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)

Question

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) (A) -538 KJ mol-1 (B) +538 KJ mol-1 (C) -102 KJ mol-1 (D) +34 KJ mol-1

Tardigrade Admin · Accepted Answer

Given reaction, Fe 2 O 3( s )+3 H 2( g ) longrightarrow 3 H 2 O (l)+2 Fe ( s ) and Hf°( H 2 O )=-286.00 kJ / mol Hf°( Fe 2 O 3)=-824 kJ / mol Therefore, HR (Heat of reaction) = Sigma Hf° (products) - Sigma Hf° (reactants) HR=-(3 × 286)-(-824) ∵(Hf°. for H 2 and . Fe =0) H R=-858+824 HR=-34 kJ / mol

Q. The standard enthalpy of formation of $H_{2} O (l)$ and $F e_{2} O_{3} (s)$ are respectively
$- 286 k J m o l^{- 1}$ and $- 824 k J m o l^{- 1}$ . What is the standard enthalpy change for the following reaction?
$F e_{2} O_{3} (s) + 3 H_{2} (g) ⟶ 3 H_{2} O (l) + 2 F e (s)$

$- 538 K J m o l^{- 1}$

$+ 538 K J m o l^{- 1}$

$- 102 K J m o l^{- 1}$

$+ 34 K J m o l^{- 1}$

$- 34 K J m o l^{- 1}$

Q. The standard enthalpy of formation of H2​O(l) and Fe2​O3​(s) are respectively −286kJmol−1 and −824kJmol−1. What is the standard enthalpy change for the following reaction? Fe2​O3​(s)+3H2​(g)⟶3H2​O(l)+2Fe(s)

−538KJmol−1

+538KJmol−1

−102KJmol−1

+34KJmol−1