| (i) \[2F{{e}_{2}}{{O}_{3}}(s)\to 4Fe(s)+3{{O}_{2}}(g);\] \[{{\Delta }_{r}}{{G}^{o}}=+1487.0kJmo{{l}^{-1}}\] |
| (ii) \[2CO(g)+{{O}_{2}}(g)\to 2C{{O}_{2}}(g);\] \[{{\Delta }_{r}}G{}^\circ =-514.4kJmo{{l}^{-1}}\] |
A) \[-112.4kJ\,\,mo{{l}^{-1}}\]
B) \[-56.2\text{ kJ mo}{{\text{l}}^{-1}}\]
C) \[-208.0\text{ kJ mo}{{\text{l}}^{-1}}\]
D) \[-168.2\text{ kJ mo}{{\text{l}}^{-1}}\]
Correct Answer: B
Solution :
(i) \[2F{{e}_{2}}{{O}_{3}}(s)\to 4Fe(s)+3{{O}_{2}}(g);\] \[{{\Delta }_{r}}{{G}^{o}}=+1487.0kJmo{{l}^{-1}}\] (ii) \[2CO(g)+{{O}_{2}}(g)\to 2C{{O}_{2}}(g);\] \[{{\Delta }_{r}}{{G}^{o}}=-514.4kJ\,\,mo{{l}^{-1}}\] Multiply above reaction with 3 (iii) \[6CO(g)+3{{O}_{2}}(g)\to 6C{{O}_{2}}(g);\] \[{{\Delta }_{r}}{{G}^{o}}=3\times -514.3=-1543.2kJ\,\,mo{{l}^{-1}}\] When we add reaction (i) and reaction (iii), we get reaction (iv) (iv) \[2F{{e}_{2}}{{O}_{3}}(s)+6CO(g)\to 4Fe(s)+6C{{O}_{2}}(g)\] Free energy change, \[{{\Delta }_{r}}{{G}^{o}}\] for the reaction will be. \[1487.0-1543.2=-56.2kJmo{{l}^{-1}}\]
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