A) -322 kJ mol-1
B) -161 kJ mol-1
C) -152 kJ mol-1
D) -76 kJ mol-1
Correct Answer: A
Solution :
Fe(s)\[\xrightarrow{{}}\] Fe2+ + 2e- ; \[\Delta G_{1}^{o}\] 2H+ + 2e- + ½O2 \[\xrightarrow{{}}\] H2O(l) ; \[\Delta G_{2}^{o}\] Fe(s) + 2H+ + ½O2 \[\xrightarrow{{}}\] Fe2+ + H2O ;\[\Delta G_{3}^{o}\] Applying, \[\Delta G_{1}^{o}+\Delta G_{2}^{o}=\Delta G_{3}^{o}\] \[\Delta G_{3}^{o}\] = (-2F ´ 0.44) + (-2F ´ 1.23) \[\Delta G_{3}^{o}\] = -(2 ´ 96500 ´ 0.44+ 2 ´ 96500 ´ 1.23) \[\Delta G_{3}^{o}\] = -322310 J \ \[\Delta G_{3}^{o}\] = -322 kJYou need to login to perform this action.
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