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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