Problem 43 - Chemistry: The Central Science in SI Units, Global Edition

Question

Consider the following reaction: $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) \quad \Delta H=-1204 \mathrm{~kJ} $$ (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(3.55 \mathrm{~g}\) of \(\mathrm{Mg}(s)\) reacts at constant pressure. (c) How many grams of \(\mathrm{MgO}\) are produced during an enthalpy change of \(-234 \mathrm{~kJ}\) ? (d) How many kilojoules of heat are absorbed when \(40.3 \mathrm{~g}\) of \(\mathrm{MgO}(s)\) is decomposed into \(\mathrm{Mg}(s)\) and \(\mathrm{O}_{2}(g)\) at constant pressure?

Step-by-Step Solution

Answer

(a) The reaction is exothermic since the enthalpy change is negative. (b) When 3.55 g of Mg reacts, the heat transferred is -88.1 kJ. (c) During an enthalpy change of -234 kJ, 7.819 g of MgO is produced. (d) The energy absorbed when 40.3 g of MgO decomposes is 602 kJ.

Step 1: (a) Identify the type of reaction

First, we need to determine whether the reaction is exothermic or endothermic. An exothermic reaction is one that releases heat and has a negative enthalpy change, while an endothermic reaction is one that absorbs heat and has a positive enthalpy change. In this case, the reaction has an enthalpy change of \(\Delta H = -1204\ \mathrm{kJ}\), which is negative. Therefore, the reaction is exothermic.

Step 2: (b) Heat transfer calculation

To calculate the amount of heat transferred when \(3.55\ \mathrm{g}\) of \(\mathrm{Mg}\) reacts, we need to determine the number of moles of \(\mathrm{Mg}\) present and use the stoichiometry of the equation to find the energy released. 1. Determine the molar mass of \(\mathrm{Mg}\): \[M_{\mathrm{Mg}} = 24.305\ \mathrm{g/mol}\] 2. Calculate the number of moles of \(\mathrm{Mg}\): \[n_{\mathrm{Mg}} = \frac{3.55\ \mathrm{g}}{24.305\ \mathrm{g/mol}} = 0.146\ \mathrm{mol}\] 3. Observe the stoichiometry: \(2\) moles of \(\mathrm{Mg}\) react to release \(-1204\ \mathrm{kJ}\) of energy. 4. Calculate the energy released: \[q_{\mathrm{Mg}} = 0.146\ \mathrm{mol} \times \frac{-1204\ \mathrm{kJ}}{2\ \mathrm{mol}} = -88.1\ \mathrm{kJ}\] So the amount of heat transferred when \(3.55\ \mathrm{g}\) of \(\mathrm{Mg}\) reacts is \(-88.1\ \mathrm{kJ}\). The negative sign indicates that the heat is released.

Step 3: (c) Mass of MgO produced

We need to find the mass of \(\mathrm{MgO}\) produced when \(-234\ \mathrm{kJ}\) of energy is released. 1. Calculate the number of moles of \(\mathrm{Mg}\) that react: \[n_{\mathrm{Mg}} = \frac{-234\ \mathrm{kJ}}{-1204\ \mathrm{kJ/mol} \times \frac{1\ \mathrm{mol\ Mg}}{2\ \mathrm{mol}}} = 0.194\ \mathrm{mol\ Mg}\] 2. Since the stoichiometry of the reaction shows \(1\) mole of \(\mathrm{Mg}\) produces \(1\) mole of \(\mathrm{MgO}\), the moles of \(\mathrm{MgO}\) produced are equal to the moles of \(\mathrm{Mg}\) reacting: \[n_{\mathrm{MgO}} = 0.194\ \mathrm{mol}\] 3. Determine the molar mass of \(\mathrm{MgO}\): \[M_{\mathrm{MgO}} = (24.305 + 16.00)\ \mathrm{g/mol} = 40.305\ \mathrm{g/mol}\] 4. Calculate the mass of \(\mathrm{MgO}\) produced: \[m_{\mathrm{MgO}} = n_{\mathrm{MgO}} \times M_{\mathrm{MgO}} = 0.194\ \mathrm{mol} \times 40.305\ \mathrm{g/mol} = 7.819\ \mathrm{g}\] Therefore, \(7.819\ \mathrm{g}\) of \(\mathrm{MgO}\) are produced during an enthalpy change of \(-234\ \mathrm{kJ}\).

Step 4: (d) Energy absorbed in MgO decomposition

We are given that \(40.3\ \mathrm{g}\) of \(\mathrm{MgO}\) is decomposed into \(\mathrm{Mg}\) and \(\mathrm{O}_{2}\) at constant pressure. We need to find how much energy is absorbed by the process. 1. Calculate the number of moles of \(\mathrm{MgO}\) present: \[n_{\mathrm{MgO}} = \frac{40.3\ \mathrm{g}}{40.305\ \mathrm{g/mol}} = 1.00\ \mathrm{mol}\] 2. Since the stoichiometry of the reaction shows that \(1\) mole of \(\mathrm{MgO}\) decomposes with the absorption of \(1204\ \mathrm{kJ}/2\ \mathrm{mol}\), we can calculate the energy absorbed: \[q_{\mathrm{MgO}} = 1.00\ \mathrm{mol\ MgO} \times \frac{1204\ \mathrm{kJ}}{2\ \mathrm{mol}} = 602\ \mathrm{kJ}\] Hence, \(602\ \mathrm{kJ}\) of heat is absorbed when \(40.3\ \mathrm{g}\) of \(\mathrm{MgO}\) is decomposed into \(\mathrm{Mg}\) and \(\mathrm{O}_{2}\) at constant pressure.

Key Concepts

Exothermic Reactions

In an exothermic reaction, energy is released in the form of heat to the surroundings. This happens because the total energy of the products is lower than the total energy of the reactants. The hallmark of an exothermic process is a negative change in enthalpy (\(\Delta H\)), indicating that heat is being released. In the example of \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\), the enthalpy change is \(-1204 \mathrm{~kJ}\). This negative sign tells us that the reaction is exothermic. When a reaction is exothermic, the container or surrounding area feels warm because heat is escaping from the system. Exothermic reactions are quite common and include processes like combustion, respiration, and many oxidation reactions. In summary, if you see a negative \(\Delta H\), you know the reaction releases heat.

Stoichiometry

Stoichiometry involves using balanced chemical equations to predict the quantities of reactants and products involved in a chemical reaction. The central idea is to use the mole ratio given by the coefficients in the balanced equation.For the reaction \(2 \mathrm{Mg} + \mathrm{O}_{2} \rightarrow 2 \mathrm{MgO}\), the coefficients tell us that 2 moles of \(\mathrm{Mg}\) react with 1 mole of \(\mathrm{O}_{2}\) to form 2 moles of \(\mathrm{MgO}\). Stoichiometry allows us to compute unknown quantities:

• If you know the amount of one substance, you can find the amount of another using the mole ratio.
• In this example, to calculate the heat released, we find the number of moles of \(\mathrm{Mg}\) reacting and then use the stoichiometric coefficients to determine associated energy changes.

This principle of stoichiometry is foundational in predicting how much product is formed or how much reactant is consumed in a reaction.

Heat Transfer

In chemical reactions, heat transfer differs between exothermic and endothermic processes. In exothermic reactions, heat is transferred from the chemical reaction to the surroundings, signifying energy release.Heat transfer can be expressed in terms of enthalpy (\(\Delta H\)). For instance, if \(3.55 \mathrm{~g}\) of \(\mathrm{Mg}\) is used, we can determine the amount of heat (\(q\)) transferred based on enthalpy and stoichiometry:1. Calculate the moles of \(\mathrm{Mg}\).2. Use the stoichiometry to find the energy released per mole and multiply this by the moles of \(\mathrm{Mg}\).The calculation for heat transfer shows that a scientific understanding of heat is crucial to grasp how energy is absorbed or released in reactions. This process is significant because it can determine whether a reaction is favorable or practical under certain conditions.

Molar Mass Calculation

Molar mass calculation is an essential skill in chemistry, as it links the mass of a substance to its moles. Every element has an atomic mass (found on the periodic table), which, when molar quantities are scaled to grams, becomes the molar mass.To calculate molar mass:

• Identify the elements present and their quantities in the compound. For example, \(\mathrm{MgO}\) contains magnesium and oxygen.
• Add the atomic masses of \(\mathrm{Mg}\) (24.305 \mathrm{~g/mol}) and \(\mathrm{O}\) (16.00 \mathrm{~g/mol}) to find \(\mathrm{MgO}\)'s molar mass as 40.305 \mathrm{~g/mol}.

Knowing how to compute the molar mass allows for conversions between grams and moles, essential for calculating reactants and products in a reaction. It simplifies the execution of stoichiometric calculations and facilitates the understanding of chemical processes on a quantifiable level.