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Enthalpy

Given that bo...

Question

Given that bond energy of $H - H$ and $C l - C l$ are $430 k J$ $m o 1^{- 1}$ and $240 k J$ $m o 1^{- 1}$ respectively and heat of formation for $H C l$ is $- 90 k J$ $m o 1^{- 1}$ . Bond enthalpy of $H C l$ is:

245 k J m o 1^{- 1}

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290 k J m o 1^{- 1}

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380 k J m o 1^{- 1}

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425 k J m o 1^{- 1}

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Solution

The correct option is D $425 k J m o 1^{- 1}$
$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g)$ $Δ H_{f}^{0} = - 90 k J m o l^{- 1}$

Now, $Δ H_{r}^{0} = (\frac{1}{2} B E_{H - H} + \frac{1}{2} B E_{C l - C l}) - B E_{H - C l}$

$- 90 = \frac{1}{2} \times 430 + \frac{1}{2} \times 240 - B E_{H - C l}$

$B E_{H - C l} = 215 + 120 + 90$

$= 425 k J m o l^{- 1}$

So, the correct answer is option D.

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Similar questions

Given that bond energies of
$H - H$ and $C l - C l$ are $430 k J m o l^{- 1}$ and $240 k J m o l^{- 1}$ respectively and $Δ_{f} H$ for $H C l$ is -90 kJ $m o l^{- 1} .$ The bond enthalpy of $H C l$ is:

Q. Given that bond energies of

H - H

and

C l - C l

are

430 k J m o l^{- 1}

and

240 k J m o l^{- 1}

respectively and

Δ_{f} H

for

H C l

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

. Bond enthalpy of

H C l

is :

Q. Given that, bond energies of

H - H

and

C l - C l

430 k J / m o l

and

240 k J / m o l

respectively.

Δ H_{f}

for

H C l

- 90 k J / m o l

. Bond enthalpy of

H C l

is:

Q. For the reaction,

A (g) + B (g) \to C (g) + D (g), Δ H^{o}

and

Δ S^{o}

are, respectively,

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

and

- 0.100 k J K^{- 1} m o 1^{- 1}

at 298 K.

The equilibrium constant for the reaction at 298 K is :

Q. The

H - H

bond energy is 430 kJ mol

^{- 1}

and

C l - C l

bond energy is 240 kJ mol

^{- 1}

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H C l

- 90

kJ. The

H - C l

bond energy is

(420 + x)

kJ mol

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Given that bond energy of H−Hand Cl−Cl are 430 kJ mo1−1 and 240 kJ mo1−1 respectively and heat of formation for HCl is −90 kJ mo1−1. Bond enthalpy of HCl is:

The correct option is D 425 kJ mo1−112H2(g)+12Cl2(g)→HCl(g) ΔH0f=−90kJmol−1Now, ΔH0r=(12BEH−H+12BECl−Cl)−BEH−Cl−90=12×430+12×240−BEH−ClBEH−Cl=215+120+90 =425kJmol−1So, the correct answer is option D.

Given that bond energy of $H - H$ and $C l - C l$ are $430 k J$ $m o 1^{- 1}$ and $240 k J$ $m o 1^{- 1}$ respectively and heat of formation for $H C l$ is $- 90 k J$ $m o 1^{- 1}$ . Bond enthalpy of $H C l$ is: