Chemical equilibium by David Richardson

Chemical Equilibrium An equilibrium reaction is one in which the rate of the forward reaction equals the rate of the verse reaction. Let’s look at the following reaction: k1 !!! ⇀ a A + b B↽ !! ! cC+dD k -1

a b rate of the forward reaction = k [A] [B] 1

rate of the reverse reaction = k

-1

d [C] [D]

At equilibrium: rate of the forward reaction = rate of the reverse reaction; therefore, k1 [A]a [B]b = k -1 [C]c [D]d k1 [C]c [D]d = k -1 [A]a [B]b K eq =

[C]c [D]d [A]a [B]b

When the dynamic state of equilibrium is reached the concentrations of A, B, C, and D are invariant (i.e., their concentrations no longer change). The reaction continues in a dynamic state, and the concentrations remain the same unless a stress is placed on the reaction, and the rate of the forward reaction is equal to the rate of the reverse reaction. The following graph illustrates the equilibrium between the reactants and the products where the concentrations of the products remain the same at equilibrium and concentrations of the reactants remain the same at

equilibrium.

The following diagram illustrates the equivalency between the rate of the forward reaction and the rate of the reverse reaction:

Let’s consider the reaction between hydrogen and iodine

H 2 (g) + I2 (g)  2 HI (g) If the initial concentrations of hydrogen and iodine gases are 0.0175 M, calculate the equilibrium concentrations of the reactants and products if the equilibrium constant is 55.64 at 425oC. 2

⎡⎣ HI(g) ⎤⎦ = 55.64 ⎡⎣ H 2(g) ⎤⎦ ⎡⎣ I 2(g) ⎤⎦ 0.0175-x

H 2 (g) +

0.0175-x

I2 (g)

 2 HI (g)

( 2x )2 ( 0.0175- x ) ( 0.0175- x )

= 55.64

4x 2 = 55.64 3.0625 x 10 −4 -0.035 x + x 2

(

4x 2 = 55.64 3.0625 x 10 -4 - 0.035 x + x 2

)

4x 2 = 1.704 x 10 −2 -1.9474 x - 55.64 x 2 0 = 1.704 x 10 −2 -1.9474 x - 51.64 x 2

-b ± b2 -4ac x= 2a

a = 51.64; b = -1.9474; and c = 0.01704

1.9474 ± 3.79237-4(51.64)(0.01704) 2(51.64) 1.9474 ±

3.79237-3.519782 103.28

1.9474 ± 0.272588 103.28

1.9474 ± 0.522 103.28

x = 2.39 x 10-2 or 1.38 x 10-2 x = 2.39 x 10-2 M is not possible, because it would lead to negative matter! However, if we select x = 1.38 x 10-2 mole Then,

[ HI(g)] = 2.76 x 10 −2 M

(g)] = ( 0.0175-x ) M = ( 0.0175 - 0.0138 ) M = 3.7 x 10 -2 M

[ I (g)] = ( 0.0175-x ) M = ( 0.0175 - 0.0138 ) M = 3.7 x 10

-2

( 0.0276 )2 ( 0.0037 )2

= 55.6

The equilibrium expression does not include solids. For example, the equilibrium expression for the following reaction: 1 S8 (s) + O 2( g)  SO 2 (g) 8

⎡⎣SO 2(g) ⎤⎦ = K eq = 4.2 x 1025 at 25o C ⎡⎣O 2(g) ⎤⎦

Water as a liquid is not included in the equilibrium expression. For example, in the following equation where water is one of the reactants

HCOOH (aq) + H 2O (l)  HCOO-(aq) + H 3O +(aq) the equilibrium expression is + ⎡⎣ HCOO-(aq) ⎤⎦ ⎡⎣ H 3O(l) ⎤⎦ = K ⎡⎣ H 2O(l) ⎤⎦ = K a ⎡⎣ HCOOH (aq) ⎤⎦

The Equilibrium Constant Involving Concentration (KC) Versus the Equilibrium Constant Involving Pressure (Kp) Let’s consider the following equilibrium reaction

N 2 (g) + 3 H 2 (g)  2 NH 3 (g) The equilibrium constant as respect to the concentration of the gaseous reactants and products, KC, at 25oC is 2

⎡⎣ NH3 (g) ⎤⎦ Kc = = 3.5 x 108 at 25oC 3 ⎡⎣ N 2 (g) ⎤⎦ ⎡⎣ H 2 (g) ⎤⎦ Let’s determine Kp for this reaction at 25oC. Using the Ideal Gas Equation, PV = nRT, we can derive a relationship between concentration and pressure. where

n p = concentration = V RT 2 PNH 3(g)

⎡⎣ NH3 (g) ⎤⎦ P (RT) = = 3 PN2(g) PH3 2(g) PN2(g) x PH3 2(g) ⎡⎣ N 2 (g) ⎤⎦ ⎡⎣ H 2 (g) ⎤⎦ x (RT) (RT)3 Kc =

2 PNH 3(g) 2 N 2(g)

2 H 2(g)

2 NH3(g)

1 (RT) 2 x 1 (RT) 4

x (RT) 2

K c = K P x (RT) 2

Therefore, the relationships between KC versus Kp for

N 2 (g) + 3 H 2 (g)  2 NH 3 (g) is

K c = K p x (RT) 2 Kp =

Kc (RT) 2

This relationship differs for different equations and depends of the stoichiometry of the chemical reaction. Now, let’s determine a numerical value for the Kp for the equilibrium between nitrogen gas/hydrogen gas and ammonia gas at twenty-five degrees Celsius.

K c = K P x (RT) 2 KP =

Kc (RT) 2

3.5 x 108 KP = (0.08205 x 298) 2 3.5 x 105 KP = (0.08205 x 298) 2 K P = 5.9 x 105

Now, let’s consider the decomposition of solid calcium carbonate to solid

calcium oxide and carbon dioxide gas.

CaCO 3 (s)  CaO (s) + CO 2( g) The equilibrium constant for this reaction is

Kc = ⎡⎣CO2 (g) ⎤⎦ The relationship between KC and Kp is

PCO2 K c = ⎡⎣CO2 (g) ⎤⎦ = RT PCO2 K p Kc = = RT RT (RT) K c = K p Consequently, for the decomposition of calcium carbonate, Kp can be found by multiplying KC by the gas constant R and the temperature (in Kelvin). Manipulating the Equilibrium Constant The equilibrium constant can be manipulated by using simple mathematics. For example, the equilibrium constant for the following reaction:

C (s) +

1 O 2(g)  CO (g) 2

⎡CO (g) ⎤⎦ K1 = ⎣ = 4.6 x 1023 at 25o C 1 ⎡⎣O2 (g) ⎤⎦ 2

However, if the synthesis equation is multipled by “2,” then the reaction would be:

2 C (s) + O2 (g)  2 CO (g) and the equilibrium expression and the equilibrium constant (at the same temperature) would be 2

⎡CO (g) ⎤⎦ K2 = ⎣ = K12 = 2.1 x 1047 at 25o C ⎡⎣O 2 (g) ⎤⎦

For the reaction:

N 2 (g) + 3 H 2 (g)  2 NH 3 (g) [ NH ] = [ N ][ H ] 2

= 3.5 x 10 8

Let’s calculate KC for

1 3 N 2 (g) + H 2 (g) 2 2

NH 3 (g)  3

[ H 2 ]2 [ N 2 ]2 K 'C = [ NH 3 ] In order to find K 'C

simply take the reciprocal and the square root of KC

[ NH ] [ N ][ H ] 2

[ NH ] [N ] [H ] 3

1 2

3 2

now, take the reciprocal of KC. 1 2

[N ] [H ] [ NH ] 2

3 2

Consequently, the relationship between KC and K’C is K 'C =

1 = KC

1 3.5 x 10 8

1 = 5.3 x 10-5 8 3.5 x 10

Determining K for a Reaction from Other K values Let’s consider the following reaction:

AgCl (s) + 2 NH 3 (aq)  Ag(NH 3 )+2 (aq) + Cl-(aq) The equilibrium constant for this reaction is

K eq

⎡⎣ Ag(NH3 ) 2+ ⎤⎦ ⎡⎣Cl(aq) ⎤⎦ = 2 ⎡⎣ NH3 (aq) ⎤⎦

The equilibrium constant can be obtained from the following two equations: (1) + AgCl (s)  Ag (aq) + Cl-(aq)

⎤⎦ = 1.8 x 10-10 Ksp = ⎡⎣Ag+(aq) ⎤⎦ ⎡⎣Cl(aq)

(2) + Ag (aq) + 2 NH 3 (aq)  Ag(NH 3 )+2 (aq)

⎡⎣ Ag(NH3 ) 2+ ⎤⎦ Kf = = 1.6 x 107 2 ⎡⎣ Ag +(aq) ⎤⎦ ⎡⎣ NH3 (aq) ⎤⎦ Now, multiply Ksp by Kf to obtain the desired equilibrium constant

⎡⎣ Ag(NH 3 ) 2+ ⎤⎦ ⎡⎣ Ag(NH 3 ) 2+ ⎤⎦ ⎡⎣Cl(aq) ⎤⎦ + ⎡ ⎤ ⎡ ⎤ x Ag Cl = (aq) (aq) 2 2 ⎣ ⎦⎣ ⎦ + ⎡⎣ Ag (aq) ⎤⎦ ⎡⎣ NH3 (aq) ⎤⎦ ⎡⎣ NH3 (aq) ⎤⎦

⎡⎣ Ag(NH 3 ) +2 ⎤⎦ ⎡⎣Cl-(aq) ⎤⎦ K f x K sp = 2 ⎡⎣ NH 3 (aq) ⎤⎦ 7

1.6 x 10 x 1.8 x 10

-10

⎡⎣ Ag(NH 3 ) 2+ ⎤⎦ ⎡⎣Cl(aq) ⎤⎦ = 2 ⎡⎣ NH 3 (aq) ⎤⎦

⎡⎣ Ag(NH 3 ) 2+ ⎤⎦ ⎡⎣Cl(aq) ⎤⎦ 2.9 x 10 = = K eq 2 ⎡⎣ NH 3 (aq) ⎤⎦ -3

Let’s use another example to obtain the equilibrium constant from other equilibrium constants. Let’s obtain the Kp at 226.8oC for Reaction (4) given the equilibrium constants for reactions (1), (2), and (3). Reaction (4) K P4

H ⋅(g) + Br ⋅

(g)

 HBr (g)

⎡⎣ HBr(g) ⎤⎦ ⎡ H 2 (g) ⎤⎦ ⎡⎣ Br2 (g) ⎤⎦ ⎡⎣ HBr(g) ⎤⎦ x ⎣ x = 2 2 2 2 ⎡⎣ H 2 (g) ⎤⎦ ⎡⎣ Br2 (g) ⎤⎦ ⎡⎣ H (g) ⎤⎦ ⎡⎣ Br (g) ⎤⎦ ⎡⎣ H (g) ⎤⎦ ⎡⎣ Br (g) ⎤⎦ 1 1 K P1 x x = K P24 K P2 K P3

⎡⎣ HBr(g) ⎤⎦ ⎡⎣ HBr(g) ⎤⎦ = = K 2P4 = K P4 2 2 ⎡⎣ H (g) ⎤⎦ ⎡⎣ Br (g) ⎤⎦ ⎡⎣ H (g) ⎤⎦ ⎡⎣ Br (g) ⎤⎦ 7.9 x 1011 = K P4 = 2.7 x 1033 -41 -15 4.8 x 10 x 2.2 x 10

The Reaction Quotient, Q 1. Q is obtained by using a similar formula as K, but Q may not be at equilibrium. 2. The reaction is at dynamic equilibrium when Q = Keq 3. If Q is greater than Keq , then the reaction is not at equilibrium and must adjust so that product(s) will go to reactant(s) until a state of dynamic equilibrium is reached 4. If Q is less than Keq , then the reaction is not at equilibrium and must adjust so that more reactant(s) will go to product(s) until a state of dynamic equilibrium is reached Let’s apply the concept of Q to the following reaction:

CH 3CH 2 CH 2 CH 3  (CH 3 )2 CHCH 3 I II Keq = 2.5 at 25oC is the reaction at equilibrium when [CH3CH2CH2CH3] = 0.97 M and [(CH3)2CHCH3] = 2.18M? If not, calculate the equilibrium concentrations. Q=

2.18 = 2.25 0.97

Q (2.25) is less than Keq (2.5) Therefore, in order for the reaction to reach equilibrium, x amount of “I” must be transformed into “II.”

0.97 M - x

2.18 + x (CH3)2CHCH3

CH3CH2CH2CH3

2.18 + x = 2.5 0.97 - x x = 0.07

Therefore, 0.07 M (assuming that the reaction takes place in a 1.00 L vessel) of I was lost and 0.07 M of II was gained. Consequently, the equilibrium concentration of CH3CH2CH2CH3 is 0.90 M (0.97 M – 0.07 M) and the equilibrium concentration of (CH3)2CHCH3 is 2.25M (2.18 M + 0.07 M). Let’s check to determine if the calculated concentrations are correct. If they are, then the values should equal 2.5 when plugged into the equilibrium expression. K=

[(CH ) CHCH ] 3 2

[CH CH CH CH ] 3

2.25 = 2.5 0.90

The values give the equilibrium expression. Let’s consider another application of Q using the following equilibrium:

N 2 (g) + O2 (g)  2 NO (g) Keq = 1.7 x 10-3 Is the reaction at equilibrium when [N2] = 0.50 M , [O2] = 0.25 M, and [NO] = 0.0042 M? If not, calculate the equilibrium concentrations.

[ NO] [ N 2 ][O2 ] 2 [0.0042] [0.50][0.25] 2

Q= Q=

= 1.41 x 10-4

Q (0.000141) is less than Keq (0.0017) 0.50 –x

0.25-x

0.0042 + 2x

N 2 (g) + O 2 (g) ! 2 NO (g) K eq = 1.7 x 10-3 =

(0.0042+2x)2 (0.50 -x)(0.25 - x)

(1.7 x 10 )( 0.50 - x )( 0.25 - x ) = (0.0042 + 2x) -3

(1.7 x 10 )( 0.125 - 0.75 x + x ) = 1.764x10 -3

(1.7 x 10 )( 0.125 - 0.75 x + x ) = 1.764 x 10 -3

−5

-5

- 0.0168x + 4x 2

+ 0.0168 x + 4 x 2

0.0002125 - 0.001275 x + 0.0017 x 2 = 1.764 x 10 −5 + 0.0168 x + 4 x 2 0 = -0.0002125 + 1.764 x 10 −5 +0.001275 x + 0.0168 x + 4 x 2 - 0.0017 x 2

−1.9486 x 10 −4 + 0.018075 x + 4 x 2 = 0 x=

-b ± b 2 - 4ac 2a

a = 4; b = 0.018075; and c = -1.9486 x 10-4 x=

−0.018075 ±

( 0.018075 )2

- 4(4)(-1.9486 x 10 −4 )

−0.018075 ±

0.000327 + 0.00311776 8

−0.018075 ± 8

0.003445

−0.018075 ± 0.0587 8

0.0406 8

x = 5.1 x 10-3 M Let’s check to make sure that the amount of material decreased from each reactant is 5.1 x 10-3 M and the amount of product formed is 1.02 x 10-2 M generate 0.0017 when plugged into the equilibrium expression. Using the equation for the equilibrium constant, i.e., K eq = 1.7 x 10-3 =

(0.0042+2x)2 (0.50 -x)(0.25 - x)

let’s see if we can obtain 1.7 x 10-3 from

( 0.0042 + 2x )2 ( 0.50 - x ) ( 0.25 - x ) (0.0042+2 x 5.1 x 10-3 )2 = (0.50 -0.0051)(0.25 - 0.0051) (0.0144)2 = 1.7x10-3 (0.495)(0.245)

The values generate 0.0017, the equilibrium constant! “Quod Erat Demonstrandum,” this translates to “which is what had to be proved"

Calculating Equilibrium Constants Equilibrium constants are calculated from the equilibrium concentrations of the reactants and products. For example, let’s calculate the equilibrium constant for the following reaction at 1000 K: 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) When1.00 mol of SO2 and 1.00 mol of O2 are placed in a 1.00 L flask, the number of moles of SO2 at equilibrium is 0.925 mole. Since the flask is 1.00 L, the equilibrium concentration of SO2 is 0.925 M. Following are the equilibrium number of moles of the products and the reactants: 0.925 0.4625 0.925 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) Therefore, the equilibrium concentrations would be: [SO3] = 0.925 mole/1.00 L = 0.925 M [O2] = 0.4625 mole/1.00 L = 0.4625 M [SO2] = 0.925 mole/1.00 L = 0.925 M 2

⎡⎣SO 3 (g) ⎤⎦ K= ⎡⎣SO 2 (g) ⎤⎦ ⎡⎣O 2 (g) ⎤⎦

( 0.925 )2 ( 0.925 ) ( 0.4625 )

Let’s calculate the KP for the following reaction at 800 K: 2 H 2 S (g)  2 H 2(g) + S2 (g)

If a tank contains hydrogen sulfide gas at 10.00 atm at 800 K, and equilibrium is achieved when the partial pressure of S2 (g) is 2.0 x 10-2 atm, calculate Kp for the reaction.

Kp =

PS2 (g) PH22 (g) PH2 2 S (g)

Originally 10.00 atm of H2S (g) At equilibrium, the number of moles of the reactants and products are 10.00 -2x

2 H 2 S (g)  2 H 2(g) + S2 (g)

x = 0.020 atm 10.00 -0.040 0.040

0.020

2 H 2S (g)  2 H 2(g) + S2 (g) (0.020) (0.040)2 Kp = (9.96)2

KP = 3.2 x 10-7 KC can be calculated from KP in the following manner: ⎡S2 (g) ⎤⎦ ⎡⎣ H 2 (g) ⎤⎦ KC = ⎣ 2 ⎡⎣ H 2S (g) ⎤⎦

Since Concentration =

P RT

KC =

⎛ PS2 (g) ⎞ ⎛ PH2 ⎞ ⎜⎝ RT ⎟⎠ ⎜⎝ RT ⎟⎠ ⎛ PH2 S (g) ⎞ ⎜⎝ RT ⎟⎠

⎛ 1 ⎞⎛ 1 ⎞ ⎜⎝ ⎟⎜ ⎟ RT ⎠ ⎝ RT ⎠ ⎛ 1 ⎞ ⎜⎝ ⎟ RT ⎠

PS2 (g) PH2(2 (g) PH2 2 S (g)

x Kp

⎛ 1 ⎞ KC = ⎜ x Kp ⎝ RT ⎟⎠ 1 ⎛ ⎞ KC = ⎜ x 3.2 x 10 -7 ⎝ 0.08205 x 800 ⎟⎠

KC = 4.8 x 10-9 Note that the equilibrium constant does not have units, because each concentration in KC or pressure in KP is divided by an activity coefficient of 1.00 atmosphere (the equilibrium concentrations are divided by 1.00 M). For example, in determining the Kp for 2 H 2 S (g)  2 H 2(g) + S2 (g)

The value of the constant doesn’t have units, because the equilibrium pressures are divided by 1.00 atm (the activity coefficient).

KP =

⎛ 0.020 atm ⎞ ⎛ 0.040 atm ⎞ ⎜⎝ ⎟⎜ ⎟ 1.00 atm ⎠ ⎝ 1.00 atm ⎠ ⎛ 9.96 atm ⎞ ⎜⎝ ⎟ 1.00 atm ⎠

= 3.2 x 10 −7

Let’s calculate the Kc for the following reaction at 25 oC: H I

The equilibrium concentration of I2 is 0.035 M when 0.050 mole of trans1,2-diiodocyclohexane (structure A) is dissolved in 1.00 L of a carbon tetrachloride solution. The temperature of the reaction is 25 oC. (0.050-0.035) mol

0.035mol

0.035 mol are the equilibrium moles

KC =

[ I ][ B] = ( 0.035 )( 0.035 ) = 0.082 2

[ A]

( 0.015 )

The relationship between Kp and KC for this reaction is ⎛ PI2 ⎞ ⎛ PB ⎞ ⎜ ⎟ I2 ][ B] ⎜⎝ RT ⎟⎠ ⎝ RT ⎠ [ KC = = ⎛ PA ⎞ [ A] ⎜⎝ ⎟ RT ⎠

⎛ 1 ⎞⎛ 1 ⎞ ⎜⎝ ⎟⎜ ⎟ PI PB RT ⎠ ⎝ RT ⎠ KC = x 2 PA ⎛ 1 ⎞ ⎜⎝ ⎟⎠ RT ⎛ 1 ⎞⎛ 1 ⎞ ⎜⎝ ⎟⎜ ⎟ RT ⎠ ⎝ RT ⎠ 0.082 = x Kp ⎛ 1 ⎞ ⎜⎝ ⎟ RT ⎠

0.082 x RT = KP 2.0 = Kp Kc for the following reaction at 1227 oC is 1.0 x 10-5:

N 2 (g) + O 2 (g)  2 NO (g) Let’s determine the equilibrium concentrations if 0.80 mole of N2 and 0.20 mole of O2 are placed initially in a 2.00 L flask at 1227oC. The initial number of moles is 0.80 mol 0.20 mol

N 2 (g) + O 2 (g)  2 NO (g) The equilibrium number of moles 0.80 – x

0.20-x

N 2 (g) + O 2 (g)  2 NO (g) The equilibrium concentrations:

[N ] = 2

( 0.80 - x ) 2.00 L

[O ] = 2

[ NO ] =

( 0.20 - x ) 2.00 L 2x 2.00 L

[ NO ]2

KC =

[ N ][ O ] 2

1.0 x 10 −5

⎛ 2x ⎞ ⎜⎝ ⎟ 2.00 L ⎠ = ⎛ 0.80 - x ⎞ ⎛ 0.20 - x ⎞ ⎜⎝ ⎟⎜ ⎟ 2.00 L ⎠ ⎝ 2.00 L ⎠

1.0 x 10 −5 =

4 x2 ( 0.80 - x ) ( 0.20 - x )

1.0 x 10 −5 =

4 x2 0.16 - 1.00 x + x 2

1.6 x 10 −6 - 1.0 x 10 −5 x + 1.0 x 10 −5 x = 4 x 2 1.6 x 10 −6 - 1.0 x 10 −5 x + 1.0 x 10 −5 x - 4 x 2 = 0 1.6 x 10 −6 - 1.0 x 10 −5 x - 4 x 2 = 0

Where a = -4; b =-1.0 x 10-5 ; and c = 1.6 x 10-6 -b ± b 2 - 4ac x= 2a

(

)

- -1.0 x 10 -5 ±

(-1.0 x 10 )

-5 2

2 ( -4 )

(

- 4 ( -4 ) 1.6 x 10 -6

)

1.0x10 -5 ± 1.0 x10 -10 + 2.56 x 10 -5 x= -8

1.0x10 -5 ± 5.06 x 10 −3 -8

1.0x10 -5 - 5.06 x 10 −3 -8

x = 6.3 x 10-4 mol

[N ] 2

[O ] 2

( 0.80 - 6.3 x 10 ) = 0.40 M = −4

2.00 L

( 0.20 - 6.3 x 10 ) = 0.10 M = −4

2.00 L

( 2 x 6.3 x 10 ) = 6.3 x 10 [ NO ] = −4

2.00 L

−4

Submitting the above concentrations into the equilibrium expression and determining if they will generate the rate constant serves as a check for the calculated concentration values. K=

[ NO ]2

[ N ][ O ] 2

( 6.3 x 10 ) =

-4 2

(0.40)(0.10)

= 1 x 10 −5

Kp and KC Revisited Let’s develop a general equation that can be used to calculate the Kp when given KC and KC when given KP. We can accomplish this by using the following general equation:

a A (g) + b B (g) ! c C (g) + d D (g)

[C(g)] [ D(g)] c

KC =

[ A(g)]a [ B(g)]b 23

⎛ PCc ⎞ ⎛ PDd ⎞ ⎜⎝ RT ⎟⎠ ⎜⎝ RT ⎟⎠ KC = ⎛ PAa ⎞ ⎛ PBb ⎞ ⎜⎝ RT ⎟⎠ ⎜⎝ RT ⎟⎠ c

⎛ 1 ⎞ ⎛ 1 ⎞ ⎜⎝ ⎟ ⎜ ⎟ PCc PDd RT ⎠ ⎝ RT ⎠ KC = a b x PAa PBb ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜⎝ ⎟ ⎜ ⎟ RT ⎠ ⎝ RT ⎠ c

⎛ 1 ⎞ ⎛ 1 ⎞ ⎜⎝ ⎟ ⎜ ⎟ RT ⎠ ⎝ RT ⎠ KC = a b x KP ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜⎝ ⎟ ⎜ ⎟ RT ⎠ ⎝ RT ⎠

KC =

⎛ 1 ⎞ ⎜⎝ ⎟ RT ⎠ ⎛ 1 ⎞ ⎜⎝ ⎟ RT ⎠

( c+d )

(a+b)

x KP

1 (RT)(c+d) KC = x KP 1 (RT)(a+b)

(RT )(a +b) KC = KP (c+ d ) (RT )

K C = (RT)(a +b)−(c+ d ) K P (c+d) is the sum of the moles of gaseous products, and (a+b) is the number of moles of gaseous reactants. The reaction between Kc and Kp can also be written in the following manner as well:

K P = (RT)(c+d)-(a+b) K C Let’s revisit the following reaction at :

2 H 2S (g)  2 H 2(g) + S2 (g) Kp = 3.2 x 10-7

K C = (RT)(a+b)-(c+d) K p The number of moles of gaseous reactants, (a+b), = 2 The number of moles of gaseous products, (c+d), = 3

K C = (RT)2 − 3 K p

K C = (RT)−1 (3.2 x 10 −7 ) 3.2 x 10 −7 3.2x10 −7 KC = = = 4.9 x 10 −9 RT 0.08205 x 800 KC and KP are equal when the number of moles of gaseous products equals the number of moles of gaseous reactants.

Le Chatelier’s Principle Le Chatelier’s Principle states that when a system is in dynamic equilibrium, and the system encounters stress, the system will appropriately adjust itself in order to relieve the stress. This simply means that the equilibrium will do what is necessary in order to maintain the value of the equilibrium constant a designated temperature. The Stresses that may be placed on an equilibrium reaction include: 1. 2. 3. 4.

A change in the temperature A change in the concentration of reactants or products A change in the volume A change in the pressure (this occurs only when there is a difference between the number of moles of gaseous reactants and products)

The Effect of Temperature Changes on Equilibria 1. Adjustment in the equilibrium is determined by whether the reaction is endothermic or exothermic. 2. If the reaction is exothermic, an increase in temperature shifts the equilibrium toward the reactants. A decrease in temperature shifts the reaction toward the products. 3. If the reaction is endothermic, an increase in temperature shifts the equilibrium toward the products. A decrease in temperature shifts the reaction toward the reactants. 4. Changes in temperature result in changes in the value of the equilibrium constant. These changes are related to the Arrhenius Equation:

K eq = A e ln K eq

ΔH oreaction RT

ΔH oreaction =+ ln A RT

K is the equilibrium constant at some designated temperatures, T. Therefore, a plot of ln Keq versus 1/T would give a straight line where the slope is â&#x20AC;&#x201C;heat of the reaction/R and the intercept is ln A. For example, letâ&#x20AC;&#x2122;s plot the equilibrium constant versus the reciprocal of the T (in Kelvin) for the following reaction:

1 1 N 2 (g) + O 2 (g) ď&#x201A;&#x201A; NO 2 2

(g)

Following is a tabulation of data that shows the changes in the equilibrium constants with changes in temperature.

A plot of ln Keq versus the reciprocal of T is

slope = -

ΔH reaction R

-1.0832 x 104 K = -

ΔH reaction R

-1.0832 x 104 K x -8.314 9.00 x 104

J = ΔH reaction K-mol

J = ΔH reaction mol

The reaction is endothermic, and increases in temperature results in an increase in the equilibrium constants for this endothermic reactions.

The Effect of Changes in the Concentrations of Reactants and Products 1. Increasing the concentration of the reactants creates a situation where Q is less than K; therefore, the system will adjust itself to produce more product so that Q will equal K, i.e., the equilibrium will shift toward the product. 2. Decreasing the concentration of the reactants creates a situation where Q is greater than K; therefore, the system will adjust itself to produce more reactants so that Q will equal K, i.e., the equilibrium will shift toward the reactants. Following is an example of changing the concentrations of an equilibrium reaction. Letâ&#x20AC;&#x2122;s consider the following reaction:

CH 3CH 2 CH 2 CH 3 ď&#x201A; (CH 3 )2 CHCH 3 n-butane

isobutane

If at 25oC, the equilibrium concentrations of n-butane and Isobutane are respectively 0.500 M and 1.25 M, what would be the equilibrium concentrations if 1.50 mol of n-butane is added to the equilibrium mixture in a 1.00 L vessel? The conditions for equilibrium are:

K eq = K eq =

[(CH3 )2CHCH3 ]

[CH3CH 2CH 2CH3 ] 1.25 = 2.5 0.50

Add 1.50 moles of n-butane; therefore, the total number of moles of nbutane would be 2.00 moles (1.50 moles + 0.50 mole = 2.00 moles)

0.500 mole + 1.50 moles

1.25 moles

not at equilibrium

2.00 moles-x

1.25 moles +x

at equilibrium

CH 3CH 2CH 2CH 3 ! (CH 3 )2CHCH 3 1.25 mol + x mol V K eq = 2.00 mol - x mol V 1.25 mol + x mol V 2.5 = 2.00 mol - x mol V V = 1.00 L

2.5 =

1.25 mol + x mol 2.00 mol - x mol

2.5 (2.00 - x) = 1.25 + x 5.0 -2.5 x = 1.25 + x 5.0 -1.25 = 2.5 x + x 3.75 = 3.5 x 3.75/3.5 = x 1.07 mol = x or 1.1 mol (two significant figures)

[CH 3CH 2CH 2CH 3 ] =

[(CH 3 )2 CHCH 3 ] =

(2.00-1.07) mol = 0.93 M 1.00 L

(1.25 + 1.07) mol = 2.32 M 1.00 L

The Effect of Changing the Volume 1. If the number of moles of products equals the number of moles of reactants, then a change in volume of the reaction vessel will have no affect on the equilibrium. 2. If the number of moles of product is greater than the number of moles of reactants, then an increase in volume of the reaction vessel will shift the equilibrium toward the formation of product. A decrease in the volume will shift the equilibrium toward the reactants. 3. If the number of moles of reactant is greater than the number of moles of products, then an increase in the volume of the reaction vessel will shift the equilibrium toward the reactants. A decrease in volume will shift the equilibrium toward the products. The Effect of Pressure Change on a Gaseous Reaction 1. If the number of moles of gaseous products equals the number of moles of gaseous reactants, then pressure changes have no effect on the equilibrium reaction. 2. If the number of moles of gaseous products is greater than the number of moles of gaseous reactants, then an increase in pressure shifts the equilibrium toward the reactants. A decrease in pressure shifts the equilibrium toward the products.

3. If the number of moles of gaseous reactants is greater than the number of moles of gaseous products, then an increase in pressure shifts the equilibrium toward the products. A decrease in pressure shifts the equilibrium toward the reactants.