Considere as seguintes reações:
(a) \mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{+2}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{+2}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})
A entalpia de combustão da reação (b) é -156,45 \mathrm{KJmol}^{-1}. Calcule \Delta \mathrm{Sr}^{0}, \Delta \mathrm{Hr}^{0} e \Delta \mathrm{Gr}^{0} para as duas reações a 298 K .
[Dados: \quad \mathrm{S}_{\mathrm{m}}{ }^{0}\left(\mathrm{Zn}^{+2}, \mathrm{aq}\right)=-112,1 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \quad \mathrm{~S}_{\mathrm{m}}{ }^{0}(\mathrm{Cu}, \mathrm{s})=33,15 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \quad \mathrm{~S}_{\mathrm{m}}{ }^{0}(\mathrm{Zn}, \mathrm{s})=41,63 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \quad \mathrm{~S}_{\mathrm{m}}{ }^{0}\left(\mathrm{Cu}^{+2}, \mathrm{aq}\right)=- (99,6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \quad \mathrm{~S}{\mathrm{m}}{ }^{0}\left(\mathrm{H}{2} \mathrm{O}, \mathrm{I}\right)=69,91 \quad \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \quad \mathrm{~S}{\mathrm{m}}{ }^{0}\left(\mathrm{CO}{2}, \mathrm{~g}\right)=213,74 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \quad \mathrm{~S}{\mathrm{m}}{ }^{0}\left(\mathrm{C}{12} \mathrm{H}{22} \mathrm{O}{11}, \quad \mathrm{~s}\right)=360,2 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}), \left.\mathrm{S}_{\mathrm{m}}{ }^{0}\left(\mathrm{O}_{2}, \mathrm{~g}\right)=205,14 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} ; \Delta \mathrm{H}_{\mathrm{f}}{ }^{0}\left(\mathrm{Zn}^{+2}, \mathrm{aq}\right)=-153,89 \mathrm{KJmol}^{-1}, \Delta \mathrm{H}_{\mathrm{f}}{ }^{0}\left(\mathrm{Cu}^{+2}, \mathrm{aq}\right)=64,77 \mathrm{KJmol}^{-1}\right]

Passo 1

Temos duas reações a 298 K:
(a) \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(aq)\to\mathrm{Zn}^{2+}(aq)+\mathrm{Cu}(s)
(b) combustão de sacarose:
\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}(s)+12\,\mathrm{O}_{2}(g)\to12\,\mathrm{CO}_{2}(g)+11\,\mathrm{H}_{2}\mathrm{O}(l)
Para (a) usaremos entalpias de formação e entropias molares dadas; para (b) calculamos \Delta S^{0} e usamos \Delta H_{\mathrm{comb}}^{0}=-156{,}45,k J\cdot mo l^{-}¹. Em ambos, \Delta G^{0}=\Delta H^{0}-T\,\Delta S^{0}.

Passo 2

Cálculo de \Delta S_{r}^{0} para (a):
\Delta S_{r}^{0} =\bigl[S^{0}(\mathrm{Zn}^{2+})+S^{0}(\mathrm{Cu})\bigr] -\bigl[S^{0}(\mathrm{Zn})+S^{0}(\mathrm{Cu}^{2+})\bigr]
=(-112,1+33,15)-(41,63+(-99,6)) =-78,95-(-57,97) =-20,98\\ \mathrm{J\,K^{-1}\,mol^{-1}}

Passo 3

Cálculo de \Delta H_{r}^{0} e \Delta G_{r}^{0} para (a):
\Delta H_{r}^{0} =\bigl[\Delta H_{f}^{0}(\mathrm{Zn}^{2+})+0\bigr] -\bigl[0+\Delta H_{f}^{0}(\mathrm{Cu}^{2+})\bigr] =(-153,89)-(+64,77) =-218,66\\ \mathrm{kJ\,mol^{-1}}
\Delta G_{r}^{0} =\Delta H_{r}^{0}-T\,\Delta S_{r}^{0} =-218,66\\ \mathrm{kJ} -298\,( -0,02098\\ \mathrm{kJ\,K^{-1}}) =-218,66+6,25 =-212,41\\ \mathrm{kJ\,mol^{-1}}

Passo 4

Cálculo de \Delta S_{r}^{0} para (b):
\Delta S_{r}^{0} =\bigl[12\,S^{0}(\mathrm{CO}_{2})+11\,S^{0}(\mathrm{H}_{2}\mathrm{O})\bigr] -\bigl[S^{0}(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11})+12\,S^{0}(\mathrm{O}_{2})\bigr]
=(12\cdot213,74+11\cdot69,91)-(360,2+12\cdot205,14) =3333,89-2821,88 =512,01\\ \mathrm{J\,K^{-1}\,mol^{-1}}

Passo 5

Para (b) \Delta H_{r}^{0}=\Delta H_{\mathrm{comb}}^{0}=-156,45,k J\cdot mo l^{-}¹ e
\Delta G_{r}^{0} =\Delta H_{r}^{0}-T\,\Delta S_{r}^{0} =-156,45\\ \mathrm{kJ} -298\,(0,51201\\ \mathrm{kJ\,K^{-1}}) =-156,45-152,18 =-308,63\\ \mathrm{kJ\,mol^{-1}}

Resposta

(a) \Delta S_{r}^{0}=-20,98\\ \mathrm{J\,K^{-1}\,mol^{-1}} \Delta H_{r}^{0}=-218,66\\ \mathrm{kJ\,mol^{-1}} \Delta G_{r}^{0}=-212,41\\ \mathrm{kJ\,mol^{-1}}
(b) \Delta S_{r}^{0}=512,01\\ \mathrm{J\,K^{-1}\,mol^{-1}} \Delta H_{r}^{0}=-156,45\\ \mathrm{kJ\,mol^{-1}} \Delta G_{r}^{0}=-308,63\\ \mathrm{kJ\,mol^{-1}}

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