Question

Chemistry! Help! Please!

1. Consider the following equation for an equilibrium system: (attached)
Which concentration(s) would be included in the denominator of the equilibrium constant expression?
a. Pb(s), CO2(g), and SO2(g)
b. PbS(s), O2(g), and C(s)
c. O2(g), Pb(s), CO2(g), and SO2(g)
d. O2(g)

2. The oxidation number of the sulfur atom in the SO^2- 4 ion is
a. +2.
b. -2.
c. +6.
d. +4.

3. In the following reaction, which is the oxidizing agent? (attached - reaction starts with AgNO2)
a. AgNO2
b. Cl2
c. KOH
d. KCl

Answer 1

1. Equilibrium expression

Answer:

d. O₂(g)

Explanation:

2PbS(s) + 3O₂(g) + C(s) ⇌ 2Pb(s) + CO₂(g) +2SO₂(g)

The general equilibrium constant expression is

`K_(eq) = \frac{\text{Products}}{\text{Reactants}}`

However, solids are not included in the expression because their activity (concentration) does not change.

For this reaction, the expression is

`K_(eq) = \frac{[\text{CO}_(2)][\text{SO}_(2)]^(2)}{ [\text{O}_(2)]^(3)}`

The only concentration term in the denominator is that of O₂(g).

2. Oxidation number

Answer:

c. +6

Explanation:

You must follow several rules to determine oxidation numbers.

The important rules for this question are:

1. The oxidation number of oxygen in a compound is usually -2.
2. The sum of all oxidation numbers in a neutral compound must equal zero.

Per Rule 1, the oxidation number of O is -2.

Write the oxidation number above the O in the formula:

`[\text{S}\stackrel{\hbox{-2}}{\hbox{O}}_(4)]^(2-)`

There are four O atoms, so their total oxidation number is -8.

Per Rule 2, the total oxidation number of S must be +6 to make the charge on the ion -2.

There is only one S atom, so it must have an oxidation number of +6.

`[\stackrel{\hbox{+6}}{\hbox{S}}\stackrel{\hbox{-2}}{\hbox{O}}_(4)]^(2-)`

3. Oxidizing agent

Answer:

b. Cl₂

Explanation:

The oxidizing agent is the substance that is reduced.

Reduction is a gain of electrons, i.e., a decrease in oxidation number.

We must identify the oxidation numbers of all atoms in the equation and discover which ones change.

I will give only the oxidation numbers that change.

`\text{Ag}\stackrel{\hbox{+3}}{\hbox{N}}\text{O}_(2) + \stackrel{\hbox{0}}{\hbox{Cl}}_(2) + \text{ 2KOH} \rightarrow \text{Ag}\stackrel{\hbox{+5}}{\hbox{N}}\text{O}_(3) + \text{2K}\stackrel{\hbox{-1}}{\hbox{Cl}} + \text{ H$_(2)$O}`

The oxidation number of N increases from +3 to +5, and that of Cl decreases from 0 to -1.

So, Cl₂ is the substance that is reduced and AgNO₂ is the substance oxidized.

Cl₂ is the oxidizing agent.