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Oxides of Nitrogen

The bond enth...

Question

The bond enthalpies of $H_{2}, X_{2}$ and HX are in the ratio of 2: 1: 2. If the enthalpy for formation of $H X$ is $- 50 k J m o l^{- 1}$ , the bond enthalpy of $H_{2}$ is:

200 k J m o l^{- 1}

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400 k J m o l^{- 1}

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100 k J m o l^{- 1}

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300 k J m o l^{- 1}

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Solution

The correct option is C $100 k J m o l^{- 1}$
$\frac{1}{2} H_{2} + \frac{1}{2} X_{2} \to H X$
Let bond enthalpy of X-X bond be x, then
$\Rightarrow \frac{1}{2} \times 2 x + \frac{1}{2} \times x - 2 x$ = - 50
$\Rightarrow x + 0.5 x - 2 x = - 50$ KJ/mol
$\Rightarrow - 0.5 x = - 50$
$\Rightarrow x = \frac{- 50}{- 0.5}$ =100KJ/mol

Hence, the correct option is C.

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Similar questions

The bond enthalpies of $H_{2}, X_{2}$ an HX are in the ratio of 2 : 1 : 2. If enthalpy of formation of HX is $- 50 k J m o l^{- 1}$ , the bond enthalpy of $X_{2}$ is

Q. The bond dissociation energies of

X_{2}

Y_{2}

and

X Y

are in the ratio of 1 : 0.5 : 1.

△ H

for the formation of

X Y

is –200 kJ

m o l^{- 1}

. The bond dissociation energy of

X_{2}

will be

A B, A_{2} a n d B_{2}

are diatomic molecules. If the bond enthalpies of

A_{2}, A B a n d B_{2}

are in the ratio 1 : 1 : 0.5 and the enthalpy of formation of AB from

A_{2} a n d B_{2}

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

, what is the bond enthalpy of

A_{2}

Q. Bond dissociation enthalpies of

H_{2} (s)

and

N_{2} (g)

are

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

and enthalpy of formation of

{N H}_{3} (g)

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

. What is enthalpy of atomization of

{N H}_{3} (g)

? What is the average bond enthalpy of

N - H

bond?

A B, A_{2}

and

B_{2}

are in the ratio

1 : 1 : 0.5

and the enthalpy of formation of

A B

from

A_{2}

and

B_{2}

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

. What is the bond enthalpy of

A_{2}

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The bond enthalpies of H2,X2 and HX are in the ratio of 2: 1: 2. If the enthalpy for formation of HX is −50kJmol−1 , the bond enthalpy of H2 is:

The correct option is C 100kJmol−112H2+12X2→HXLet bond enthalpy of X-X bond be x, then⇒12×2x+12×x−2x= - 50⇒x+0.5x−2x=−50KJ/mol⇒−0.5x=−50⇒x=−50−0.5=100KJ/molHence, the correct option is C.

The bond enthalpies of $H_{2}, X_{2}$ and HX are in the ratio of 2: 1: 2. If the enthalpy for formation of $H X$ is $- 50 k J m o l^{- 1}$ , the bond enthalpy of $H_{2}$ is: