When 4 g of iron is burnt to ferric oxide at | vedclass

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(B) The chemical equation for the formation of ferric oxide $(Fe_2O_3)$ is: $2Fe(s) + \frac{3}{2}O_2(g) \rightarrow Fe_2O_3(s)$.Given that $4 \ g$ of

Vedclass · Accepted Answer

$-819.8$

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(B) The chemical equation for the formation of ferric oxide $(Fe_2O_3)$ is: $2Fe(s) + \frac{3}{2}O_2(g) ightarrow Fe_2O_3(s)$.Given that $4 \ g$ of iron releases $29.28 \ kJ$ of heat.The molar mass of iron $(Fe)$ is $56 \ g \ mol^{-1}$.For $2 \ moles$ of iron,the mass is $2 	imes 56 = 112 \ g$.Since heat is evolved,the enthalpy change $(\Delta H)$ is negative.For $4 \ g$ of $Fe$,$\Delta H = -29.28 \ kJ$.For $112 \ g$ of $Fe$ (which corresponds to $1 \ mole$ of $Fe_2O_3$),the enthalpy of formation is:$\Delta H_f = \frac{-29.28 \ kJ}{4 \ g} 	imes 112 \ g = -819.8 \ kJ \ mol^{-1}$.

When $4 \ g$ of iron is burnt to ferric oxide at constant pressure,$29.28 \ kJ$ of heat is evolved. What is the enthalpy of formation of ferric oxide in $kJ \ mol^{-1}$? (At. Wt. of $Fe = 56$)

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