The lead-acid storage battery is the oldest rechargeable battery in existence. It was invented in 1859 by French physician Gaston Plante and still retains application today, more than 150 years later.

There are two reactions that take place during discharge of the lead-acid storage battery. In one step, sulfuric acid decomposes to form sulfur trioxide and water:

H2SO4*(l)* → SO3*(g)* + H2O*(l)* = Δ*H*+113.kJ

In another step, lead, lead(IV) oxide, and sulfur trioxide react to form lead(II) sulfate:

Pb*(s)* + PbO2*(s)* + 2SO3*(g)* → 2PbSO4*(s)* = Δ*H*−775.kJ

Calculate the net change in enthalpy for the formation of **one mole** of lead(II) sulfate from lead, lead(IV) oxide, and sulfuric acid from these reactions.
Round your answer to the nearest kJ.

To calculate the net change in enthalpy for the formation of **one mole** of lead(II) sulfate from lead, lead(IV) oxide, and sulfuric acid, we need to analyze the given reactions and combine their enthalpy changes.

### Step 1: Write down the reactions and their enthalpy changes.
1. **Decomposition of sulfuric acid:**
   \[
   \text{H}_2\text{SO}_4(l) \rightarrow \text{SO}_3(g) + \text{H}_2\text{O}(l) \quad \Delta H = +113 \, \text{kJ}
   \]
   
2. **Reaction forming lead(II) sulfate:**
   \[
   \text{Pb}(s) + \text{PbO}_2(s) + 2\text{SO}_3(g) \rightarrow 2\text{PbSO}_4(s) \quad \Delta H = -775 \, \text{kJ}
   \]

### Step 2: Consider the formation of **one mole** of lead(II) sulfate.
The second reaction produces **two moles** of PbSO₄, but we need the enthalpy change for the formation of **one mole** of PbSO₄. Therefore, we'll divide the enthalpy change of this reaction by 2:
\[
\Delta H = \frac{-775 \, \text{kJ}}{2} = -387.5 \, \text{kJ}
\]

### Step 3: Combine the enthalpy changes.
To calculate the overall enthalpy change for the formation of **one mole** of PbSO₄, we need to combine the enthalpy change of the sulfuric acid decomposition and half of the second reaction's enthalpy:

\[
\Delta H_{\text{net}} = +113 \, \text{kJ} + (-387.5 \, \text{kJ})
\]

\[
\Delta H_{\text{net}} = -274.5 \, \text{kJ}
\]

### Step 4: Round to the nearest kJ.
\[
\Delta H_{\text{net}} = -275 \, \text{kJ}
\]

### Final Answer:
The net change in enthalpy for the formation of one mole of lead(II) sulfate is **−275 kJ**.

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Do you need further clarification or details?

Here are 5 related questions you may find interesting:
1. How does sulfuric acid contribute to the operation of a lead-acid battery?
2. What is the role of lead(IV) oxide in the lead-acid battery?
3. How does the enthalpy change affect the efficiency of the battery?
4. Can we calculate the overall energy provided by a lead-acid battery?
5. How does the lead-acid battery compare to more modern batteries in terms of efficiency and energy density?

**Tip:** In thermochemistry, the sign of enthalpy (\( \Delta H \)) indicates whether the reaction is exothermic (negative) or endothermic (positive).

The lead-acid storage battery is the oldest rechargeable battery in existence. It was invented in 1859 by French physician Gaston Plante and still retains application today, more than 150 years later.

There are two reactions that take place during discharge of the lead-acid storage battery. In one step, sulfuric acid decomposes to form sulfur trioxide and water:

H2SO4*(l)* → SO3*(g)* + H2O*(l)* = ΔH +113.kJ

In another step, lead, lead(IV) oxide, and sulfur trioxide react to form lead(II) sulfate:

Pb*(s)* + PbO2*(s)* + 2SO3*(g)* → 2PbSO4*(s)* = ΔH −775.kJ

Calculate the net change in enthalpy for the formation of one mole of lead(II) sulfate from lead, lead(IV) oxide, and sulfuric acid from these reactions. Round your answer to the nearest kJ.

Learn how to calculate the net change in enthalpy for the formation of one mole of lead(II) sulfate from lead, lead(IV) oxide, and sulfuric acid. This problem involves concepts from thermochemistry, including stoichiometry and enthalpy changes, suitable for high school students in Grades 10-12.

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