Heat of Formation

The standard ...

Question

The standard heats of formation of $N O_{2}$ (g) and $N_{2} O_{4}$ (g) are $8.0$ and $2.0 k c a l m o l^{- 1}$ respectively, the heat of dimerization of $N O_{2}$ in $k c a l$ is

10.0

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- 6.0

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- 12.0

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- 14.0

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Solution

The correct option is D $- 14.0$
$(i) \frac{1}{2} N_{2} + O_{2} \to N O_{2}; △ H = 8.0 k c a l (i i) N_{2} + 2 O_{2} \to N_{2} O_{4}; △ H = 2.0 k c a l 2 N O_{2} \to N_{2} O_{4}; △ H = ?$
Multiply $(i)$ by 2 and subtract $(i)$ from $(i i)$ we get:
$2 N O_{2} \to N_{2} O_{4}$
$△ H = 2 - (8 \times 2) = - 14 k c a l$

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Similar questions

Q. The standard heats of formation of

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and

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are 8.0 and 2.0 Kcal

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The standard heats of formation in K.cal mol $^{- 1}$ of NO $_{2 (g)}$ and N $_{2}$ O $_{4 (g)}$ are 8.0 and 2.0 respectively. The heat of dimerisation of $N O_{2}$ in K.cal is $2 N O_{2 (g)} \to N_{2} O_{4 (g)}$

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Heat of Formation

Standard XII Chemistry

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The standard heats of formation of NO2 (g) and N2O4 (g) are 8.0 and 2.0 kcal mol−1 respectively, the heat of dimerization of NO2 in kcal is

The correct option is D −14.0(i)12N2+O2→NO2;△H=8.0 kcal(ii)N2+2O2→N2O4;△H=2.0 kcal2NO2→N2O4;△H=? Multiply (i) by 2 and subtract (i) from (ii) we get: 2NO2→N2O4 △H=2−(8×2)=−14kcal

The standard heats of formation of $N O_{2}$ (g) and $N_{2} O_{4}$ (g) are $8.0$ and $2.0 k c a l m o l^{- 1}$ respectively, the heat of dimerization of $N O_{2}$ in $k c a l$ is