Question

Equilibrium and Le Chateleir's Principal

Instructions: Write the equilibrium expression for each chemical equation.

1. 2 H2Sg⇌2H2g+S2g
2. NH4HSs⇌NH3g+H2Sg

Oxidation-Reduction Reactions

Instructions: Assign an oxidation state to each element, ion, or molecule.

1. Cl^-   
2. Br₂
3. K  
4. Ca²⁺

Instructions: Assign an oxidation state to each atom in each compound.

5. NaCl $biQAAAAASUVORK5CYII=$
6. CH4
7. SbCl5 $1NN0AAAAASUVORK5CYII=$
8. Al2O3

Instructions: Assign an oxidation state to each atom in each polyatomic ion.

9. NH4+
10. Cl04-
11. Cr2072- $cVKNTpnQMAAAAASUVORK5CYII=$   
12. Mn04-

Instructions: Identify the oxidation state of P in each ion.

13. PO43- $g87TAFVyASmWwAAAABJRU5ErkJggg==$    
14. HPO42- $wPXnKhLP2UwaAAAAAASUVORK5CYII=$
15. H2PO4- $ZdTus3+QIquxe42GQAAAAASUVORK5CYII=$
16. H3PO4 $9oAAAAASUVORK5CYII=$   

Answer 1

i) 2 H2S(g) ====== 2H2(g) + S2(g)

Keq = [H2(g)]^2*[S2(g)]/[H2S(g)]^2

ii) NH4HS(S) ======= NH3(g) + H2S(g)

Keq = [NH3(g)]* [H2S(g)]

iii) oxidation state Cl in Cl- is -1

oxidation state of Br in Br2 is 0

oxidation state of K is 0

oxidation state of Ca in Ca2+ is +2

iv) In NaCl oxidation state of Na is +1 and that of Cl is -1

In CH4 oxidation state of C is -4 and that of H is +1

In SbCl5 oxidation state of Sb is +5 and that of Cl is -1

In Al2O3 oxidation state of Al is +3 and that of O is -2

v) In NH4+ oxidation state of N is -3 and that of H is +1

In ClO4- oxidation state of Cl is +7 and that of O is -2

In Cr2O7 2- oxidation state of Cr is +6 and that of O is -2

In MnO4- oxidation state of Mn is +7 and that of O is -2

vi) In PO43- oxidation state of P is +5

In HPO42- it is +5

In H2PO4- it is +5

In H3PO4 it is +5