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Heat of Formation

Find the enth...

Question

Find the enthalpy of formation of hydrogen flouride on the basis of following data:
Bond energy of $H - H b o n d = 434 k J m o l^{- 1}$
Bond energy of $F - F b o n d = 158 k J m o l^{- 1}$
Bond energy of $H - F b o n d = 565 k J m o l - 1$

Δ H = - 269 k J

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Δ H = + 269 k J

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Δ H = - 4497 k J

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None of these

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Solution

The correct option is A $Δ H = - 269 k J$
The balanced chemical reaction for the formation of hydrogen fluoride is:

$\frac{1}{2} H_{2} (g) + \frac{1}{2} F_{2} (g) ⟶ H F (g)$

$Δ_{f} H = BE of reactant - BE of product$

$= (\begin{matrix} \frac{1}{2} BE of H_{2} + \frac{1}{2} BE of F_{2} \end{matrix}) - (BE of HF)$

$= \frac{1}{2} \times 434 + \frac{1}{2} \times 158 - 565$

$= 217 + 79 - 565$

$= - 269 k J$
Hence, the enthalpy of formation of hydrogen fluoride is -269 kJ

Suggest Corrections

Similar questions

H - H = 434 k J m o l^{- 1}

F - F = 158 k J m o l^{- 1}

H - F = 565 k J m o l^{- 1}

(k J m o l^{- 1})

C \equiv \equiv C

C_{2} H_{2}

C_{2} H_{2}

225 k J m o l^{- 1}

C - - H

350 k J m o l^{- 1}

2 C (s) ⟶ 2 C (g); Δ H = 1410 k J m o l^{- 1}

H_{2} (g) ⟶ 2 H (g); Δ H = 330 k J m o l^{- 1}

25^{o} C

\frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) \to O H (g); Δ H = 10.06 k c a l

H_{2} (g) \to 2 H (g); Δ H = 104.18 k c a l

O_{2} (g) \to 2 O (g); Δ H = 118.32 k c a l

O - H

O H

E_{H - H} = 104.2 k c a l m o l^{- 1}

E_{F - F} = 36.6 k c a l m o l^{- 1}

E_{H - F} = 134.6 k c a l m o l^{- l}

X_{H} = 2.1.

C l - C l

C H_{4 (g)} + C l_{(g)} \to C H_{3} C l_{(g)} + H C l_{(g)}; Δ H = - 100.3 k J

C - H, C - C l, H - C l

413, 326

431

k J m o l^{- 1}

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