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

Δ H_f^0(Al_2O_3)= 1675 kJ mol^ 1 Δ H_f^0(CaO)= 635 kJ mol^ 1 3CaO+2Al→ 3Ca+Al_2O_3 Al and Ca are in its standard state so enthalpy of formation of these elem ...

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Standard XII

Chemistry

Internal Energy

The standard ...

Question

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 (Rounded off to the Nearest Integer).

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Solution

$Δ H_{f}^{0} (A l_{2} O_{3}) = - 1675 k J m o l^{- 1}$

$Δ H_{f}^{0} (C a O) = - 635 k J m o l^{- 1}$

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

$A l$ and $C a$ are in its standard state so enthalpy of formation of these elements is zero.

$∴ Δ H_{f}^{0} = Δ H_{f}^{0} (A l_{2} O_{3}) - 3 Δ H_{f}^{0} (C a O)$

$Δ H_{r}^{0} = - 1675 - 3 (- 635)$
$Δ H_{r}^{0} = 230 k J$

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The standard enthalpies of formation of Al2O3 and CaO are −1675 kJmol−1 and −635 kJ mol−1 respectively. For the reaction, 3CaO+2Al→3Ca+Al2O3 the standard reaction enthalpy, ΔrH0=______kJ (Rounded off to the Nearest Integer).

ΔH0f(Al2O3)=−1675 kJ mol−1 ΔH0f(CaO)=−635 kJ mol−1 3CaO+2Al→3Ca+Al2O3 Al and Ca are in its standard state so enthalpy of formation of these elements is zero. ∴ΔH0f=ΔH0f(Al2O3)−3ΔH0f(CaO) ΔH0r=−1675−3(−635) ΔH0r=230 kJ

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 (Rounded off to the Nearest Integer).