The H - H bond energy is 430 kJ mol -1 and Cl - Cl bond energy is 240 kJ mol -1- - Hf of HCl is -90 kJ- The H - Cl bond energy is 420-x kJ mol -1- where x is-

12H_2(g)+12Cl_2(g)→ HCl Δ H_F= 90 KJ Δ H_f= B.E for reactants B.E for products. 90=4302+2402 B.E of HCl 90=215+120 #160 ...

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Standard XII

Chemistry

Hess' Law

The H - H b...

Question

The $H - H$ bond energy is 430 kJ mol $^{- 1}$ and $C l - C l$ bond energy is 240 kJ mol $^{- 1}$ . $Δ H_{f}$ of $H C l$ is $- 90$ kJ. The $H - C l$ bond energy is $(420 + x)$ kJ mol $^{- 1}$ , where $x$ is:

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Solution

$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l$
$Δ H_{F} = - 90 K J$
$Δ H_{f} =$ B.E for reactants - B.E for products.
$- 90 = \frac{430}{2} + \frac{240}{2} -$ B.E of $H C l$
$- 90 = 215 + 120 -$ B.E of $H C l$
$- 90 = 335 -$ B.E of $H C l$
B.E. of $H C l$ $= 335 + 90 = 425 K J - m o l e^{- 1}$
$∴ X = 5$ .

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Similar questions

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 :

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. Bond energies of H - H and CI - CI are

430 k J m o l^{- 1}

and

242 k J m o l^{- 1}

respectively.

Δ H_{f}

for HCl is

91 k J m o l^{- 1}

. What will be the bond energy of H - Cl bond (per mole value)?

Q. 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

- 90 k J

m o 1^{- 1}

. Bond enthalpy of

H C l

is:

Q. Heat evolved in the reaction

H_{2} + C l_{2} \to 2 H C l

182

KJ. Bond energies of

H - H

and

C l - C l

are

430

and

242

KJ/mol respectively.

Thus

H - C l

bond energy is :

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The H−H bond energy is 430 kJ mol−1 and Cl−Cl bond energy is 240 kJ mol−1. ΔHf of HCl is −90 kJ. The H−Cl bond energy is (420+x) kJ mol−1, where x is:

12H2(g)+12Cl2(g)→HClΔHF=−90KJΔHf= B.E for reactants - B.E for products.−90=4302+2402− B.E of HCl−90=215+120− B.E of HCl−90=335− B.E of HClB.E. of HCl =335+90=425KJ−mole−1∴X=5.

The $H - H$ bond energy is 430 kJ mol $^{- 1}$ and $C l - C l$ bond energy is 240 kJ mol $^{- 1}$ . $Δ H_{f}$ of $H C l$ is $- 90$ kJ. The $H - C l$ bond energy is $(420 + x)$ kJ mol $^{- 1}$ , where $x$ is: