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 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.
Heat is released to the surroundings, so $\Delta H$ is negative 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.
For the reaction $\text{H}_2(g)+\text{I}_2(g)\rightarrow 2\text{HI}(g)$, $\Delta H=-10\ \text{kJ}$. What does this indicate about heat flow?
Heat is absorbed from 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 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.
Explanation
This question tests the skill of interpreting a small negative ΔH in terms of heat flow direction. The reaction has ΔH = -10 kJ, indicating a weakly exothermic process where heat is released to the surroundings. This reflects a slight net energy release in forming HI. Thus, 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 due to the misconception of ignoring the negative sign and assuming endothermicity. A transferable strategy is to carefully note the sign of ΔH: even small negative values indicate exothermicity and heat release.
A reaction mixture becomes noticeably colder while the reaction proceeds at constant pressure. Which statement about $\Delta H$ is most consistent with this observation?
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.
Heat is released by the surroundings, so $\Delta H$ is positive for the reaction.
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.
Explanation
This question tests the skill of inferring the sign of ΔH from an observational change in temperature during a reaction at constant pressure. The reaction mixture becoming colder indicates that the system is absorbing heat from the surroundings, characteristic of an endothermic reaction. This absorption causes the temperature drop as energy is drawn into the system to facilitate the reaction. Therefore, choice A correctly states that heat is absorbed from the surroundings, so ΔH is positive for the reaction. A tempting distractor is choice B, which claims heat is released to the surroundings so ΔH is negative, but this is incorrect because it stems from the misconception that a temperature decrease means heat release rather than absorption by the system. A transferable strategy is to link temperature changes to heat flow: cooling of the system suggests endothermic (positive ΔH), while warming suggests exothermic (negative ΔH).
For the reaction $\text{CaCO}_3(s)\rightarrow \text{CaO}(s)+\text{CO}_2(g)$, $\Delta H=+178\ \text{kJ}$ as written. What does this imply about heat flow?
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.
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.
Explanation
This question tests the skill of interpreting a positive ΔH value in terms of heat flow for a decomposition reaction. The reaction has ΔH = +178 kJ, meaning it is endothermic, as the positive sign indicates the system absorbs heat from the surroundings. In this process, the enthalpy of the products is higher than that of the reactants, requiring energy input to break down CaCO₃. Therefore, choice D correctly states that heat is absorbed from the surroundings, so ΔH is positive for the reaction. A tempting distractor is choice A, which also says heat is released to the surroundings so ΔH is negative, but this is incorrect because it misapplies the sign convention, confusing positive ΔH with exothermic behavior. A transferable strategy is to check the sign of ΔH first: positive means endothermic and heat absorption by the system, negative means exothermic and heat release.
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?
No heat is transferred, so $\Delta H$ is zero for the reaction.
Heat is absorbed by the surroundings, so $\Delta H$ is negative for the reaction.
Heat is absorbed from the surroundings, so $\Delta H$ is positive for the reaction.
Heat is released by the surroundings, so $\Delta H$ is positive for the reaction.
Heat is released to the surroundings, so $\Delta H$ is negative 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 $\text{C}(s)+\text{O}_2(g)\rightarrow \text{CO}_2(g)$, $\Delta H=-394\ \text{kJ}$. Which statement correctly describes the enthalpy change?
Heat is released by the surroundings, so $\Delta H$ is positive for the reaction.
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.
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 describing the enthalpy change for a formation reaction with negative ΔH. The reaction has ΔH = -394 kJ, meaning it is exothermic, with the system releasing heat to the surroundings as CO₂ forms. This large negative value highlights the stability of CO₂. Thus, choice A correctly states that heat is released to the surroundings, so ΔH is negative for the reaction. A tempting distractor is choice B, which says heat is absorbed from the surroundings so ΔH is positive, but this is incorrect stemming from the misconception that formation of gases is endothermic. A transferable strategy is to recall that many formation reactions of stable compounds are exothermic with negative ΔH.
For the reaction $2\text{NO}_2(g)\rightarrow 2\text{NO}(g)+\text{O}_2(g)$, $\Delta H=+114\ \text{kJ}$. Which statement is correct?
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 absorbed from the surroundings, so $\Delta H$ is positive for the reaction.
Heat is released by the surroundings, so $\Delta H$ is positive for the reaction.
Heat is released to the surroundings, so $\Delta H$ is negative for the reaction.
Explanation
This question tests the skill of analyzing a positive ΔH for a decomposition reaction in terms of heat absorption. The reaction has ΔH = +114 kJ, signifying an endothermic process where the system absorbs heat from the surroundings to break down NO₂. This positive value indicates higher enthalpy in the products. Thus, choice A correctly states that heat is absorbed from the surroundings, so ΔH is positive for the reaction. A tempting distractor is choice B, which says heat is released to the surroundings so ΔH is negative, but this is incorrect because it stems from the misconception that all decompositions are exothermic. A transferable strategy is to evaluate the reaction type and ΔH sign: decompositions often require energy input, making them endothermic with positive ΔH.
The dissolution of ammonium nitrate in water is described as endothermic. Which statement about $\Delta H$ for the dissolution process is correct?
Heat is absorbed by the surroundings, so $\Delta H$ is negative 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.
No heat is transferred, so $\Delta H$ is zero for the process.
Heat is absorbed from the surroundings, so $\Delta H$ is positive for the process.
Explanation
This question tests the skill of associating the term 'endothermic' with the sign of ΔH for a dissolution process. The dissolution is described as endothermic, meaning the process absorbs heat from the surroundings, resulting in a positive ΔH. This absorption is necessary to overcome lattice energy in ammonium nitrate, cooling the solution. Therefore, choice C 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 of misidentifying endothermic processes as exothermic. A transferable strategy is to remember that endothermic processes, like some dissolutions, have positive ΔH and involve the system drawing in heat from the surroundings.
For the reaction $\text{NH}_4\text{Cl}(s)\rightarrow \text{NH}_3(g)+\text{HCl}(g)$, $\Delta H=+176\ \text{kJ}$. What does this imply?
Heat is released to the surroundings, so $\Delta H$ is negative for the reaction.
No heat is transferred, so $\Delta H$ is zero for the reaction.
Heat is absorbed by the surroundings, so $\Delta H$ is negative for the reaction.
Heat is absorbed from the surroundings, so $\Delta H$ is positive for the reaction.
Heat is released by the surroundings, so $\Delta H$ is positive for the reaction.
Explanation
This question tests the skill of analyzing a positive ΔH for a dissociation reaction. The reaction has ΔH = +176 kJ, meaning it is endothermic, absorbing heat from the surroundings to break down NH₄Cl. This requires energy to separate into gases. Therefore, choice B correctly states that heat is absorbed from the surroundings, so ΔH is positive for the reaction. A tempting distractor is choice A, which claims heat is released to the surroundings so ΔH is negative, but this is incorrect stemming from the misconception that all salt decompositions release energy. A transferable strategy is to identify gas-producing decompositions as often endothermic with positive ΔH.
For the reaction $\text{H}_2(g)+\tfrac{1}{2}\text{O}_2(g)\rightarrow \text{H}_2\text{O}(l)$, $\Delta H=-286\ \text{kJ}$. Which statement is correct?
Heat is released by the surroundings, so $\Delta H$ is positive for the reaction.
Heat is absorbed from the surroundings, so $\Delta H$ is positive for the reaction.
Heat is released to the surroundings, so $\Delta H$ is negative for the reaction.
No heat is transferred, so $\Delta H$ is zero for the reaction.
Heat is absorbed by the surroundings, so $\Delta H$ is negative for the reaction.
Explanation
This question tests the skill of interpreting the formation enthalpy of water as negative. The reaction has ΔH = -286 kJ, signifying exothermicity with heat released to the surroundings. This reflects the stability of the water molecule. Thus, choice A correctly states that heat is released to the surroundings, so ΔH is negative for the reaction. A tempting distractor is choice B, which says heat is absorbed from the surroundings so ΔH is positive, but this is incorrect due to the misconception that gas-to-liquid transitions are endothermic. A transferable strategy is to recall standard formation enthalpies: negative for stable compounds like water, indicating heat release.