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Write Brief Answer · Q35

Q.For the reaction Ag2O(s)→2Ag(s)+12O2(g)Ag_2O(s) \rightarrow 2Ag(s) + \tfrac{1}{2}O_2(g): ΔH=30.56\Delta H = 30.56 kJ mol−1^{-1} and ΔS=6.66\Delta S = 6.66 JK−1^{-1}mol−1^{-1} (at 1 atm). Calculate the temperature at which ΔG\Delta G is equal to zero. Also predict the direction of the reaction

(i) at this temperature and
(ii) below this temperature.
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Step 1. ΔG=0\Delta G=0 occurs when ΔH=TΔS\Delta H=T\Delta S, i.e. T=ΔH/ΔST=\Delta H/\Delta S.

Step 2. ΔH=30.56\Delta H=30.56 kJ mol−1=30,560^{-1}=30{,}560 J mol−1^{-1}; ΔS=6.66\Delta S=6.66 J K−1^{-1}mol−1^{-1}.

Step 3. T=30,560/6.66≈4588.6T=30{,}560/6.66\approx4588.6 K ≈4589\approx4589 K.

Step 4. Since ΔH>0\Delta H>0 and ΔS>0\Delta S>0 (Table 7.5's third row), the reaction is non-spontaneous (ΔG>0\Delta G>0) BELOW this temperature and spontaneous ABOVE it. (i) AT T≈4589T\approx4589 K, ΔG=0\Delta G=0 exactly, so the system is at equilibrium (neither direction is favoured). …

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