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Answer the following questions · Q9

Q.ix. Calculate standard enthalpy of reaction, Fe2O3(s)+3CO(g)→2 Fe(s)+3CO2(g)\mathrm{Fe_2O_3(s) + 3CO(g) \rightarrow 2\,Fe(s) + 3CO_2(g)}, from the following data. ΔfH0\Delta_f H^0(Fe2_2O3_3) = -824 kJ/mol, ΔfH0\Delta_f H^0(CO) = -110 kJ/mol, ΔfH0\Delta_f H^0(CO2_2) = -393 kJ/mol Ans. : (-25 kJ)

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Step 1. ΔrH0 = Σ ΔfH0(products) - Σ ΔfH0(reactants). Fe(s) is an element, so ΔfH0(Fe)=0.

Step 2. Products: 2 mol Fe (0 kJ each) + 3 mol CO2 (-393 kJ/mol each) = 0 + 3(-393) = -1179 kJ. …

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