Introduction to Enthalpy of Reaction
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AP Chemistry › Introduction to Enthalpy of Reaction
The combustion of methane is represented by $\text{CH}_4(g)+2\text{O}_2(g)\rightarrow \text{CO}_2(g)+2\text{H}_2\text{O}(l)$ with $\Delta H=-890\ \text{kJ}$. Which statement is correct?
Heat is released to the surroundings, so $\Delta H$ is negative for the reaction.
Heat is absorbed by the surroundings, so $\Delta H$ is negative for the reaction.
Heat is released by the surroundings, so $\Delta H$ is positive for the reaction.
No heat is transferred, so $\Delta H$ is zero for the reaction.
Heat is absorbed from the surroundings, so $\Delta H$ is positive for the reaction.
Explanation
This question tests the skill of relating a negative ΔH for combustion to the concepts of exothermicity and heat flow. The combustion reaction has ΔH = -890 kJ, signifying an exothermic process where the system releases a large amount of heat to the surroundings. This is typical for combustion reactions, where bonds in the products are stronger, leading to energy release. Thus, choice C correctly interprets that heat is released to the surroundings, so ΔH is negative for the reaction. A tempting distractor is choice D, which states heat is absorbed from the surroundings so ΔH is positive, but this is incorrect due to the misconception that combustion is endothermic. A transferable strategy is to recognize that most combustion reactions are exothermic with negative ΔH, involving heat release from the system.
A reaction is observed to warm the container it is in, and the process is occurring at constant pressure. Which statement about $\Delta H$ is correct?
Heat is absorbed by the surroundings, so $\Delta H$ is negative for the reaction.
Heat is released by the surroundings, so $\Delta H$ is positive for the reaction.
No heat is transferred, so $\Delta H$ is zero for the reaction.
Heat is released to the surroundings, so $\Delta H$ is negative for the reaction.
Heat is absorbed from the surroundings, so $\Delta H$ is positive for the reaction.
Explanation
This question tests the skill of deducing ΔH from temperature increase at constant pressure. The container warming indicates heat release to the surroundings, typical of an exothermic reaction with negative ΔH. This heat transfer raises the temperature of the surroundings. Therefore, choice C correctly states that heat is released to the surroundings, so ΔH is negative for the reaction. A tempting distractor is choice A, which says heat is absorbed from the surroundings so ΔH is positive, but this is incorrect because it misinterprets warming as absorption, a common observational misconception. A transferable strategy is to interpret warming of the surroundings as evidence of an exothermic reaction with negative ΔH for the system.
For the reaction $\text{CO}_2(s)\rightarrow \text{CO}_2(g)$ (sublimation of dry ice), $\Delta H=+26\ \text{kJ}$. Which statement is correct?
Heat is absorbed by the surroundings, so $\Delta H$ is negative for the process.
Heat is absorbed from the surroundings, so $\Delta H$ is positive for the process.
No heat is transferred, so $\Delta H$ is zero for the process.
Heat is released to the surroundings, so $\Delta H$ is negative for the process.
Heat is released by the surroundings, so $\Delta H$ is positive for the process.
Explanation
This question tests the skill of relating positive ΔH to phase change processes like sublimation. The sublimation has ΔH = +26 kJ, indicating endothermicity as the system absorbs heat from the surroundings to change CO₂ from solid to gas. This absorption overcomes intermolecular forces. Therefore, choice B correctly states that heat is absorbed from the surroundings, so ΔH is positive for the process. A tempting distractor is choice A, which claims heat is released to the surroundings so ΔH is negative, but this is incorrect due to the misconception that sublimation releases energy like condensation. A transferable strategy is to associate endothermic phase changes (like sublimation) with positive ΔH and heat absorption.
For the reaction $\text{Ba(OH)}_2\cdot 8\text{H}_2\text{O}(s)+2\text{NH}_4\text{SCN}(s)\rightarrow \text{Ba(SCN)}_2(aq)+2\text{NH}_3(aq)+10\text{H}_2\text{O}(l)$, the process is described as endothermic. Which statement about $\Delta H$ is correct?
No heat is transferred, so $\Delta H$ is zero for the reaction.
Heat is released by the surroundings, so $\Delta H$ is positive for the reaction.
Heat is absorbed by the surroundings, so $\Delta H$ is negative for the reaction.
Heat is released to the surroundings, so $\Delta H$ is negative for the reaction.
Heat is absorbed from the surroundings, so $\Delta H$ is positive for the reaction.
Explanation
This question tests the understanding of enthalpy change (ΔH) in chemical reactions, specifically relating the term 'endothermic' to the sign of ΔH and heat flow. The reaction is described as endothermic, meaning the system absorbs heat from the surroundings, resulting in a positive ΔH value. Endothermic processes require energy input, which increases the system's enthalpy. Therefore, the correct statement is that heat is absorbed from the surroundings, so ΔH is positive for the reaction. A tempting distractor might state that heat is absorbed by the surroundings with a negative ΔH, which incorrectly equates surroundings' heat gain with endothermic behavior for the system. To approach similar problems, define terms from the system's viewpoint: endothermic means positive ΔH and heat absorption by the system.
For the reaction $\text{KBr}(l)\rightarrow \text{KBr}(s)$ (freezing), $\Delta H=-20\ \text{kJ}$. Which statement is correct?
Heat is absorbed by the surroundings, so $\Delta H$ is negative for the process.
No heat is transferred, so $\Delta H$ is zero for the process.
Heat is released to the surroundings, so $\Delta H$ is negative for the process.
Heat is released by the surroundings, so $\Delta H$ is positive for the process.
Heat is absorbed from the surroundings, so $\Delta H$ is positive for the process.
Explanation
This question tests the understanding of enthalpy change (ΔH) in chemical reactions, specifically identifying whether a phase change is exothermic or endothermic based on the sign of ΔH. The given freezing process has ΔH = -20 kJ, indicating an exothermic process where the system releases heat to the surroundings. Freezing releases energy as intermolecular forces strengthen, leading to a negative ΔH value. Therefore, the correct statement is that heat is released to the surroundings, so ΔH is negative for the process. A tempting distractor might claim that heat is absorbed from the surroundings with a positive ΔH, which confuses freezing with melting. To approach similar problems, recall that the sign of ΔH for a reverse process is opposite, aiding in verification.
For the reaction $\text{CaO}(s)+\text{H}_2\text{O}(l)\rightarrow \text{Ca(OH)}_2(s)$, $\Delta H=-65\ \text{kJ}$. Which statement is correct?
Heat is released by the surroundings, so $\Delta H$ is positive for the reaction.
Heat is absorbed by the surroundings, so $\Delta H$ is negative for the reaction.
No heat is transferred, so $\Delta H$ is zero for the reaction.
Heat is released to the surroundings, so $\Delta H$ is negative for the reaction.
Heat is absorbed from the surroundings, so $\Delta H$ is positive for the reaction.
Explanation
This question tests the understanding of enthalpy change (ΔH) in chemical reactions, specifically identifying whether a reaction is exothermic or endothermic based on the sign of ΔH. The given reaction has ΔH = -65 kJ, indicating an exothermic process where the system releases heat to the surroundings. In exothermic reactions, energy is released as bonds form or rearrange, leading to a negative ΔH value. Therefore, the correct statement is that heat is released to the surroundings, so ΔH is negative for the reaction. A tempting distractor might claim that heat is absorbed from the surroundings with a positive ΔH, which mistakenly applies the characteristics of endothermic reactions. To approach similar problems, always remember that ΔH is defined from the system's perspective, with negative values for heat-releasing processes.
A reaction is described as exothermic at constant pressure. Which statement correctly connects this description to $\Delta H$?
$\Delta H$ is zero because exothermic reactions reach equilibrium.
$\Delta H$ is negative because heat is absorbed from the surroundings.
$\Delta H$ is positive because heat is absorbed from the surroundings.
$\Delta H$ is negative because heat is released to the surroundings.
$\Delta H$ is positive because heat is released to the surroundings.
Explanation
This question tests the connection between the term 'exothermic' and enthalpy change. Exothermic reactions release heat to the surroundings, which means energy flows out of the system. By thermodynamic convention, when heat leaves the system, ΔH is negative. This negative value indicates that the products have lower enthalpy than the reactants because energy has been released. Choice E incorrectly pairs exothermic behavior with positive ΔH and heat absorption, completely reversing the correct relationships. To remember this, think: exothermic means energy exits, and exiting energy gives negative ΔH.
A student observes that a reaction mixture becomes colder than the surrounding air while the reaction proceeds at constant pressure. Which statement about $\Delta H$ for the reaction is most consistent with this observation?
$\Delta H$ is positive because heat is absorbed from the surroundings.
$\Delta H$ is positive because heat is released to the surroundings.
$\Delta H$ is negative because heat is released to the surroundings.
$\Delta H$ is zero because the temperature changes only reflect kinetics.
$\Delta H$ is negative because heat is absorbed from the surroundings.
Explanation
This question tests the ability to connect observable temperature changes to enthalpy changes. When the reaction mixture becomes colder than its surroundings, heat must be flowing from the surroundings into the reaction system. This absorption of heat indicates an endothermic process, which corresponds to a positive ΔH value. The temperature drop occurs because thermal energy from the immediate environment is being consumed by the reaction. Choice C incorrectly associates a temperature drop with negative ΔH and heat release, reversing the correct relationship. To determine the sign of ΔH from temperature observations, remember that if the system cools, it's absorbing heat (positive ΔH).
When $\text{CaCl}_2(s)$ dissolves in water, the solution warms noticeably. Which statement best describes $\Delta H$ for the dissolution process?
$\Delta H$ is negative because heat is absorbed from the surroundings.
$\Delta H$ is negative because heat is released to the surroundings.
$\Delta H$ is zero because dissolving does not involve chemical bonds.
$\Delta H$ is positive because heat is absorbed from the surroundings.
$\Delta H$ is positive because heat is released to the surroundings.
Explanation
This question tests the ability to deduce enthalpy change from observable temperature changes during dissolution. When CaCl₂ dissolves and the solution warms noticeably, heat is being released from the dissolving process to the surroundings. This heat release indicates an exothermic dissolution with negative ΔH. The warming occurs because the energy released when water molecules solvate the Ca²⁺ and Cl⁻ ions exceeds the energy required to break apart the ionic lattice. Choice A incorrectly associates solution warming with positive ΔH and heat absorption, reversing the correct relationship. When a dissolution process warms the solution, it's always exothermic with negative ΔH.
The decomposition reaction $2\text{HgO}(s)\rightarrow 2\text{Hg}(l)+\text{O}_2(g)$ has $\Delta H=+181\ \text{kJ}$ (as written). Which statement best describes the process?
The reaction absorbs heat from the surroundings because $\Delta H$ is negative.
The reaction releases heat to the surroundings because $\Delta H$ is negative.
The reaction releases heat to the surroundings because $\Delta H$ is positive.
The reaction transfers no heat because it produces a gas.
The reaction absorbs heat from the surroundings because $\Delta H$ is positive.
Explanation
This question tests understanding of positive enthalpy values in decomposition reactions. The reaction shows ΔH = +181 kJ, indicating that 181 kJ of heat must be absorbed to decompose 2 moles of HgO. The positive sign means this is an endothermic process where heat flows from the surroundings into the system. This energy input is needed to break the Hg-O bonds and overcome the stability of the solid compound. Choice C incorrectly pairs positive ΔH with heat release, reversing the sign convention. For any reaction with positive ΔH, heat must be absorbed from the surroundings to make the reaction proceed.