A) \[-2221.1kJ\text{ }mo{{l}^{-1}}\]
B) \[-339.3\text{ }kJ\text{ }mo{{l}^{-1}}\]
C) \[-439.3\text{ }kJ\text{ }mo{{l}^{-1}}\]
D) \[-523.2\text{ }kJ\text{ }mo{{l}^{-1}}\]
Correct Answer: B
Solution :
For the oxidation of ammonia at 298 K the standard enthalpy change\[(\Delta H)\]and standard entropy change\[(\Delta S)\]are\[-382.64\text{ }kJ\text{ }mo{{l}^{-1}}\] and\[-145.6\text{ }J{{K}^{-1}}mo{{l}^{-1}}\]respectively. The relationship of Gibbs free energy\[(\Delta G)\]with\[\Delta H\]and\[\Delta S\]is represented in form of following equation. \[\Delta G=\Delta H-T\Delta S\] Given that \[\Delta H=-382.64\text{ }kJ\text{ }mo{{l}^{-1}}\] \[\Delta S=-145.6\text{ }J\text{ }{{K}^{-1}}mo{{l}^{-1}}\] \[=-145.6\times {{10}^{-3}}kJ\,{{K}^{-1}}\] or \[\Delta G=-382.64-(298\times -145.6\times {{10}^{-3}})\] \[=-339.3\text{ }kJ\,mo{{l}^{-1}}\]
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