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Standard XII

Chemistry

Heat of Formation

The standard ...

Question

The standard enthalpy of formation of $N H_{3}$ is $- 46.0$ $k J m {o l}^{- 1}$ . lf the enthalpy of formation of $H_{2}$ from its atoms 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 $+ 352 k J m {o l}^{- 1}$
Enthalpy of formation of $N H_{3} = - 46 k J / m o l e$

$N_{2} + 3 H_{2} \to 2 N H_{3}; Δ H_{r} = - 2 \times 46 k J$

Bond breaking is endothermic and bond formation is exothermic

Assuming $x$ is the bond energy of $N - H$ bond $(k J m o 1^{- 1})$

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

$x = 352 k J / m o l$

Hence, the correct option is $B$

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Similar questions

Q. Bond dissociation enthalpies of

H_{2} (g) a n d N_{2} (g)

are

436.0 k J m o l^{- 1}

and

941.3 k J m o l^{- 1}

respectively. Enthalpy of formation of

N H_{3}

(g) is

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

. What is the enthalpy of atomization of

N H_{3}

Q. Given are the bond enthalpies:

ϵ_{N \equiv N} = 945 {kJ mol}^{- 1}, ϵ_{H - H} = 436 {kJ mol}^{- 1}

and

ϵ_{(N - H)} = 391 {kJ mol}^{- 1}

. The enthalpy change of the reaction

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

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

Q. Calculate the lattice enthalpy of

K C l

from the following data at standard states-
Enthalpy of sublimation of

K = 89 k J m o l^{- 1}

Enthalpy of dissociation of

C l_{2} = 244 k J m o l^{- 1}

Ionisation energy of

K = 425 k J m o l^{- 1}

Electron gain enthalpy of

C l = - 355 k J m o l^{- 1}

Enthalpy of formation of

K C l = - 438 k J m o l^{- 1}

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The standard enthalpy of formation of NH3 is −46.0 kJmol−1. lf the enthalpy of formation of H2 from its atoms is −436 kJmol−1 and that of N2 is -712 kJmol−1, the average bond enthalpy of N−H bond in NH3 is:

The correct option is B +352 kJmol−1Enthalpy of formation of NH3=−46 kJ/mole N2+3H2→2NH3; ΔHr=−2×46 kJBond breaking is endothermic and bond formation is exothermicAssuming x is the bond energy of N−H bond (kJmo1−1)712+(3×436)−6x=−46×2x=352 kJ/molHence, the correct option is B

The standard enthalpy of formation of $N H_{3}$ is $- 46.0$ $k J m {o l}^{- 1}$ . lf the enthalpy of formation of $H_{2}$ from its atoms 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: