Applied Le Chatelieir's Principle Lesson 4

Applied Le Chatelieir's

Principle

Lesson 4

Lab 19A

Le Chatelier’s Principle

Copy the five Equilibrium Equations into your lab handout

Part I

Thymol Blue (blue) + H+ ⇌ Thymol Blue (yellow)

Fe3+(aq) + SCN-

(aq) ⇌ FeSCN2+(aq)

yellow brown

Fe3+(aq) + 3OH- → Fe(OH)3(s)

Part II

Co(H2O)62+ + 2Cl- ⇌ Co(H2O)4Cl2

2- + 2H2O(l)

pink blue

Part III

Energy + 2CrO42- + 2H+ ⇌ Cr2O7

2- + H2O(l)

yellow orange

Ba2+ + CrO42- →

BaCrO4(s)

Part IV

Cu(H2O)42+ + 2NH3

⇌ Cu(OH)2(s) + 2NH4+ + 2H2O

light blue powder blue

Cu(OH)2(s) + 4NH3 ⇌ Cu(NH3)42+ + 2OH-

powder blue violet

Part V

Application of Le Chatelier’s Principle

Energy + 2CrO42- + 2H+ ⇄ Cr2O7

2- + H2O(l)

yellow orangeKnow the following:

Any acid is H+

Any base is OH-

Acids react with bases H+ + OH- → H2O(l)

Stress Shift [CrO42-] [Cr2O7

2-] colour

Add H+

Add CrO42-

Add Cr2O72-



2- + H2O(l)


Any acid is H+

Any base is OH-



2-] colour

Add H+

Add CrO42-

Add Cr2O72-



2- + H2O(l)


Any acid is H+

Any base is OH-



2-] colour

Add H+

Add CrO42-

Add Cr2O72-



2- + H2O(l)


Any acid is H+

Any base is OH-



2-] colour

Add H+ right decrease increase orange

Add CrO42-

Add Cr2O72-



2- + H2O(l)


Any acid is H+

Any base is OH-



2-] colour


Add CrO42- right increase increase ?

Add Cr2O72-

Application of Le Chatelier’s Principle-1


2- + H2O(l)


Any acid is H+

Any base is OH-



2-] colour


Add CrO42- right increase increase ?

Add Cr2O72- left increases increases ?



2- + H2O(l)

yellow orange



2-] colour



2- + H2O(l)

yellow orange



2-] colour

Add NaOH



2- + H2O(l)

yellow orange


NaOH → Na+ + OH-


2-] colour

Add NaOH



2- + H2O(l)

yellow orange


NaOH → Na+ + OH-

spectator ion stress ion- lowers H+


2-] colour

Add NaOH



2- + H2O(l)

yellow orange


NaOH → Na+ + OH-

spectator ion stress ion- lowers H+


2-] colour

Add NaOH left increases decreases yellow



2- + H2O(l)

yellow orange

BaCrO4 has low solubilityBaCr2O7 has high solubility


2-] colour



2- + H2O(l)

yellow orange



2-] colour

Ba(NO3)2



2- + H2O(l)

yellow orange

BaCrO4 has low solubility- Ba2+ reacts with CrO42-

BaCr2O7 has high solubility- no reaction

Ba(NO3)2 → Ba2+ + 2NO3-


2-] colour

Ba(NO3)2



2- + H2O(l)

yellow orange


Ba(NO3)2 → Ba2+ + 2NO3-

stress spectator


2-] colour

Ba(NO3)2



2- + H2O(l)

yellow orange ↓ Ba2+ reacts with this ion


Ba(NO3)2 → Ba2+ + 2NO3-

stress spectator


2-] colour

Ba(NO3)2



2- + H2O(l)

yellow orange ↓ Ba2+ reacts with this ion


Ba(NO3)2 → Ba2+ + 2NO3-

stress spectator


2-] colour

Ba(NO3)2 left decreases decreases yellow



2- + H2O(l)

yellow orange


2-] colour

HCl



2- + H2O(l)

yellow orange

HCl → H+ + Cl-


2-] colour

HCl



2- + H2O(l)

yellow orange

HCl → H+ + Cl-

stress spectator


2-] colour

HCl



2- + H2O(l)

yellow orange

HCl → H+ + Cl-

stress spectator


2-] colour

HCl right decreases increases orange

Application of Le Chatelier’s Principle- 2

Energy + Co(H2O)62+ + 2Cl- ⇄ Co(H2O)4Cl2

2- + 2H2O(l)

pink blue

Stress stress ion spectator ion shift colour

HCl



2- + 2H2O(l)

pink blue

HCl → H+ + Cl-


HCl



2- + 2H2O(l)

pink blue

HCl → H+ + Cl-

spectator stress


HCl


Co(H2O)62+ + 2Cl- ⇄ Co(H2O)4Cl2 + 2H2O(l)

pink blue

HCl → H+ + Cl-

spectator stress


HCl Cl- H+ right blue



2- + 2H2O(l)

pink blue


Increase temperature



2- + 2H2O(l)

pink blue


Increase temperature



2- + 2H2O(l)

pink blue


↑ temp none none right blue



2- + 2H2O(l)

pink blue


AgNO3



2- + 2H2O(l)

pink blue

AgNO3 → Ag+ + NO3-


AgNO3



2- + 2H2O(l)

pink blue


↑ stress- low solubility (data book-4) with Cl-


AgNO3



2- + 2H2O(l)

pink blue


↑ stress- low solubility (data book-4) with Cl-


AgNO3 Ag+ NO3- left pink

Here is a the first question from your lab. I will give you a good answer so that you understand the level of detail that is required in the response

1. Explain the colour changes observed when Fe(NO3)3, KSCN, and

NaOH were added to the thiocyanatoiron equilibrium system. Start by writing the equilibrium equation.

Fe3+(aq) + SCN-

(aq) ⇄ Fe(SCN)2+(aq)

yellow brown

Adding Fe(NO3)3 increases the [Fe3+] and causes the reaction to shift

right and turn brown. Adding KSCN increases the [SCN-] and causes the reaction to shift right and turn brown. Adding NaOH increases the [OH-] which reacts with Fe3+, lowering the [Fe3+] causing the equilibrium to shift left and turn yellow.

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Applied Le Chatelieir's Principle Lesson 4

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