SATHEE: Chemical Thermodynamics Question 360

Chemical Thermodynamics Question 360

Question: Given that

(i) $Δ_{f} H^{\circ}$ of $N_{2} O$ is $82 k J m o l^{- 1}$

(ii) Bond energies of $N \equiv N, N = N, O = O$ and $N = O$ are 946, 418, 498 and $607 k J m o l^{- 1}$ respectively, The resonance energy of $N_{2} O$ is:

Options:

A) $- 88 k J$

B) $- 66 k J$

C) $- 62 k J$

D) $-, 44, k J$

Show Answer

Answer:

Correct Answer: A

Solution:

$N_{2} (g) + \frac{1}{2} O_{2} \to N_{2} O (g)$

$N \equiv N (g) + \frac{1}{2} (O = O) \to \underset{\cdot \cdot}{\ddot{N}} = \overset{+}{}, = \ddot{\underset{\cdot \cdot}{O}} (g)$

$Δ H_{f}^{^{\circ}} =$ [Energy required for breaking of bonds]-[Energy released for forming of bonds] $= (Δ H_{N \equiv N} + \frac{1}{2} Δ H_{O = O} - (Δ H_{N = N} + Δ H_{N = O})$

$= (946 + \frac{1}{2} \times 498) - (418 + 607)$

$= 170 k J, m o l^{- 1}$ Resonance energy $= 82 - 170$

$= - 88 k J, m o l^{- 1}$

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Chemical Thermodynamics Question 360

Question: Given that

Options:

Answer:

Solution:

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