You visited us 1 times! Enjoying our articles? Unlock Full Access!

Gibbs Free Energy & Spontaneity

Given the bon...

Question

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

-93 KJ

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

102 KJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

90 KJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

105 KJ

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B -93 KJ
It takes 945 kJ of energy to break $N \equiv N$ .

It takes 436 $\times$ 3=1308 kJ to break $3 H_{2}$ molecules.

It releases 2 $\times$ 3 $\times$ 391=2346 kJ to form $2 N H_{3}$ molecules.

Enthalpy of the reaction $=$ Enthalpy of reactants $-$ Enthalpy of products $= 945 + 1308 - 2346 = - 93$ kJ

$∴$ Enthalpy of reaction is $- 93$ kJ.

Suggest Corrections

Similar questions

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)

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. 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)

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)

Q. 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.

Join BYJU'S Learning Program

Given the bond energies N≡N, H−H and N−H bonds are 945, 436 and 391 KJ mole−1 respectively, the enthalpy of the following reaction N2(g)+3H2(g)→2NH3(g) is:

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