A mixture of the two oxides contains FeO and Fe2O3 in the mole ratio 2:1, is changed into 1:2 mole ratio by oxidation of mixture. Then how much of heat (in Kcal) is released at constant pressure and temperature per mole of the initial mixture? Report your answer after dividing with 9. Given:CompoundEnthalpy of formation (kcal/mol)FeOs−65.0Fe2O3(s)−175.0

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Q.

A mixture of the two oxides contains ${FeO and Fe}_{2} O_{3}$ in the mole ratio 2:1, is changed into 1:2 mole ratio by oxidation of mixture. Then how much of heat (in Kcal) is released at constant pressure and temperature per mole of the initial mixture? Report your answer after dividing with 9. Given:

Compound	Enthalpy of formation (kcal/mol)
$F e O (s)$	−65.0
${Fe}_{2} O_{3} (s)$	−175.0

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Detailed Solution

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$2 F e O + \frac{1}{2} O_{2} \to F e_{2} O_{3}; Δ H = (- 175) - 2 (- 65) = - 45 k c a l$

Initial	$2 a$	$a$
Final	$2 a - 2 x$	$a + x$

$\frac{2 a - 2 x}{a + x} = \frac{1}{2} \Rightarrow x = \frac{3}{5} a$ and heat released = 45x kcal

$∴$ Heat released per mole of initial mixture $= \frac{45 x}{3 a} = 9 k c a l$

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