CameraIcon

SearchIcon

MyQuestionIcon

1

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

Fluorides of Xenon

Calculate the...

Question

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

Open in App

Solution

$Δ H = [Δ H_{(N \equiv N)} + 3 \times Δ H_{(H - H)}] - [6 Δ H_{(N - H)}]$
$945 + 3 \times 435.0 - 6 \times 389.0 = - 83.64 k J$
Heat formation of ${N H}_{3} = \frac{Δ H}{2} = - \frac{83.64}{2} = - 41.82 k J {m o l}^{- 1}$

Suggest Corrections

thumbs-up

0

Similar questions

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

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

m o l^{- 1}

m o l^{- 1}

m o l^{- 1}

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

N - H

226, 104

93

k c a l

Q. Using the bond enthalpy data given below, calculate the enthalpy of formation of acetone (g)
Bond energy

C - H = 413.4 k J {m o l}^{- 1}

C - C = 347.0 k J {m o l}^{- 1}

C = O = 728.0 k J {m o l}^{- 1}

O = O = 495.0 k J {m o l}^{- 1}

H - H = 435.8 k J {m o l}^{- 1}

;

Δ H_{s u b} C (s) = 718.4 k J {m o l}^{- 1}

Q. Calculate the resonance energy of

N_{2} O

from the following data:

△_{f} H^{\circ}

N_{2} O = 82 k J m o l^{- 1}

, standard bond enthalpies of

N = N, N = N, O = O

N = O

946, 418, 498

607 k J m o l^{- 1}

Q. Based on the values of bond energies

(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}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Xenon Oxyfluorides

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Fluorides of Xenon

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon