1. Tardigrade
  2. Question
  3. Chemistry
  4. The standard enthalpy of formation of H 2 O (l) and Fe 2 O 3(s) are respectively -286 kJ mol -1 and -824 kJ mol -1. What is the standard enthalpy change for the following reaction? Fe 2 O 3(s)+3 H 2(g) longrightarrow 3 H 2 O (l)+2 Fe (s)

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Q. The standard enthalpy of formation of $H _{2} O (l)$ and $Fe _{2} O _{3}(s)$ are respectively
$-286\, kJ\, mol ^{-1}$ and $-824\, kJ\, mol ^{-1}$. What is the standard enthalpy change for the following reaction?
$Fe _{2} O _{3}(s)+3 H _{2}(g) \longrightarrow 3 H _{2} O (l)+2 Fe (s)$

KEAMKEAM 2016Thermodynamics

Solution:

Given reaction,

$Fe _{2} O _{3}( s )+3 H _{2}( g ) \longrightarrow 3 H _{2} O (l)+2 Fe ( s )$

and $H_{f}^{\circ}\left( H _{2} O \right)=-286.00\, kJ / mol$

$H_{f}^{\circ}\left( Fe _{2} O _{3}\right)=-824\, kJ / mol$

Therefore, $H_{R}$ (Heat of reaction)

$=\Sigma H_{f}^{\circ}$ (products) $-\Sigma H_{f}^{\circ}$ (reactants)

$H_{R}=-(3 \times 286)-(-824)$

$\because\left(H_{f}^{\circ}\right.$ for $H _{2}$ and $\left. Fe =0\right)$

$H _{R}=-858+824$

$H_{R}=-34\, kJ / mol$