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Standard XII

Chemistry

Heat of Formation

Calculate the...

Question

Calculate the heat of formation of ammonia from the following data:
$N_{2} (g) + 3 H_{2} (g) ⟶ 2 {N H}_{3} (g)$
The bond energies of $N \equiv N, H - H$ and $N - H$ bonds are $226, 104$ and $93$ $k c a l$ respectively.

- 5 k c a l

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- 10 k c a l

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- 15 k c a l

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- 20 k c a l

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Solution

The correct option is A $- 20 k c a l$
Heat of formation of ammonia is calculated as energy of reactant minus the energy of product
Heat of formation = $[226 + 3 (104)] - 2 (3 * 93)$
Heat of formation = $538 - 558 = - 20 k c a l$

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Similar questions

Given the bond energies $N \equiv N$ , $H - H$ and $N - H$ bonds are $945, 436$ and $391 k J m o l e^{- 1}$ respectively, the enthalpy of the following reaction $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ is

Q. Calculate the enthalpy of formation of ammonia from the following bond energy data:

(N - N)

bond

= 398 k J {m o l}^{- 1}

;

(H - H)

bond

= 435 k J {m o l}^{- 1}

and

(N \equiv N)

bond

= 945.36 k J {m o l}^{- 1}

Q. The bond energies of

N \equiv N

H - H

and

N - H

bonds are

945,

436

and

391

KJ/mole respectively. The enthalpy of the reaction

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

Q. The enthalpy of formation of ammonia when calculated from the following bond energy data is:

[B.E. of

N - H, H - H, N \equiv N

is 389 kJ

m o l^{- 1}

, 435 kJ

m o l^{- 1}

, 945.36 kJ

m o l^{- 1}

respectively]

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Calculate the heat of formation of ammonia from the following data:N2(g)+3H2(g)⟶2NH3(g)The bond energies of N≡N,H−H and N−H bonds are 226,104 and 93 kcal respectively.

The correct option is A −20 kcalHeat of formation of ammonia is calculated as energy of reactant minus the energy of productHeat of formation = [226+3(104)]−2(3∗93)Heat of formation = 538−558=−20kcal

Calculate the heat of formation of ammonia from the following data:
$N_{2} (g) + 3 H_{2} (g) ⟶ 2 {N H}_{3} (g)$
The bond energies of $N \equiv N, H - H$ and $N - H$ bonds are $226, 104$ and $93$ $k c a l$ respectively.

The correct option is A $- 20 k c a l$
Heat of formation of ammonia is calculated as energy of reactant minus the energy of product
Heat of formation = $[226 + 3 (104)] - 2 (3 * 93)$
Heat of formation = $538 - 558 = - 20 k c a l$