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Standard XII

Chemistry

Heat of Formation

N 2 g → 2N g ...

Question

$N_{2} (g) \to 2 N (g)$ $Δ H_{r x n}^{o} = 945.2$ kJ
Calculate the enthalpy of formation of $N (g)$ .

-

945.2 kJ/mol

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0.0 kJ/mol

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472.6 kJ/mol

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945.2 kJ/mol

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Solution

The correct option is C 472.6 kJ/mol
Enthalpy of formation : The enthalpy of formation of 1 mol of prouduct.
$N_{2} (g) \to 2 N (g)$
$Δ_{r x n}^{o} =$ Enthalpy of reaction $= 2 \times$ Enthalpy of formation of $N_{(g)}$
Hence ,Enthalpy of formation , $Δ_{f o r m a t i o n}^{0} o f N_{(g)} = \frac{945.2}{2} = 472.6$ kJ/mol

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N2(g)→2N(g) ΔHorxn=945.2 kJCalculate the enthalpy of formation of N(g).

The correct option is C 472.6 kJ/molEnthalpy of formation : The enthalpy of formation of 1 mol of prouduct. N2(g)→2N(g) Δorxn= Enthalpy of reaction =2× Enthalpy of formation of N(g)Hence ,Enthalpy of formation ,Δ0formationofN(g)=945.22=472.6 kJ/mol

$N_{2} (g) \to 2 N (g)$ $Δ H_{r x n}^{o} = 945.2$ kJ
Calculate the enthalpy of formation of $N (g)$ .