Question

The question asks what is Using the provided table and the equation below, determine the heat of formation (in kJ/mol) for PbS.

2 PbS (s) + 3 O₂ (g) → 2 SO₂ (g) + 2 PbO (s) ∆H° = -828.4 kJ/mol

Answer 1

Step-by-step explanation:

To calculate the heat of formation (ΔHf) of PbS, we need to use Hess's law which states that the enthalpy change of a reaction is independent of the pathway between the initial and final states.

We can use the given equation and the given table of standard enthalpies of formation (ΔHf°) to calculate the ΔHf of PbS.

First, let's write the balanced chemical equation for the formation of PbS from its elements:

Pb (s) + S (s) → PbS (s)

The enthalpy change for this reaction can be expressed as:

ΔHf° (PbS) = [ΔHf° (PbO) + ΔHf° (SO₂)] - [2ΔHf° (PbS) + 3ΔHf° (O₂)]

We are given the value of ΔHf° for PbO, which is -217.3 kJ/mol. The value of ΔHf° for SO₂ is not given explicitly, but we can calculate it using the following equation:

2 S (s) + 2 O₂ (g) → 2 SO₂ (g) ΔHf° = -296.8 kJ/mol

Note that this is the reverse of the given equation, so we need to change the sign of the ΔHf° value.

Now, we can substitute the values into the equation to calculate ΔHf° (PbS):

ΔHf° (PbS) = [-217.3 kJ/mol + (-296.8 kJ/mol)] - [2ΔHf° (PbS) + 3(0 kJ/mol)]

Simplifying, we get:

ΔHf° (PbS) = -514.1 kJ/mol + 2ΔHf° (PbS)

Rearranging and solving for ΔHf° (PbS), we get:

ΔHf° (PbS) = (-514.1 kJ/mol)/(2) = -257.05 kJ/mol

Therefore, the heat of formation of PbS is -257.05 kJ/mol.

Answer 2

The heat of formation of PbS is **-1856.8 kJ/mol**.

Here's how:

**1. Identify the known and unknown values:**

* **Equation:** 2 PbS (s) + 3 O₂ (g) → 2 SO₂ (g) + 2 PbO (s) ∆H° = -828.4 kJ/mol

* **Known values:**

* ΔH° for the overall reaction: -828.4 kJ/mol

* ΔH° for O₂ (g): 0 kJ/mol

* ΔH° for SO₂ (g): -296.9 kJ/mol

* ΔH° for PbO (s): -217.3 kJ/mol

* **Unknown value:** ΔH° for PbS (s)

**2. Apply the Hess's Law equation:**

Hess's Law states that the overall enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps that make up the reaction, regardless of the path taken. In this case, we can express the heat of formation of PbS using the known enthalpies of formation of the other compounds and the enthalpy change of the given reaction:

ΔH°_f(PbS) = ΔH°(overall) + Σ nΔH°_f(products) - Σ mΔH°_f(reactants)

where:

* ΔH°_f(PbS) is the heat of formation of PbS

* ΔH°(overall) is the enthalpy change of the given reaction (-828.4 kJ/mol)

* n is the stoichiometric coefficient of each product (2 for SO₂ and 2 for PbO)

* ΔH°_f(products) are the heats of formation of the products (-296.9 kJ/mol for SO₂ and -217.3 kJ/mol for PbO)

* m is the stoichiometric coefficient of each reactant (2 for PbS and 3 for O₂)

* ΔH°_f(reactants) are the heats of formation of the reactants (0 kJ/mol for O₂)

**3. Substitute and solve:**

ΔH°_f(PbS) = -828.4 kJ/mol + (2 * -296.9 kJ/mol) + (2 * -217.3 kJ/mol) - (3 * 0 kJ/mol)

ΔH°_f(PbS) = -828.4 kJ/mol - 593.8 kJ/mol - 434.6 kJ/mol

ΔH°_f(PbS) = -1856.8 kJ/mol

Therefore, the heat of formation of PbS is **-1856.8 kJ/mol**.