Extra Credit 18
- Page ID
- 96900
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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, KP, 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 (Ksp = 5
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, Ksp, 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.