Calculate the magnitude of resonance energy of N 2 O in kJ mol -1 from the following data- H f 0 of N 2 O is 82 kJ mol -1Bond energies-N - N 946 kJmol -1N - N 418 kJ mol -1O - O 498 kJ mol -1N - O 607 kJ mol -1

Given, the heat of formation of N_2O is Δ H_f(exp)=82 kJmol^ 1. Again, N_2+1/2O_2→ N_2O Δ H_f(cal)=B.E. of reactants B.E of products nbsp; ⇒Δ H_f(cal)=[946+1 ...

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Standard XII

Chemistry

Enthalpy

Calculate the...

Question

Calculate the magnitude of resonance energy of $N_{2} O (i n k J m o l^{- 1})$ from the following data:
$Δ H_{f}^{0}$ of $N_{2} O$ is $82 k J m o l^{- 1}$
Bond energies:
$N \equiv N 946 k J m o l^{- 1}$
$N = N 418 k J m o l^{- 1}$
$O = O 498 k J m o l^{- 1}$
$N = O 607 k J m o l^{- 1}$

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Solution

Given, the heat of formation of $N_{2} O$ is $Δ H_{f (e x p)} = 82 k J m o l^{- 1}$ .
Again, $N_{2} + \frac{1}{2} O_{2} \to N_{2} O$
$Δ H_{f (c a l)} = B . E . o f r e a c t a n t s - B . E o f p r o d u c t s$
$\Rightarrow Δ H_{f (c a l)} = [946 + \frac{1}{2} (498) - (418 + 607)] = 170 k J / m o l$
Resonance energy is a stabilization energy.
So, resonance energy = $Δ H_{f (e x p)} - Δ H_{f (c a l)} = 82 - 170 = - 88 k J m o l^{- 1}$

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Similar questions

Q. Calculate the magnitude of resonance energy of

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

from the following data:

Δ H_{f}^{0}

N_{2} O

82 k J m o l^{- 1}

Bond energies:

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}

Q. Calculate the resonance energy of

N_{2} O

from the following data:

△_{f} H^{o} o f 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}

Q. Calculate resonance energy of

N_{2} O

from the following data
Hence,

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

Assume structure

N^{Θ} = N^{⨁} = O

BE of

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

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

mol

^{- 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}

Q. Calculate resonance energy of

N_{2} O

from the following data
Hence,

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

Assume structure

N^{Θ} = N^{⨁} = O

BE of

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

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

mol

^{- 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}

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

and

N = O

bonds are

946, 418, 498

and

607 k J m o l^{- 1}

respectively.

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Calculate the magnitude of resonance energy of N2O (in kJ mol−1) from the following data: ΔH0f of N2O is 82 kJmol−1 Bond energies: N≡N 946 kJmol−1 N=N 418 kJmol−1 O=O 498 kJmol−1 N=O 607 kJmol−1

Given, the heat of formation of N2O is ΔHf(exp)=82 kJmol−1. Again, N2+12O2→N2O ΔHf(cal)=B.E. of reactants−B.E of products ⇒ΔHf(cal)=[946+12(498)−(418+607)]=170 kJ/mol Resonance energy is a stabilization energy. So, resonance energy = ΔHf(exp)−ΔHf(cal)=82−170=−88 kJmol−1

Calculate the magnitude of resonance energy of $N_{2} O (i n k J m o l^{- 1})$ from the following data:
$Δ H_{f}^{0}$ of $N_{2} O$ is $82 k J m o l^{- 1}$
Bond energies:
$N \equiv N 946 k J m o l^{- 1}$
$N = N 418 k J m o l^{- 1}$
$O = O 498 k J m o l^{- 1}$
$N = O 607 k J m o l^{- 1}$