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?

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.

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.

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 \(-234\ \mathrm{kJ}\).

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.