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

Heat of Formation

Bond energies...

Question

Bond energies can be obtained by using the following relation:
$Δ H (r e a c t i o n) = \sum$ Bond energy of bonds, broken in the reactants $- \sum$ Bond energy of bonds, formed in the products
Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms.The heat of formation of $N O$ from its elements is $+ 90 k J m o l^{- 1}$ , What is the approximate bond dissociation energy of the bond in $N O$ ?
$B E_{N \equiv N} = 941 k J m o l^{- 1}$ $B E_{O = O} = 499 k J m o l^{- 1}$

630 k J m o l^{- 1}

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

700 k J m o l^{- 1}

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

860 k J m o l^{- 1}

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

810 k J m o l^{- 1}

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

Open in App

Solution

The correct option is A $630 k J m o l^{- 1}$
As we know,
As we know,
$Δ H = H_{P} - H_{R}$
so for
$O_{2} (g) + N_{2} (g) ⟶ 2 N O (g) Δ H = 180 k J / m o l e$
$Δ H_{B . D .} = 1 / 2 (941 + 499 - 180) = 630 k J / m o l e$

Suggest Corrections

Similar questions

Δ H (r e a c t i o n) = \sum

- \sum

Δ H

= \sum

- \sum

N - H

O - H

F - H

N O

+ 90 k J m o l^{- 1}

N O

B E_{N == N} = 941 k J m o l^{- 1}

B E_{O == O} = 499 k J m o l^{- 1}

+ 90 k j {m o l}^{- 1}

{m o l}^{- 1}

{m o l}^{- 1}

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}

Join BYJU'S Learning Program