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HE III Wednesday, November 30 7:00 – 8:15. 180 Bevier AQ5 Bhagi AQ6 Bhagi AQ8 Bhagi AQJ Blair. 160 English AQ2 Abadeer AQ3 Abadeer. 66 Library AQ1 Abadeer AQ4 Abadeer AQB Barrick. 101 Armory AQ7 Barrick AQ9 Barrick AQA Barrick. Conflict Exam 5:30 – 6:45 161 Noyes Lab.

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HE III

Wednesday, November 30

7:00 – 8:15

180 Bevier

AQ5 Bhagi

AQ6 Bhagi

AQ8 Bhagi

AQJ Blair

160 English

AQ2 Abadeer

AQ3 Abadeer

66 Library

AQ1 Abadeer

AQ4 Abadeer

AQB Barrick

101 Armory

AQ7 Barrick

AQ9 Barrick

AQA Barrick

Conflict Exam

5:30 – 6:45

161 Noyes Lab


Equilibrium Calculations

K

at a given temperature, it is constant

K = [products]eqp

For solution chemistry

[reactants]eqr

K = Pproductsp

For gases

Preactantsr

or

Kc = [products]eqp

[reactants]eqr

K = Kc (RT)n

R = 0.08207


Equilibrium Calculations

it changes until it = K

Q

is not constant

Q = [products]ip

For solution chemistry

[reactants]ir

Q = Pprod ip

For gases

Preact ir

if Q > K

reverse reaction

too much product initially

if Q < K

forward reaction

too much reactant initially

if Q = K

system at equilibrium


Equilibrium Calculations

Write the expression for K for the reaction:

CaCO3 (s)  CaO (s) + CO2 (g)

K = 0.0045

at 800oC

K

= PCO2

1 mol of CaCO3, 1 mol of CaO and 1 atm CO2

Q =

CaCO3

1

make more

  • forward

CaO

b) reverse

1 mol of CaCO3 and 1 atm CO2

0

Q =


Equilibrium Calculations

reaction table

ICE table

+ I2 (g)

 2HI (g)

H2 (g)

Q = 02/(0.5)2

at 453oC,

calculate K

at equilibrium,

[H2] = 0.107 M

= 0.50

– x

x = 0.393

=

(0.786)2

=

54.3

= (2x)2

K =

[HI]2eq

[I2]eq

[H2]eq

(0.50–x)

(0.50–x)

(0.107)2

[I2] (M)

[HI] (M)

[H2] (M)

Initial

0.50

0.50

0.00

Change

- x

- x

+2x

2x

0.50 – x

0.50 – x

Equilibrium


Calculations

K =

54.3

+ I2 (g)

 2HI (g)

H2 (g)

K =

x = .782

54.3 =

(.224 + 2x)2

Q =

(.224)2

= .195

< K

x = .355

(.623 – x)

(.414 – x)

(.414)

(.623)

x =

-b ±

- 57.2x

50.3x2

+ 13.96

= 0

b2 – 4ac

2a

ax2

bx

c

[H2] (M)

[HI] (M)

[I2] (M)

Initial

.623

.414

.224

Change

- x

-x

+2x

Equilibrium

.623 - x

.414 - x

.224 + 2x


Equilibrium Calculations

CH3OH (aq)

CH2O (aq) + H2 (aq)

K = 3.7 x 10-10 at 298 K

Starting with a 1.00 M solution of CH3OH,

what are the equilibrium concentrations ?

[CH2O] (M)

[H2] (M)

[CH3OH] (M)

0

Initial

1.00

0

-x

+x

+x

Change

Equilibrium

1.00 - x

x

x


Equilibrium Calculations

CH3OH (aq)

CH2O (aq) + H2 (aq)

1.92 x 10-5

x 100 =

0.00182%

1.00

1.00 – 0.0000192

= 1.00

x = 1.92 x 10-5

K = 3.7 x 10-10

= x2

= x2

1.00

1.00 - x

1.00

very small

1.00 – x

[CH2O] (M)

[H2] (M)

[CH3OH] (M)

0

Initial

1.00

0

-x

+x

+x

Change

Equilibrium

1.00 - x

x

x


N2 O4 (g)  2 NO2 (g) Kc = 4.0 x 10-7

  • If 0.1 mol of N2 O4 (g) is reacted in a 1.0 L , what

  • is the equilibrium concentration of NO2 (g) ?

  • 2.6 x 10-5M

  • 1.2 x 10-2M

  • 2.0 x 10-4M

  • 4.0 x 10-7M

  • 7.8 x 10-3M


FeSCN2+ (aq)  Fe3+ (aq) + SCN- (aq)

K = 8.3 x 10-5

Calculate the equilibrium concentrations

when starting with 2.0 M FeSCN2+

8.3 x 10-5 = x2

x2

x = 0.0129

2.0 - x

2.0

0.0129

x 100% =

0.644%

2.0

5% approximation O.K.


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