The standard heat of formation of F e 2 O 3 s is -824 kJ mol -1- The heat change for reaction 4 F e s -3 O 2 g - 2 F e 2 O 3 s is-A- -1648-4 kJB- -824-2 kJC- 648-4 kJD- 1648-4 kJ

The correct option is A 1648.4 kJΔ H^∘=∑Δ H^∘_f(product) ∑Δ H^∘_f(reactants) =[2 ×Δ H^∘_f Fe_2O_3(s)] [4 Δ H^∘_fFe(s)+3Δ H^∘_fO_2(g)] =2( 824.2 kJ) [4 × z ...

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Standard XII

Chemistry

Hess' Law

The standard ...

Question

The standard heat of formation of $F e_{2} O_{3} (s)$ is -824 kJ $m o l^{- 1}$ . The heat change for reaction $4 F e (s) + 3 O_{2} (g) \to 2 F e_{2} O_{3} (s)$ is:

648.4 kJ

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1648.4 kJ

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-1648.4 kJ

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- 824.2 kJ

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Solution

The correct option is C -1648.4 kJ
$Δ H^{\circ} = \sum Δ H_{f}^{\circ} (p r o d u c t) - \sum Δ H_{f}^{\circ} (r e a c t a n t s) = [2 \times Δ H_{f}^{\circ} F e_{2} O_{3} (s)] - [4 Δ H_{f}^{\circ} F e (s) + 3 Δ H_{f}^{\circ} O_{2} (g)] = 2 (- 824.2 k J) - [4 \times z e r o + 3 \times z e r o] = - 1648.4 k J$

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Similar questions

Q. Given

2 F e_{2} O_{3} (s) \to 4 F e (s) + 3 O_{2} (g); Δ G^{\circ} = - 1487 K J m o l^{- 1}

2 C O (g) + O_{2} (g) \to 2 C O_{2} (g); Δ G^{\circ} = - 514.4 K J m o l^{- 1}

Free energy change

Δ G^{o}

for the reaction

2 F e_{2} O_{3} (s) + 6 C O (g) \to 4 F e (s) + 6 C O_{2} (g)

will be

Q. Given
(i)

2 F e_{2} O_{3} (s) \to 4 F e (s) + 3 O_{2} (g);

Δ_{r} G^{o} = + 1487.0

m o l^{- 1}

(ii)

2 C O (g) + O_{2} (g) \to 2 C O_{2} (g);

Δ_{r} G^{o} = - 514.4

m o l^{- 1}

Free energy change,

Δ_{r} G^{o}

for the reaction,

2 F e_{2} O_{3} (s) + 6 C O (g) \to 4 F e (s) + 6 C O_{2} (g)

will be?

Q. Determine the heat of the following reaction:

F e O (s) + {F e}_{2} O_{3} (s) ⟶ {F e}_{3} O_{4} (s)

Given informations:

2 F e (s) + O_{2} (g) ⟶ 2 F e O (s); Δ H^{o} = - 544 k J

4 F e (s) + 3 O_{2} (g) ⟶ 2 {F e}_{2} O_{; 3} (s); Δ H^{o} = - 1648.4 k J

{F e}_{3} O_{4} (s) ⟶ 3 F e (s) + 2 O_{2} (g); Δ H^{o} = + 1118.4 k J

Q. Determine the heat of the following reaction:

F e O (s) + F e_{2} O_{3} (s) \to F e_{3} O_{4} (s)

Given informations:

2 F e (s) + O_{2} (g) \to 2 F e O (s); Δ H^{\circ} = - 544 k J 4 F e (s) + 3 O_{2} (g) \to 2 F e_{2} O_{3} (s); Δ H^{\circ} = - 1648.4 k J F e_{3} O_{4} (s) \to 3 F e (s) + 2 O_{2} (g); Δ H^{\circ} = + 1118.4 k J

For oxidation of iron,

4 F e (s) + {3 O}_{2} (g) \to 2 {F e}_{2} O_{3} (s)

entropy change is - 549.4

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

at 298 K. In spite of negative entropy change of this reaction, why is the reaction spontaneous?

(

△, H^{⊝}

for this reaction is

- 1648 \times 10^{3} J {m o l}^{- 1}

)

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The standard heat of formation of Fe2O3(s) is -824 kJ mol−1. The heat change for reaction 4Fe(s)+3O2(g)→2Fe2O3(s) is:

The correct option is C -1648.4 kJΔH∘=∑ΔH∘f(product)−∑ΔH∘f(reactants)=[2×ΔH∘fFe2O3(s)]−[4ΔH∘fFe(s)+3ΔH∘fO2(g)]=2(−824.2kJ)−[4×zero+3×zero]=−1648.4kJ

The standard heat of formation of $F e_{2} O_{3} (s)$ is -824 kJ $m o l^{- 1}$ . The heat change for reaction $4 F e (s) + 3 O_{2} (g) \to 2 F e_{2} O_{3} (s)$ is: