTION
[C] c [D] d. P C c P D d. [A] a [B] b. P A a P B b. Homogeneous equilibria. Chemical equilibria in which reactants and products are in the same phase. K c =. K p =. Equilibrium constant expressed in terms of molar concentrations. Equilibrium constant expressed - PowerPoint PPT PresentationTRANSCRIPT
Homogeneous equilibria
• Chemical equilibria in which reactants and products are in the same phase
Kc =[C]c[D]d
[A]a[B]b
Equilibrium constant expressedin terms of molar concentrations
Kp =PC
cPDd
PAaPB
b
Equilibrium constant expressedin terms of gas partial pressures
Heterogeneous equilibria
• Chemical equilibria in which reactants and products are not in the same phase
Heterogeneous equilibria
• Chemical equilibria in which reactants and products are not in the same phase
• The concentrations of pure solids and pure liquids do not, and can not, change
• Solids and liquids are left out of the equilibrium constant expression
Kc = [Ag+]1[Cl-]1
Predicting reaction direction
K
Predicting reaction direction
• If Q < K, reaction moves spontaneously towards products until equilibrium point
• If Q > K, reaction moves spontaneously towards reactants until equilibrium point
• If reaction is at equilibrium, reaction stays at equilibrium
Predicting reaction direction
K
Q>KQ<K
Predicting reaction direction
2 NOCl(g) 2 NO(g) + Cl2(g) → ← If Kc = 4.4 x 10-4 at 500 K, determine the direction of the reactionif [NO] = .04 n/L, [Cl2] = .02 n/L and [NOCl] = 1.63 n/L
Q =[NO]2[Cl2]
[NOCl]2=
[.04]2[.02]
[1.63]2= 1.2 x 10-5
Q < K
Predicting reaction direction
2 NOCl(g) 2 NO(g) + Cl2(g) → ← If Kc = 4.4 x 10-4 at 500 K, determine the direction of the reactionif [NO] = .04 n/L, [Cl2] = .02 n/L and [NOCl] = 1.63 n/L
Q =[NO]2[Cl2]
[NOCl]2=
[.04]2[.02]
[1.63]2= 1.2 x 10-5
Q < K
Reaction will proceed spontaneously towards products until the equilibrium point is reached
Reaction atequilibrium
Reaction tendsto form products
Reaction tendsto form products
K
Q
Q
Q
KK
[C]c[D]d
Q = [A]a[B]b
aA + bB cC + dD → ←
K
Q
Q
Q
KK
aA + bB = cC + dD
aA + bB > cC + dD aA + bB < cC + dD
aA + bB cC + dD → ← [C]c[D]d
Q = [A]a[B]b
N2 + 3 H2 2 NH3→←
Calculating equilibrium concentrations
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
N2 + 3 H2 2 NH3→←
Calculating equilibrium concentrations
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Initial concentration
Equilibrium concentration
H2N2 NH3
.5 .8
.15
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Initial concentration
Equilibrium concentration
H2N2 NH3
.5 .8
.15
0
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Initial concentration
Equilibrium concentration
H2N2 NH3
.5 .8
.15
0
Change in concentration +.15
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Initial concentration
Equilibrium concentration
H2N2 NH3
.5 .8
.15
0
Change in concentration +.15-.225
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Initial concentration
Equilibrium concentration
H2N2 NH3
.5 .8
.15
0
Change in concentration +.15-.225
.15 n/L NH3(3 n/L H2/2 n/L NH3)
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Initial concentration
Equilibrium concentration
H2N2 NH3
.5 .8
.15
0
Change in concentration +.15-.225-.075
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Initial concentration
Equilibrium concentration
H2N2 NH3
.5 .8
.15
0
Change in concentration +.15-.225-.075
.15 n/L NH3(1 n/L N2/2 n/L NH3)
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Initial concentration
Equilibrium concentration
H2N2 NH3
.5 .8
.15
0
Change in concentration +.15-.225-.075
.575.425
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Initial concentration
Equilibrium concentration
H2N2 NH3
.5 .8
.15
0
Change in concentration +.15-.225-.075
.575.425
EQUILIBRIUM TABLE
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Equilibrium concentration
H2N2 NH3
.15.575.425
FROM EQUILIBRIUM TABLE
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Equilibrium concentration
H2N2 NH3
.15.575.425
FROM EQUILIBRIUM TABLE
K =[NH3]2
[N2] [H2]3
N2 + 3 H2 2 NH3→←
Suppose at a certain temperature you combined .5 n/L N2 and .8 n/L H2, and at equilibrium had produced .15 n/L NH3. What is the equilibrium constant for the reaction at this temperature?
Equilibrium concentration
H2N2 NH3
.15.575.425
FROM EQUILIBRIUM TABLE
K =[NH3]2
[N2] [H2]3=
[.15]2
[.425][.575]3= .278