Based on the values of bond energies given- - fHo of N2H4g is- Given- N - N - 159 kJ mol-1 - H - H - 436 kJ mol-1 N - N - 941 kJ mol-1 - N - H - 398 kJ mol-1

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

Heat of Reaction

Based on the ...

Question

Based on the values of bond energies given, $Δ_{f} H^{o}$ of $N_{2} H_{4} (g)$ is:
Given: $N - N = 159 {kJ mol}^{- 1}; H - H = 436 {kJ mol}^{- 1}$
$N \equiv N = 941 {kJ mol}^{- 1}; N - H = 398 {kJ mol}^{- 1}$

700 {kJ mol}^{- 1}

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

62 {kJ mol}^{- 1}

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

- 98 {kJ mol}^{- 1}

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

- 711 {kJ mol}^{- 1}

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

Open in App

Solution

Suggest Corrections

Similar questions

(B . E .)

Δ_{f} H^{\circ}

N_{2} H_{4} (g)

N - N = 159 {kJ mol}^{- 1}; H - H = 436 {kJ mol}^{- 1} N \equiv N = 941 {kJ mol}^{- 1}; N - H = 398 {kJ mol}^{- 1}

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

ϵ_{(N - H)} = 391 {kJ mol}^{- 1}

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

Δ_{f} H^{0}

N_{2} H_{4} (g)

N - N = 159 k J m o l^{- 1}; H - H = 436 k J m o l^{- 1}

N \equiv N = 941 k J m o l^{- 1}; {N - H = 398 k J m o l}^{- 1}

N_{2} O

△ H^{\circ} (N_{2} O) = 82 k J m o l^{- 1}

N^{Θ} = N^{⨁} = O

(N \equiv N) b o n d = 950 k J m o l^{- 1}

(N = N) b o n d = 420 k J

^{- 1}

O = O) b o n d = 500 k J m o l^{- 1}

(O = N) b o n d = 610 k J m o l^{- 1}

N_{2} O (i n k J m o l^{- 1})

Δ H_{f}^{0}

N_{2} O

82 k J m o l^{- 1}

N \equiv N 946 k J m o l^{- 1}

N = N 418 k J m o l^{- 1}

O = O 498 k J m o l^{- 1}

N = O 607 k J m o l^{- 1}

Join BYJU'S Learning Program