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Heat of Formation

Use the bond ...

Question

Use the bond enthalpies in the table to determine $Δ H^{o}$ for the formation of hydrazine, $N_{2} H_{4} (g)$ from Nitrogen and Hydrogen in standard state according to the equation : $N_{2} (g) + 2 H_{2} (g) \to N_{2} H_{4} (g)$
Bond Bond enthalpies
$N - N$ $159 k J m o l^{- 1}$
$N = N$ $418 k J m o l^{- 1}$
$N \equiv N$ $941 k J m o l^{- 1}$
$H - H$ $436 k J m o l^{- 1}$
$H - N$ $389 k J m o l^{- 1}$

Δ H^{o} = - 425 k J

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Δ H^{o} = - 98 k J

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Δ H^{o} = + 98 k J

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Δ H^{o} = + 711 k J

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Solution

The correct option is D $Δ H^{o} = + 98 k J$
$Δ H_{r x n}^{o} = Δ H_{f}^{o} [N_{2} H_{4}] = [(941 + 2 \times 436) - (159 + 4 \times 389)]$
$Δ H^{o} = + 98 k J$

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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}

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}

{C l}_{2} (g) ⟶ 2 C l (g); Δ H^{o} = 242.3 k J {m o l}^{- 1}

I_{2} (g) ⟶ 2 I (g); Δ H^{o} = 151.0 k J {m o l}^{- 1}

{I C l} (g) ⟶ I (g) + C l (g); Δ H^{o} = 211.3 k J {m o l}^{- 1}

I_{2} (s) ⟶ I_{2} (g); Δ H^{o} = 62.76 k J {m o l}^{- 1}

I_{2} (s)

{C l}_{2} (g)

I C l (g)

Δ H^{o}

N a_{2} O (s) + S O_{3} (g) \to N a_{2} S O_{4} (g)

N a (s) + H_{2} O (l) \to N a O H (s) + \frac{1}{2} H_{2} (g); Δ H^{o} = - 146 k J

N a_{2} S O_{4} (s) + H_{2} O (l) \to 2 N a O H (s) + S O_{3} (g); Δ H^{o} = + 418 k J

2 N a O (s) + 2 H_{2} (g) \to 4 N a (s) + 2 H_{2} O (l); Δ H^{o} = + 259 k J

C - C l

C H_{3} C l (g) + {C l}_{2} (g) ⟶ C H_{2} {C l}_{2} (g) + H C l (g)

Δ H^{o} = - 104 k J

C - H

C l - C l

H - C l

414

243

431 k J {m o l}^{- 1}

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