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Standard XII

Chemistry

Properties of Enthalpy

The heat of a...

Question

The heat of atomization of a compound $X Y_{3}$ in gaseous state is $E k J m o l^{- 1}$ . What is the bond energy of $X - Y$ bond in $k J m o l^{- 1}$ unit?

E

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3 E

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\frac{E}{3}

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\frac{E}{6}

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Solution

The correct option is C $\frac{E}{3}$
Given, the heat of atomization of a compound $X Y_{3}$ in gaseous state is $E k J m o l^{- 1}$ .
$∴ X Y_{3} (g) \to X (g) + 3 Y (g); Δ H = E k J m o l^{- 1}$
Hence Bond energy of $X - Y$ bond = $\frac{E}{3}$ , as $X Y_{3}$ contains three $X - Y$ bonds.

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Similar questions

Q. The heat of atomization of a compound

X Y_{3}

in gaseous state is

E k J m o l^{- 1}

. What is the bond energy of

X - Y

bond in

k J m o l^{- 1}

unit?

Q. The average energy required to break a P - P bond in

P_{4} (s)

into gaseous atoms is

53.2 k c a l m o l^{- 1} .

The bond dissociation energy of

H_{2} (g)

104.2 k c a l m o l^{- 1}; △_{f} H^{o}

P H_{3} (g)

from

P_{4} (s)

5.5 k c a l m o l^{- 1} .

The P - H bond energy in

k c a l m o l^{- 1}

is [Neglect presence of van der Waals forces in

P_{4} (s)

]
Given:

P_{4} (s)

contains

6 P - P

single bonds.

Q. What is the heat of formation of liquid methyl alcohol in

k J m o l^{- 1},

use the following data.
Heat of vaporization of liquid methyl alcohol =

38 k J m o l^{- 1}

.
Heat of formation of gaseous atoms from the elements in their standard states:

H = 218 k J m o l^{- 1}, C = 715 k J m o l^{- 1}, O = 249 k J m o l^{- 1}

Average bond energies :

C - H = 415 k J m o l^{- 1} C - O = 356 k J m o l^{- 1} O - H = 463 k J m o l^{- 1}

Q. In a planar tetra-atomic molecule

X Y_{3}

X

is at the centroid of the equilateral triangle formed by the atoms,

Y

. If the

X - Y

bond distance is

1 A^{\circ}

. What is the distance between the centres of any two

Y

atoms?

Q. Using the data provided, calculate the multiple bond energy

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

of a

C \equiv \equiv C

bond in

C_{2} H_{2}

. Given that the heat of formation of

C_{2} H_{2}

225 k J m o l^{- 1}

(take the bond energy of a

C - - H

bond as

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}

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The heat of atomization of a compound XY3 in gaseous state is E kJ mol−1. What is the bond energy of X−Y bond in kJmol−1 unit?

The correct option is C E3Given, the heat of atomization of a compound XY3 in gaseous state is E kJ mol−1. ∴XY3(g)→X(g)+3Y(g); ΔH=E kJ mol−1 Hence Bond energy of X−Y bond =E3, as XY3 contains three X−Y bonds.

The heat of atomization of a compound $X Y_{3}$ in gaseous state is $E k J m o l^{- 1}$ . What is the bond energy of $X - Y$ bond in $k J m o l^{- 1}$ unit?

The correct option is C $\frac{E}{3}$
Given, the heat of atomization of a compound $X Y_{3}$ in gaseous state is $E k J m o l^{- 1}$ .
$∴ X Y_{3} (g) \to X (g) + 3 Y (g); Δ H = E k J m o l^{- 1}$
Hence Bond energy of $X - Y$ bond = $\frac{E}{3}$ , as $X Y_{3}$ contains three $X - Y$ bonds.