Extra Credit 18

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Q13.4.7

When heated, iodine vapor dissociates according to this equation:

\[\ce{I2}(g)⇌\ce{2I}(g)\]

At 1274 K, a sample exhibits a partial pressure of I₂ of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, K_P, for the decomposition at 1274 K.

Solution:

\[K_p=\frac {P_{I_{(g)}}^2} {P_{I_{2_(g)}}}\]

\[K_p=\frac {0.1378^2} {0.1122}=0.1692\]

Thus, \(K_p\) of this equation is 0.1692 at 1274K.

Q14.3.19

Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F⁻ or CN⁻, is the stronger base?

Solution:

According to E1,

\(HF + H_2O\rightleftharpoons F^- +H_3O^+\) \(K_{a_{HF}}= 3.5\times 10^{-4} \)

\(HCN + H_2O\rightleftharpoons CN^- +H_3O^+\) \(K_{a_{HCN}}= 4.9\times 10^{-10} \)

Apparently, \(K_{a_{HF}}> K_{a_{HCN}}\), which means HF is stronger than HCN.

Since a stronger acid produces a weaker conjugate base,

Thus, \(CN^-\) is a stronger base.

Q15.1.X

Calculate the molar solubility of AgBr in 0.035 M NaBr (K_sp = 5 × 10^–13).

Solution:

\[K_{sp}=[Ag^+][Cl^-]\]

\[AgCl_{(s)} \leftrightharpoons Ag^+ _{(aq)} +Cl^- _{(aq)}\]

ICE Table	\[AgCl_{(s)}\]	\[Ag^+_{(aq)}\]	\[Cl^-_{(aq)}\]
Initial	/	0	0.035
Change	/	x	x
Equilibrium	/	x	0.035+x

\[K_{sp}=[Ag^+][Cl^-] = x\times (0.035+x) = 5 \times 10^{-13}\]

Let's assume that 0.035+x is approximately equal to 0.035, since \(K_{sp}\) is very small

\[K_{sp}= x\times 0.035=5 \times 10^{-13}\]

\(x= 1.75\times 10^{-14}\ll 0.035\) the assumption is valid

Thus, the molar solubility of AgBr in 0.035 M NaBr is \(1.75\times 10^{-14} M\).

(Using Wolfram

\[x=1.42857\times 10^{-11}\]

Thus, the molar solubility of AgBr in 0.035 M NaBr is \(1.42857\times 10^{-11} M\).)

Q16.4.16

Given that the \(ΔG^\circ_\ce{f}\) for Pb²⁺(aq) and Cl⁻(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, K_sp, for PbCl₂(s).

Solution:

\[PbCl_{(s)}=Pb^{2+}_{(aq)} + Cl^-_{aq}\]

\[\Delta G^\circ_{rxn}=\Delta G^\circ_{Pb^{2+}_{(aq)}} + \Delta G^\circ_{Cl^{-}_{(aq)}} -\Delta G^\circ_{PbCl_{(s)}}\]

According to T1 and information from the question,

we could find \[\Delta G^\circ_{rxn}= (-24.3) + (-131.2) - (-359.4)= 203.9 KJ/mol =203900 J/mol\]

Since the reaction reaches equilibrium when PbCl dissolve in the water, we could assume that \(\Delta G^\circ_{rxn}=0\)

\[\Delta G^\circ =-RTlnK_{sp}\]

Thus, \[K_{sp}=1.81\times10^{-36}\]

Q5.4.32

Calculate the enthalpy of combustion of propane, C₃H₈(g), for the formation of H₂O(g) and CO₂(g). The enthalpy of formation of propane is −104 kJ/mol.

Solution:

\[C_3H_8 + 5O_2\rightarrow 4H_2O + 3CO_2\]

\[\Delta H_{rxn} =4\Delta H_{f_{(H_2O)}} +3\Delta H_{f_{(CO_2)}}-\Delta H_{f_{(C_3H_8)}} - 5\Delta H_{f_{(O_2)}}\]

According to T1 and the information given in the question,

we could find

\[\Delta H_{rxn}=4\times(-285.8) + 3\times(-393.5) - (-104) - 5\times(0) = -2219.7 KJ/mol\]

Thus, the enthalpy of combustion of propane is -2219.7 KJ/mol.