A) \[81.98\text{ }kJ\text{ }mo{{1}^{-1}}\]
B) \[-\text{ }81.98\text{ }kJ\text{ }mo{{1}^{-1}}\]
C) \[\text{ }819.8\text{ }kJ\text{ }mo{{1}^{-1}}\]
D) \[-819.8\,kJ\,mo{{l}^{-1}}\]
Correct Answer: D
Solution :
\[\underset{\text{2}\,\text{mol}}{\mathop{2Fe}}\,(s)+\frac{3}{2}{{O}_{2}}(g)\xrightarrow{{}}\underset{\text{1}\,\text{mol}}{\mathop{F{{e}_{2}}{{O}_{3}}(S)}}\,\] Moles of \[Fc=\frac{4}{56}=\frac{1}{14}mol\] Moles of \[F{{e}_{2}}{{O}_{3}}\]formed \[=\frac{1}{28}\,mol\] Heat released = 29.28 kJ \[\Delta {{H}^{o}}f(F{{e}_{2}}{{O}_{3}})=-\frac{29.28}{1/28}kJ\,mo{{l}^{-1}}\] \[=-819.84\,kJ\,mo{{l}^{-1}}\]
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