The standard enthalpy of formation of $N H_{3}$ is $46.0 k J m o l^{- 1}$ . If the enthalpy of formation of $H_{2}$ from its atom is $- 436 k J m o l^{- 1}$ and that of $N_{2}$ is $- 712 k J m o l^{- 1}$ , the average bond enthalpy of $N - H$ bond in $N H_{3}$ is

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

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+ 352 k J m o l^{- 1}

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+ 1056 k J m o l^{- 1}

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- 1102 k J m o l^{- 1}

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Solution

The correct option is B $+ 352 k J m o l^{- 1}$
$N_{2} + 3 H_{2} \to 2 N H_{3} Δ H = 2 \times - 46.0 k J m o l^{- 1}$

Let $x$ be the bond enthalpy of N - H bond then

[Note: Enthalpy of formation or bond formation enthalpy is given which is negative but the given reaction involves bond breaking hence values should be taken as positive.]

$Δ H = \sum B o n d e n e r g i e s o f p r o d u c t s - \sum B o n d e n e r g i e s o f r e a c t a n t s$

$2 \times - 46 = 712 + 3 \times (436) - 6 x$

$- 92 = 2020 + 92$

$6 x = 2112$

$x = + 352 k J / m o l$

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Similar questions

Q. The standard enthalpy of formation of

N H_{3}

46.0

kJ/mol. If the enthalpy of formation of

H_{2}

from its atoms is

- 436

kJ/mol and that of

N_{2}

- 712

kJ/mol, the average bond enthalpy of

N - H

bond in

N H_{3}

is:

The standard enthalpy of formation of $N H_{3}$ is $46.0 k J m o l^{- 1}$ . If the enthalpy of formation of $H_{2}$ from its atom is $- 436 k J m o l^{- 1}$ and that of $N_{2}$ is $- 712 k J m o l^{- 1}$ , the average bond enthalpy of $N - H$ bond in $N H_{3}$ is