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JEE MainChemistryToughInorganic Chemistry

Calculate ΔrG∘\Delta_{r}G^{\circ} for (NH4CI, s) at 310 K.
Given: ΔrH∘\ \Delta_{r}H^{\circ} (NH4CI, s)= -314.5
ΔrCp=0\Delta_{r}C_{p} = 0
SN2(g)o=192JK−1mol−1S_{N_{2}(g)}^{o} = 192{JK}^{- 1}{mol}^{- 1}
SH2(g)∘=130.5JK−1mol−1S_{H_{2}(g)}^{\circ} = 130.5{JK}^{- 1}{mol}^{- 1}
SCl2∘(g)=233JK−1mol−1S_{{Cl}_{2}}^{\circ}(g) = 233{JK}^{- 1}{mol}^{- 1}
S∘NH4Cl(s)=99.5JK−1mol−1S^{\circ}{NH}_{4}Cl(s) = 99.5{JK}^{- 1}{mol}^{- 1}
All given data are at 300 K.

Show answer and solution
ΔrS∘(NH4Cl,s) at 300K\Delta_{r}S^{\circ}\left( {NH}_{4}Cl,s \right)\ at\ 300K
=SNH4Cl(s)∘−[12SN2∘+2SH2∘+12SCl2∘]= S_{{NH}_{4}Cl(s)}^{\circ} - \left\lbrack \frac{1}{2}S_{N_{2}}^{\circ} + 2S_{H_{2}}^{\circ} + \frac{1}{2}S_{{Cl}_{2}}^{\circ} \right\rbrack
=−374JK−1mol−1= - 374{JK}^{- 1}{mol}^{- 1}
ΔfCP=0\Delta_{f}C_{P} = 0
∴ΔfS310∘=ΔfS300∘\therefore\Delta_{f}S_{310}^{\circ} = \Delta_{f}S_{300}^{\circ}
=−374JK−1mol−1= - 374{JK}^{- 1}{mol}^{- 1}
ΔfH310∘=ΔfH300∘=−314.5\Delta_{f}H_{310}^{\circ} = \Delta_{f}H_{300}^{\circ} = - 314.5
ΔfG310∘=ΔfH∘−310ΔS∘=−314.5−310(−374)1000=−198.56kJ/mol\Delta_{f}G_{310}^{\circ} = \Delta_{f}H^{\circ} - 310\Delta S^{\circ} = - 314.5 - \frac{310( - 374)}{1000} = - 198.56kJ/mol

Source: Grass-G / CH-23 Chemistry Paper 31 Dec. FST.docx

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