The standard enthalpies of formation of Al2O3 and CaO are -1675 kJ mol-1and -635 kJ mol-1 respectively- For the reaction 3CaO-2Al- 3Ca-Al2O3 the standard reaction enthalpy- -rH0-kJ- Round off to the Nearest Integer

We know, Δ_rH^0=Δ_rH_f^0(Products) Δ_rH_f^0(Reactants) =Δ_rH_f^0(Al_2O_3) 3×Δ_rH_f^0(CaO) Δ_rH^0= 1675 3( 635) Δ_rH^0= 1675+ 1905 Δ_rH^0= 230 kJ ...

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Internal Energy

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The standard enthalpies of formation of $A l_{2} O_{3}$ and $C a O$ are $- 1675 k J m o l^{- 1}$ and $- 635 k J m o l^{- 1}$ respectively. For the reaction $3 C a O + 2 A l \to 3 C a + A l_{2} O_{3}$ the standard reaction enthalpy, $Δ_{r} H^{0} =$ ________kJ. (Round off to the Nearest Integer)

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A l_{2} O_{3}

C a O

- 1675 k J m o l^{- 1}

- 635 k J m o l^{- 1}

3 C a O + 2 A l \to 3 C a + A l_{2} O_{3}

Δ_{r} H^{0}

H_{2} O (l)

{F e}_{2} O_{3} (s)

- 286

k J

{m o l}^{- 1}

- 824

k J

{m o l}^{- 1}

{F e}_{2} O_{3} (s) + 3 H_{2} (g) ⟶ 3 H_{2} O (l) + 2 F e (s)

{F e}_{2} O_{3} + 2 A l \to {A l}_{2} O_{3} + 2 F e

S - F

k J m o l^{- 1}

S F_{6}

S F_{6} (g), S (g) a n d F (g)

m o l^{- 1}

\frac{1}{2}

A l_{2} O_{3}

2 A l + \frac{3}{2} O_{2} ⟶ {A l}_{2} O_{3}

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