JEE MAIN - Chemistry (2018 - 15th April Evening Slot - No. 20)
Given
(i) 2Fe2O3(s) $$ \to $$ 4Fe(s) + 3O2(g);
$$\Delta $$rGo = + 1487.0 kJ mol-1
(ii) 2CO(g) + O2(g) $$ \to $$ 2CO2(g);
$$\Delta $$rGo = $$-$$ 514.4 kJ mol-1
Free energy change, $$\Delta $$rGo for the reaction
2Fe2O3(s) + 6CO(g) $$ \to $$ 4Fe(s) + 6CO2(g) will be :
(i) 2Fe2O3(s) $$ \to $$ 4Fe(s) + 3O2(g);
$$\Delta $$rGo = + 1487.0 kJ mol-1
(ii) 2CO(g) + O2(g) $$ \to $$ 2CO2(g);
$$\Delta $$rGo = $$-$$ 514.4 kJ mol-1
Free energy change, $$\Delta $$rGo for the reaction
2Fe2O3(s) + 6CO(g) $$ \to $$ 4Fe(s) + 6CO2(g) will be :
$$-$$ 112.4 kJ mol-1
$$-$$ 56.2 kJ mol-1
$$-$$ 168.2 kJ mol-1
$$-$$ 208.0 kJ mol-1
Explanation
Given
(i) 2Fe2O3(s) $$ \to $$ 4Fe(s) + 3O2(g);
$$\Delta $$rGo = + 1487.0 kJ mol-1
(ii) 2CO(g) + O2(g) $$ \to $$ 2CO2(g);
$$\Delta $$rGo = $$-$$ 514.4 kJ mol-1
Multiply reaction (ii) with 3 and we get,
(iii) 6CO(g) + 3O2 (g) $$\buildrel \, \over \longrightarrow $$ 6CO2(g) ;
$$\Delta $$rGo = 3 $$ \times $$ $$-$$ 514.4 = $$-$$ 1543.2 kJ mol$$-$$1
Now add reaction (i) with (iii), we get
2Fe2O3(s) + 6CO(g) $$\buildrel \, \over \longrightarrow $$ 4Fe(s) + 6CO2(g)
$$\therefore\,\,\,$$ Free energy change of the reaction,
$$\Delta $$rGo = 1487.0 $$-$$ 1543.2 = $$-$$ 56.2 kJ/mol
(i) 2Fe2O3(s) $$ \to $$ 4Fe(s) + 3O2(g);
$$\Delta $$rGo = + 1487.0 kJ mol-1
(ii) 2CO(g) + O2(g) $$ \to $$ 2CO2(g);
$$\Delta $$rGo = $$-$$ 514.4 kJ mol-1
Multiply reaction (ii) with 3 and we get,
(iii) 6CO(g) + 3O2 (g) $$\buildrel \, \over \longrightarrow $$ 6CO2(g) ;
$$\Delta $$rGo = 3 $$ \times $$ $$-$$ 514.4 = $$-$$ 1543.2 kJ mol$$-$$1
Now add reaction (i) with (iii), we get
2Fe2O3(s) + 6CO(g) $$\buildrel \, \over \longrightarrow $$ 4Fe(s) + 6CO2(g)
$$\therefore\,\,\,$$ Free energy change of the reaction,
$$\Delta $$rGo = 1487.0 $$-$$ 1543.2 = $$-$$ 56.2 kJ/mol
Comments (0)
