11th Class Chemistry Thermodynamics

  • question_answer 10) For oxidation of iron: \[4Fe(s)+3{{O}_{2}}(g)\to 2F{{e}_{2}}{{O}_{3}}(s);\] entropy change is\[-549.4J{{K}^{-1}}mo{{l}^{-1}}\] at 298K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous? \[{{\Delta }_{r}}{{H}^{{}^\circ }}\]for this reaction is\[-1648\times {{10}^{3}}J\,mo{{l}^{-1}}\].

    Answer:

    Information shadow: \[\Delta {{S}_{system}}=-549.4J{{K}^{-1}}mo{{l}^{-1}}\] \[{{\Delta }_{r}}{{H}^{{}^\circ }}=-1648\times {{10}^{3}}J\,mo{{l}^{-1}}\] \[T=298K\](at standard state) Problem solving strategy: \[\Delta {{S}_{Total}}=\Delta {{S}_{System}}+\Delta {{S}_{surroundings}}\] \[\Delta {{S}_{Total}}>0\]for spontaneous process. Working it out: \[\Delta {{S}_{surroundings}}=\frac{\Delta {{H}_{surroundings}}}{T}\] \[=\frac{+1648\times {{10}^{3}}}{298}=5530J{{K}^{-1}}mo{{l}^{-1}}\] \[\Delta {{S}_{Total}}=-549.4+5530\] \[=4980.6J{{K}^{-1}}mo{{l}^{-1}}\] \[=+ve\] Hence, the reaction under consideration is spontaneous.  


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