AP Chemistry 4.7 Types of Chemical Reactions - Exam Style questions - FRQs- New Syllabus

(a)
\( \mathrm{I_2} \)
Iodine has the longest bond length because iodine atoms have the largest atomic radius among the halogens listed. Since the atoms are larger, the distance between the two nuclei in \( \mathrm{I_2} \) is greatest.

Atomic radius increases down Group \( 17 \), so the bond length of the diatomic halogen molecules also increases down the group.

(b)
The favorable reaction is:
\( \mathrm{2Br^- + Cl_2 \rightarrow Br_2 + 2Cl^-} \)

Relevant half-reactions:
\( \mathrm{Cl_2 + 2e^- \rightarrow 2Cl^-} \qquad E^\circ = 1.36\ \mathrm{V} \)
\( \mathrm{2Br^- \rightarrow Br_2 + 2e^-} \qquad E^\circ = -1.07\ \mathrm{V} \)

Therefore,
\( E^\circ_{\text{cell}} = 1.36 + (-1.07) = 0.29\ \mathrm{V} \)

Since \( E^\circ \) is positive, the reaction is thermodynamically favorable under standard conditions.

(c)
Molecules of \( \mathrm{Br_2} \) are nonpolar, so the only intermolecular force between \( \mathrm{Br_2} \) molecules is London dispersion force.
Molecules of \( \mathrm{BrCl} \) are polar, so they experience both London dispersion forces and dipole-dipole forces.

However, \( \mathrm{Br_2} \) has a larger electron cloud and is more polarizable than \( \mathrm{BrCl} \). Thus, the dispersion forces in \( \mathrm{Br_2} \) are stronger overall than the combined intermolecular forces in \( \mathrm{BrCl} \), so \( \mathrm{Br_2} \) has the higher boiling point.

(d)
Before equilibrium is established, only \( \mathrm{BrCl}(g) \) is present, so:

\( P = \dfrac{nRT}{V} \)

\( P = \dfrac{(0.100\ \mathrm{mol})(0.0821\ \mathrm{L\ atm\ mol^{-1}\ K^{-1}})(298\ \mathrm{K})}{2.00\ \mathrm{L}} \)

\( P = 1.22\ \mathrm{atm} \)

(e)
For \( \mathrm{2\,BrCl}(g) \rightleftharpoons \mathrm{Br_2}(g) + \mathrm{Cl_2}(g) \),

\( K_{eq} = \dfrac{[\mathrm{Br_2}][\mathrm{Cl_2}]}{[\mathrm{BrCl}]^2} \)

or, in terms of partial pressures,

\( K_{eq} = \dfrac{P_{\mathrm{Br_2}}P_{\mathrm{Cl_2}}}{\left(P_{\mathrm{BrCl}}\right)^2} \)

(f)
Initial pressure of pure \( \mathrm{BrCl} \) is \( 1.22\ \mathrm{atm} \). Since \( 42\% \) decomposes:

Pressure of \( \mathrm{BrCl} \) decomposed:
\( (0.42)(1.22) = 0.51\ \mathrm{atm} \)

From \( \mathrm{2\,BrCl}(g) \rightleftharpoons \mathrm{Br_2}(g) + \mathrm{Cl_2}(g) \), if \( 2x = 0.51 \), then

\( x = 0.255 \approx 0.26\ \mathrm{atm} \)

So at equilibrium:
\( P_{\mathrm{BrCl}} = 1.22 – 0.51 = 0.71\ \mathrm{atm} \)
\( P_{\mathrm{Br_2}} = 0.26\ \mathrm{atm} \)
\( P_{\mathrm{Cl_2}} = 0.26\ \mathrm{atm} \)

Therefore,

\( K_{eq} = \dfrac{(0.26)(0.26)}{(0.71)^2} \)

\( K_{eq} = \dfrac{0.0676}{0.5041} \approx 0.13 \)

Since \( K_{eq} < 1 \), the equilibrium favors \( \mathrm{BrCl} \), which matches the fact that only part of the sample decomposes.

(g)
Use

\( \Delta H^\circ = \sum (\text{bond energies})_{\text{broken}} – \sum (\text{bond energies})_{\text{formed}} \)

For \( \mathrm{2\,BrCl}(g) \rightarrow \mathrm{Br_2}(g) + \mathrm{Cl_2}(g) \):

Bonds broken: \( 2(\mathrm{Br-Cl}) \)
Bonds formed: \( \mathrm{Br-Br} + \mathrm{Cl-Cl} \)

So,

\( 1.6 = 2(\mathrm{Br-Cl}) – (193 + 243) \)

\( 1.6 = 2(\mathrm{Br-Cl}) – 436 \)

\( 437.6 = 2(\mathrm{Br-Cl}) \)

\( \mathrm{Br-Cl} = 218.8 \approx 219\ \mathrm{kJ/mol} \)

Therefore, the bond energy of the \( \mathrm{Br-Cl} \) bond is \( 219\ \mathrm{kJ/mol} \).