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Chapter 5: Problem 81

Without doing any calculations, predict the sign of \(\Delta H\) for each of the following reactions: (a) $\mathrm{NaCl}(s) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{Cl}^{-}(\mathrm{g})$ (b) \(2 \mathrm{H}(g) \longrightarrow \mathrm{H}_{2}(g)\) (c) \(\mathrm{Na}(g) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{e}^{-}\) (d) \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(l)\)

Short Answer

Expert verified
(a) \(\Delta H > 0\) (endothermic) (b) \(\Delta H < 0\) (exothermic) (c) \(\Delta H > 0\) (endothermic) (d) \(\Delta H > 0\) (endothermic)

Step by step solution

01

In this reaction, solid sodium chloride (NaCl) is dissociating into its gaseous ions, Na+ and Cl-. The ionic bond between Na and Cl in the solid state has to be broken, which requires energy input. Breaking the bond is an endothermic process. #Step 2: Determine the sign of ΔH#

Since the reaction requires the input of energy to break the ionic bond, the reaction is endothermic and \(\Delta H\) will be positive. (b) \(2 \mathrm{H}(g) \longrightarrow \mathrm{H}_{2}(g)\) #Step 1: Analyze the reaction#
02

In this reaction, two hydrogen atoms (H) in the gaseous state are combining to form a hydrogen molecule (H2). This process involves the formation of a covalent bond between two hydrogen atoms, which releases energy. Bond formation is an exothermic process. #Step 2: Determine the sign of ΔH#

Since the reaction involves the formation of a covalent bond and releases energy, the reaction is exothermic, and \(\Delta H\) will be negative. (c) \(\mathrm{Na}(g) \longrightarrow \mathrm{Na}^{+}(g)+\mathrm{e}^{-}\) #Step 1: Analyze the reaction#
03

In this reaction, a gaseous sodium atom (Na) is losing an electron to become a gaseous Na+ ion. The process of losing an electron requires energy to overcome the attraction between the nucleus and the outer electron. This process is an endothermic process known as ionization. #Step 2: Determine the sign of ΔH#

Since the reaction requires the input of energy for ionization, the reaction is endothermic, and \(\Delta H\) will be positive. (d) \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(l)\) #Step 1: Analyze the reaction#
04

In this reaction, solid iodine (I2) is converting into liquid iodine. This process involves the breaking of intermolecular forces, particularly London dispersion forces, between iodine molecules in the solid state. The breaking of these intermolecular forces requires the input of energy. The process is an endothermic process. #Step 2: Determine the sign of ΔH#

Since the reaction requires the input of energy for breaking intermolecular forces, the reaction is endothermic, and \(\Delta H\) will be positive.

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Most popular questions from this chapter

Without doing any calculations, predict the sign of \(\Delta H\) for each of the following reactions: (a) \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) (b) \(2 \mathrm{~F}(g) \longrightarrow \mathrm{F}_{2}(g)\) (c) $\mathrm{Mg}^{2+}(g)+2 \mathrm{Cl}^{-}(g) \longrightarrow \mathrm{MgCl}_{2}(s)$ (d) \(\mathrm{HBr}(g) \longrightarrow \mathrm{H}(g)+\mathrm{Br}(g)\)

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{I}_{2}(s)\) or $1 \mathrm{~mol} \mathrm{I}_{2}(g)$ at the same temperature, (b) \(2 \mathrm{~mol}\) of iodine atoms or \(1 \mathrm{~mol}\) of \(\mathrm{I}_{2},(\mathbf{c}) 1 \mathrm{~mol} \mathrm{I}_{2}(g)\) and $1 \mathrm{~mol} \mathrm{H}_{2}(g)$ at \(25^{\circ} \mathrm{C}\) or \(2 \mathrm{~mol} \mathrm{HI}(g)\) at $25^{\circ} \mathrm{C},(\mathbf{d}) 1 \mathrm{~mol} \mathrm{H}_{2}(g)\( at \)100^{\circ} \mathrm{C}$ or \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(300^{\circ} \mathrm{C}\).

The hydrocarbons cyclohexane $\left(\mathrm{C}_{6} \mathrm{H}_{12}(l), \Delta H_{f}^{\circ}=-156\right.$ \(\mathrm{kJ} / \mathrm{mol}\) ) and 1-hexene $\left(\mathrm{C}_{6} \mathrm{H}_{12}(l), \Delta H_{f}^{\circ}=-74 \mathrm{~kJ} / \mathrm{mol}\right)$ have the same empirical formula. (a) Calculate the standard enthalpy change for the transformation of cyclohexane to 1-hexene. (b) Which has greater enthalpy, cyclohexane or 1-hexene? (c) Without doing a further calculation and knowing the answer to (b), do you expect cyclohexane or 1-hexene to have the larger combustion enthalpy?

(a) Under what condition will the enthalpy change of a process equal the amount of heat transferred into or out of the system? (b) During a constant-pressure process, the system releases heat to the surroundings. Does the enthalpy of the system increase or decrease during the process? (c) In a constant-pressure process, \(\Delta H=0\). What can you conclude about \(\Delta E, q,\) and \(w ?\)

Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\). From the following enthalpy of reaction data and data in Appendix C, calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s);\) $$ \begin{aligned} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) & \\ \Delta H^{\circ}=&-127.2 \mathrm{~kJ} \end{aligned} $$
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