Question

\(\mathrm{HgO}\) decomposes when exposed to heat: \(2 \mathrm{HgO}(\mathrm{s}) \rightarrow 2 \mathrm{Hg}(\ell)+\mathrm{O}_{2}(\mathrm{~g}) \Delta H=181.6 \mathrm{~kJ}\) What mass of \(\mathrm{O}_{2}\) can be made with \(100.0 \mathrm{~kJ}\) ?

Short answer

17.60 g of \(\mathrm{O}_2\) can be made with 100.0 kJ.

Step-by-step solution

Understand the Given Reaction

The reaction \(2 \mathrm{HgO}(\mathrm{s}) \rightarrow 2 \mathrm{Hg}(\ell)+\mathrm{O}_{2}(\mathrm{~g}) \) is endothermic with an enthalpy change \(\Delta H = 181.6 \,\mathrm{kJ}.\) This means that 181.6 kJ of heat is required to decompose 2 moles of HgO and produce 1 mole of \(\mathrm{O}_2.\)

Find Moles of \(\mathrm{O}_{2}\) Produced

Since 181.6 kJ produces 1 mole of \(\mathrm{O}_{2},\) we need to find out how many moles are produced with 100.0 kJ. Set up the proportion:\[\frac{1 \, \text{mole}}{181.6 \mathrm{~kJ}} = \frac{x \, \text{moles}}{100.0 \mathrm{~kJ}}\]Solving for \(x:\)\[ x = \frac{100.0}{181.6} \, \text{moles} \]

Calculate Moles of \(\mathrm{O}_{2}\)

Perform the division from Step 2:\[x = 0.550 \, \text{moles of } \mathrm{O}_{2}\] This is the number of moles of \(\mathrm{O}_{2}\) produced with 100.0 kJ.

Determine Mass of \(\mathrm{O}_{2}\) Produced

Using the molar mass of \(\mathrm{O}_2,\) which is 32.00 g/mol, calculate the mass of \(\mathrm{O}_2\) produced:\[\text{Mass} = 0.550 \, \text{moles} \times 32.00 \, \text{g/mol}\]This gives:\[\text{Mass} = 17.60 \, \text{g}\]

Key concepts

Endothermic Reaction

In chemical reactions, energy plays a crucial role, often in the form of heat. An **endothermic reaction** is a type of reaction that absorbs heat from the surroundings. Essentially, this means it requires an input of energy to proceed. In the given exercise, the decomposition of mercury(II) oxide (HgO) is an endothermic reaction.
Here, ΔH = 181.6 \(\mathrm{kJ}\), meaning that 181.6 kJ of energy is needed to decompose 2 moles of mercury(II) oxide. One can imagine this as the surroundings offering up their heat to break the bonds within the \(\mathrm{HgO}\)molecules, allowing the substance to break down into its constituent parts: liquid mercury and oxygen gas.
The opposite of endothermic reactions are **exothermic reactions**, which release heat, making the surroundings warmer. But with endothermic reactions like this one, rather than releasing energy, they require it to maintain and drive the process forward.

Enthalpy Change

**Enthalpy change**, represented as \(\Delta H\),is a key concept in thermodynamics that quantifies the heat change during a chemical reaction at constant pressure. It tells us whether a substance absorbs or releases energy:

• If \(\Delta H > 0\), the reaction absorbs energy, making it endothermic.
• If \(\Delta H < 0\), the reaction releases energy, making it exothermic.

In this exercise, the \(\Delta H\) for the reaction of mercury(II) oxide decomposition is positive, at 181.6 kJ. This value indicates the energy absorbed to transform 2 moles of \(\mathrm{HgO}\) into 1 mole of oxygen gas and 2 moles of liquid mercury. Understanding this helps us predict how much energy we need to add if we desire to produce a specific amount of product, like oxygen in this case.

Molar Mass Calculation

**Molar mass calculation** is a fundamental concept in chemistry, essential for converting between mass and moles. Molar mass is the mass of one mole of a substance, given in grams per mole (g/mol). For chemical calculations, you must know the molar mass of all the involved substances.
In the mercury(II) oxide decomposition problem, we calculated how many grams of oxygen (\(\mathrm{O}_2\)) are produced from a given amount of energy. The molar mass of \(\mathrm{O}_2\) is 32.00 g/mol. Knowing this allows us to convert the moles of oxygen calculated (from the energy provided) into grams efficiently. Here are the steps we used:

• Determine the moles of \(\mathrm{O}_2\) produced using the relationship between energy and moles.
• Multiply the moles of \(\mathrm{O}_2\) by its molar mass (32.00 g/mol) to find the mass in grams.

Making these calculations turns abstract numerical data into measurable quantities we can work with in a lab setting, such as the 17.60 grams of \(\mathrm{O}_2\) the problem yields.