Problem 50 - Chemistry The Molecular Science

Question

Given the thermochemical expression $$ \begin{aligned} \mathrm{CaO}(\mathrm{s})+3 \mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{CaC}_{2}(\mathrm{~s})+\mathrm{CO}(\mathrm{g}) \\ \Delta_{\mathrm{r}} H^{\circ} &=464.8 \mathrm{~kJ} / \mathrm{mol} (a) \(2 \mathrm{~mol} \mathrm{O}_{2}\) reacts (b) \(0.115 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{3}\) forms (c) \(4.73 \mathrm{~g}\) NO reacts \end{aligned} $$ calculate the quantity of energy transferred when (a) \(34.8 \mathrm{~mol} \mathrm{CO}(\mathrm{g})\) is formed by this reaction. (b) A metric ton ( \(1000 \mathrm{~kg}\) ) of \(\mathrm{CaC}_{2}(\mathrm{~s})\) is manufactured. (c) \(0.432 \mathrm{~mol}\) carbon reacts with \(\mathrm{CaO}(\mathrm{s})\).

Step-by-Step Solution

Answer

(a) 16,178.24 kJ, (b) 7,248,081.58 kJ, (c) 66.9312 kJ of energy transferred.

Step 1: Understand the given reaction

The reaction provided is: \({\text{CaO}( ext{s}) + 3 \text{C}( ext{s}) \rightarrow \text{CaC}_2(\text{s}) + \text{CO}( ext{g})}\), with a standard reaction enthalpy change \(\Delta_r H^{\circ} = 464.8 \text{kJ/mol}.\) This means for 1 mole of CaC extsubscript{2} formed, or 1 mole of CO formed, 464.8 kJ of energy is involved.

Step 2: Calculate energy for (a) 34.8 mol CO formation

Since 1 mole of CO formation corresponds to 464.8 kJ of energy, for \(34.8\) moles of CO, the energy \(Q\) is calculated as:\[ Q = 34.8 \, \text{mol} \times 464.8 \, \text{kJ/mol} = 16178.24 \, \text{kJ} \]

Step 3: Convert metric ton to moles for (b) 1000 kg CaC₂

First, find the molar mass of \(\text{CaC}_2\). Calcium has a molar mass of \(40.08 \, \text{g/mol}\) and carbon \(12.01 \, \text{g/mol}\). Thus, \( \text{CaC}_2 = 40.08 + 2 \times 12.01 = 64.10 \, \text{g/mol}.\)\[ \text{Mass of CaC}_2 = 1000 \, \text{kg} = 1000000 \, \text{g} \]Convert to moles:\[ \text{moles of CaC}_2 = \frac{1000000 \, \text{g}}{64.10 \, \text{g/mol}} \approx 15600.62 \, \text{mol} \]

Step 4: Calculate energy for (b) calcium carbide production

Using the moles from Step 3 and energy change per mole:\[ Q = 15600.62 \, \text{mol} \times 464.8 \, \text{kJ/mol} \approx 7,248,081.58 \, \text{kJ} \]

Step 5: Calculate energy for (c) 0.432 mol carbon reaction

Consider that 3 moles of carbon are needed per reaction cycle (Equation's stoichiometry: 3 C per 1 reaction).Fraction of reaction given by \( \frac{0.432}{3} \approx 0.144. \)Therefore, the energy transferred \( Q \) is:\[ Q = 0.144 \times 464.8 \, \text{kJ} = 66.9312 \, \text{kJ} \]

Key Concepts

Reaction enthalpy

In the world of thermochemistry, reaction enthalpy (\( \Delta_{r} H^{\circ} \)) is a pivotal concept. It represents the heat change that takes place at constant pressure when a chemical reaction occurs. This value is crucial because it tells us how much energy, in the form of heat, is absorbed or released during a reaction. The standard reaction enthalpy is calculated under standard conditions, or 1 atmosphere of pressure and typically at 298 K (25 °C).

• This energy change helps us understand whether a reaction is endothermic (absorbing heat) or exothermic (releasing heat).
• An endothermic reaction has a positive \( \Delta_{r} H^{\circ} \), indicating that it absorbs heat from the surroundings.
• An exothermic reaction has a negative \( \Delta_{r} H^{\circ} \), indicating that it releases heat to the surroundings.

In this specific exercise, the reaction enthalpy is given as 464.8 \(\text{kJ/mol}\). This means that either 464.8 \(\text{kJ}\) of energy is absorbed or released per mole of the reaction occurring. Understanding this value is critical as it underpins our calculations for how energy is transferred when reactants convert to products in specified amounts.

Molar mass calculation

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms present in a compound. This is essential for converting between mass and the number of moles, which are critical units in chemistry.
An accomplished calculation example from the exercise is with calcium carbide (CaC₂). We determined its molar mass by adding:

• Calcium's atomic mass: 40.08 g/mol
• Carbon's atomic mass: 12.01 g/mol, and since CaC₂ contains two carbon atoms, we multiply this by 2

Thus, the molar mass of \( \text{CaC}_2 \) is \( 40.08 + 2 \times 12.01 = 64.10 \, \text{g/mol} \). Calculating the molar mass correctly allows you to convert the mass of a compound to moles, which is necessary for energy and stoichiometric calculations.

Stoichiometry

Stoichiometry involves the quantitative relationships between the amounts of reactants and products in a chemical reaction. It's based on the balanced chemical equation and the concept of the mole.
For example, in the complete reaction of \( \text{CaO}(\text{s}) + 3\text{C}(\text{s}) \rightarrow \text{CaC}_2(\text{s}) + \text{CO}(\text{g}) \), stoichiometry tells us:

• 1 mole of \( \text{CaO} \) reacts with 3 moles of carbon \( \text{C} \)
• The reaction produces 1 mole of \( \text{CaC}_2 \) and 1 mole of carbon monoxide \( \text{CO} \).

This stoichiometric ratio helps us determine how much of each substance is needed or produced in a reaction. Problems often require using these ratios to calculate how much product can be formed from given reactants or how much reactant is necessary to achieve a desired amount of product.

Energy calculation

Calculating energy changes in a reaction often involves understanding both reaction enthalpy and stoichiometry. The main aim is to determine the total energy absorbed or released as substances react.
Using the enthalpy change \( \Delta_{r} H^{\circ} \), and the amounts derived from stoichiometric calculations allows us to compute energy transfers. For instance:

• If you want to know the energy involved in forming 34.8 moles of CO from the given reaction, multiply the number of moles by the enthalpy change: \[ Q = 34.8 \times 464.8 \, \text{kJ/mol} = 16178.24 \, \text{kJ}. \]
• Similarly, knowing the amount of \( \text{CaC}_2 \) produced from a metric ton of reactants allows calculation of the total energy exchanged in producing such heart of moles of product.

These energy calculations provide insights into the practical limits of chemical production, cost estimation, and safety measures required for reactions and industrial applications.