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There are essentially two types of problems which can be analyzed by ICE tables.PowerPoint Presentation

There are essentially two types of problems which can be analyzed by ICE tables.

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There are essentially two types of problems which can be analyzed by ICE tables.

Type 1

A. The initial or equilibrium concentration of some

substances must be determined.

B. Initial or equilibrium concentrations of some substances

are given, but not both. Change is therefore treated as

an unknown (x)

C. The equilibrium constant is given.

Type 2

A. The equilibrium constant or concentration must be

determined.

B. Initial and equilibrium concentrations of at least one

substance are given so that change can be calculated

directly.

C. All other initial and equilibrium concentrations of

substances are determined directly from the table.

A analyzed Type 1 problem was reviewed in the previous Powerpoint

(Equilibrium 5 – ICE tables A). Here is a second example.

Hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas as shown: H2 + F2 → 2 HF

The equilibrium constant, Kc , for the reaction at 200 C is

1.00 x 10 2. If 4.00 mol of each reactant are added to a 2.00 L flask, what would be the concentration of all three substancesat equilibrium?

ICE Table

H2 + F2 → 2 HF

Initial 2.00 2.00 0

Change -x -x 2x

Equil (M) 2.00 -x 2.00 - x 2x

ICE Table analyzed

H2 + F2 → 2 HF

Initial 2.00 2.00 0

Change -x -x 2x

Equil (M) 2.00 -x 2.00 - x 2x

Note certain characteristics of the ICE table:

1. Only initial concentrations of some substances are

provided. The change and equilibrium values are

therefore based on an unknown value (x).

2. Since the equilibrium constant is large (1.00 x 10 2), the

value of x is significant relative to the initial

concentration (2). The equilibrium concentration of the

reactants cannot be simplified to approximately 2.

Hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas as shown: H2 + F2 → 2 HF

The equilibrium constant, Kc , for the reaction at 200 C is

1.00 x 10 2. If 4.00 mol of each reactant are added to a 2.00 L flask, what would be the concentration of all three substancesat equilibrium?

H2 + F2 → 2 HF

Initial 2.00 2.00 0

Change -x -x 2x

Equil (M) 2.00 -x 2.00 - x 2x

Kc = 1.00 x 10 2 = [[ HF ] 2= [ 2x ] 2 = [ 2x ] 2

[ H2 ] [ F2 ] [2.0 – x] [2.0 – x] [2.00-x] 2

Hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas as shown: H2 + F2 → 2 HF

The equilibrium constant, Kc , for the reaction at 200 C is

1.00 x 10 2. If 4.00 mol of each reactant are added to a 2.00 L flask, what would be the concentration of all three substancesat equilibrium?

Kc = 1.00 x 10 2 = [[ HF ] 2= [ 2x ] 2 = [ 2x ] 2

[ H2 ] [ F2 ] [2.0 – x] [2.0 – x] [2.00-x] 2

The equation is a perfect square and can be easily simplified.

10 = 2x 20 – 10x = 2x 20 = 12x x = 1.67

2-x

H fluoride gas as shown: H2 + F2 → 2 HF

Initial 2.00 2.00 0

Change -x -x 2x

Equil (M) 2.00 -x 2.00 - x 2x

Kc = 1.00 x 10 2 = [[ HF ] 2= [ 2x ] 2 = [ 2x ] 2

[ H2 ] [ F2 ] [2.0 – x] [2.0 – x] [2.00-x] 2

10 = 2x 20 – 10x = 2x 20 = 12x x = 1.67

2-x

HF= 2(1.67) = 2.79 M

H2 = 2-1.67 = 0.33 M

F2 = 2-1.67 = 0.33 M

Type 2 Problems fluoride gas as shown: H

A. The equilibrium constant or concentrations must be

determined.

B. Initial and equilibrium concentrations of some

substances are both given.

C. All other initial and equilibrium concentrations of

substances are determined directly from the table.

For the reaction: H2 + I2 → 2HI

Initial concentrations of hydrogen and iodine were both

0.4 M. At equilibrium and 100 C, the concentration of hydrogen iodine was 0.13 M. Determine the concentrations of hydrogen and iodine at equilibrium and determine the equilibrium constant for the reaction?

For the reaction: H fluoride gas as shown: H2 + I2 → 2HI

Initial concentrations of hydrogen and iodine were both 0.40 M

At equilibrium and 100 C, the concentration of hydrogen

iodide was 0.13 M. Determine the concentrations of hydrogen and iodine at equilibrium and determine the equilibrium

constant for the reaction?

H2 + I2 → 2HI

Initial 0.4 0.4 0

Change

Equil 0.13

H fluoride gas as shown: H2 + I2 → 2HI

Initial 0.4 0.4 0

Change

Equil 0.13

Note certain characteristics of the ICE table. Although not

complete, the table includes some values for initial

concentrations and some values for equilibrium conditions.

As a result, the table can be completed with no unknown

values.

Since HI went from 0 initially to 0.13 at equilibrium, the change

for HI must be 0.13.

H2 + I2 → 2HI

Initial 0.4 0.4 0

Change 0.13

Equil 0.13

The ratios of change for all substances in a reaction must match the coefficients in the balanced chemical equation.

Since HI has a change of 0.13, the change of H2 and I2 is one-half

of 0.13 or 0.065.

H2 + I2 → 2HI

Initial 0.4 0.4 0

Change - 0.065 - 0.0650.13

Equil 0.13

The equilibrium concentrations equal the sum of the initial concentrations and the change.

Since both H2 and I2 decreased by 0.065, the equilibrium

concentrations must be 0.4-0.065 = 0.335

H2 + I2 → 2HI

Initial 0.4 0.4 0

Change - 0.065 - 0.0650.13

Equil 0.335 0.335 0.13

Note that the table is completed without any unknown values.

This is because partial information was available about both

initial and equilibrium concentration.

For the reaction: H concentrations and the change.2 + I2 → 2HI

Initial concentrations of hydrogen and iodine were both 0.40 M

At equilibrium and 100 C, the concentration of hydrogen

iodide was 0.13 M. Determine the concentrations of hydrogen and iodine at equilibrium and determine the equilibrium

constant for the reaction?

H2 + I2 → 2HI

Initial 0.4 0.4 0

Change - 0.065 - 0.0650.13

Equil 0.335 0.3350.13

H2 = I2 = 0.335 M

For the reaction: H concentrations and the change.2 + I2 → 2HI

Initial concentrations of hydrogen and iodine were both 0.40 M

At equilibrium and 100 C, the concentration of hydrogen

iodide was 0.13 M. Determine the concentrations of hydrogen and iodine at equilibrium and determine the equilibrium

constant for the reaction?

H2 + I2 → 2HI

Initial 0.4 0.4 0

Change- 0.065 - 0.0650.13

Equil 0.335 0.3350.13

H2 = I2 = 0.335 M

Kc = [HI] 2 = [0.13] 2 = 0.15

[H2] [I2] [0.335] 0.335]

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