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?

Short Answer

Expert verified
(a) The reaction is exothermic, as ΔH has a negative value (-1204 kJ). (b) The heat transferred when 3.55g of Mg reacts is -88.9 kJ. (c) 15.7 grams of MgO are produced during an enthalpy change of -234 kJ. (d) 602 kilojoules of heat are absorbed when 40.3g of MgO is decomposed into Mg(s) and O₂(g) at constant pressure.

Step by step solution

01

(a) Identify reaction type

To determine whether the reaction is exothermic or endothermic, we examine the sign of the enthalpy change, ΔH. If the sign is negative, the reaction is exothermic (releases heat), and if the sign is positive, the reaction is endothermic (absorbs heat). The given ΔH is -1204 kJ, which is negative, so the reaction is exothermic.
02

(b) Calculate heat when 3.55g Mg reacts

First, convert mass of Mg to moles using its molar mass (24.31 g/mol): \(moles \, Mg = \frac{3.55 \,g}{24.31 \frac{g}{mol}} = 0.146 \,mol\) Next, using the balanced equation and ΔH, we can calculate the heat transferred during the reaction of 0.146 moles of Mg. For the balanced equation, 2 moles of Mg react to release 1204 kJ of heat. Therefore, \(heat \, transferred = 0.146\, moles \times \frac{1204 \,kJ}{2 \,moles} = -88.9 \,kJ\) The heat transferred is -88.9 kJ (negative sign indicates heat released, as the reaction is exothermic).
03

(c) Calculate MgO mass with ΔH = -234 kJ

First, find moles of Mg that reacted to produce ΔH of -234 kJ: \(moles \, Mg = \frac{-234 \,kJ}{-1204 \,\frac{kJ}{2 \, moles}} = 0.389 \, moles\) Now, using stoichiometry, we have 1 mole of MgO produced for one mole of Mg reacted: \(moles \, MgO = 0.389 \, moles\) Finally, calculate mass of MgO produced using its molar mass (40.31 g/mol): \(mass \, MgO = 0.389 \, moles \times 40.31 \frac{g}{mol} = 15.7 \,g\) Thus, 15.7 grams of MgO are produced during an enthalpy change of -234 kJ.
04

(d) Calculate heat with 40.3g MgO decomposed

First, convert mass of MgO to moles using its molar mass (40.31 g/mol): \(moles \, MgO = \frac{40.3\,g}{40.31 \frac{g}{mol}} = 1 \, mol\) Next, using the balanced equation and ΔH, we can calculate the heat absorbed during the decomposition of 1 mole of MgO. We reverse the given reaction to obtain the decomposition reaction: $$ 2 \mathrm{MgO}(s) \longrightarrow 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \quad \Delta H=1204 \mathrm{~kJ} $$ For the decomposition reaction, 2 moles of MgO decompose into Mg(s) and O₂(g) with an absorption of 1204 kJ of heat. Therefore, \(heat \, absorbed = 1\, mole \times \frac{1204 \,kJ}{2 \,moles} = 602 \,kJ\) Thus, 602 kilojoules of heat are absorbed when 40.3g of MgO is decomposed into Mg(s) and O₂(g) at constant pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exothermic and Endothermic Reactions
Understanding the difference between exothermic and endothermic reactions is crucial for students studying thermochemistry. An exothermic reaction is one that releases energy, usually in the form of heat, to its surroundings. The signature of such a reaction is a negative value for the enthalpy change \( \Delta H \), such as the reaction where magnesium \( \mathrm{Mg} \) combines with oxygen \( \mathrm{O}_{2} \) to form magnesium oxide \( \mathrm{MgO} \). In contrast, an endothermic reaction absorbs energy from the surroundings, indicated by a positive \( \Delta H \). This fundamental concept helps students predict the energy flow during chemical reactions and is important in fields such as environmental science, engineering, and materials science.

Enthalpy, represented by \( H \), is essentially the total energy of a compound. When discussing exothermic and endothermic reactions, it is the change in enthalpy \( \Delta H \) that is of interest. Knowing whether a reaction is exothermic or endothermic can help predict whether a reaction will occur spontaneously and what effects it might have on its environment - for instance, in the design of hand warmers or cold packs. So remember, exothermic reactions give off heat (\(-\Delta H\)), and endothermic reactions absorb heat (\(+\Delta H\)).
Enthalpy Change Calculations
The calculation of enthalpy change lies at the heart of thermochemistry problems, allowing us to measure the heat involved in a chemical reaction. Enthalpy change, \( \Delta H \), indicates the amount of heat absorbed or released when a substance undergoes a chemical reaction at constant pressure. It is usually expressed in kilojoules per mole \( \mathrm{kJ/mol} \).

When striving for a comprehensive understanding, students should be comfortable using the stoichiometry of the reaction to relate the \( \Delta H \) value mentioned in the balanced chemical equation to the actual amount of reactants or products. For example, in the provided exercise, our reactant magnesium \( \mathrm{Mg} \) reacts with oxygen to produce magnesium oxide \( \mathrm{MgO} \) with a \( \Delta H \) of -1204 \( \mathrm{kJ} \). To find the heat transferred when a different mass of \( \mathrm{Mg} \) reacts, we convert this mass to moles and then use the given \( \Delta H \) value, proportionally scaling it based on the stoichiometry of the balanced equation. This application of enthalpy change is crucial for fields such as energy management, food chemistry, and any process or industry where heat transfer plays a role.

Applying the concepts accurately is key: for each mole of reactant or product, there is a specific amount of heat transferred. This relationship is the cornerstone of solving any enthalpy change calculation.
Stoichiometry in Chemical Reactions
Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It's like a recipe for chemistry - telling us how much of each reactant we need to produce a certain amount of product. To master stoichiometry, one must be proficient in balancing chemical equations, interpreting mole ratios, and converting between grams, moles, and molecules.

For instance, the reaction presented in the exercise highlights the stoichiometry between magnesium \( \mathrm{Mg} \) and magnesium oxide \( \mathrm{MgO} \) - for every two moles of \( \mathrm{Mg} \) that react, two moles of \( \mathrm{MgO} \) are produced. When calculating how many grams of \( \mathrm{MgO} \) are produced from a heat change, we use the molar mass and stoichiometric coefficients to find the moles of \( \mathrm{MgO} \), and then convert it to grams.

Stoichiometry also helps in understanding limiting reactants and yields, important for industrial applications like pharmaceutical manufacturing and materials science. It's a versatile tool that is essential in predicting the outcomes of reactions and in scaling up processes from the laboratory to commercial production. Always remember that stoichiometry is based on the conservation of mass and the balanced chemical equation - every atom of the reactants must be accounted for in the products.

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Most popular questions from this chapter

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