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Classification of Hydrocarbons

Calculate the...

Question

Calculate the enthalpy change for the reaction, $H_{2} + F_{2} \to 2 H F$ given that
Bond energy of H – H bond = 434 kJ/mol
Bond energy of F – F bond = 158 kJ/mol
Bond energy of H – F bond = 565 kJ/mol

538 kJ

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–538 kJ

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27 kJ

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-27

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Solution

The correct option is B –538 kJ
$Δ H = 434 + 158 - 2 \times 565$
$Δ H = - 538$

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Similar questions

X e F_{2 (g)} + H_{2} (g) \to 2 H F_{(g)} + X e_{(g)}, Δ H^{o} = - 430 k J

\to H - H = 435 k J / m o l \to H - F = 565 k J / m o l

X e - F

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

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

H - F b o n d = 565 k J m o l - 1

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.

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.

H | C | H = H | C | H + H - H \to H - H | C | H - H | C | H - H

H - H

431.37 k J m o l^{- 1}

C = C

606.10 k J m o l^{- 1}

C - C

336.49 k J m o l^{- 1}

C - H

410.50 k J m o l^{- 1}

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