You visited us 1 times! Enjoying our articles? Unlock Full Access!

Heat Capacity

For the react...

Question

For the reaction, $O (g) + O_{2} (g) \to O_{3} (g); Δ H^{o} = - 107.2 k J$ . Hence, the average bond energy in $O_{3}$ is:
(Assume $O = O$ bond energy $498.8 k J m o l^{- 1}$ )

606.0 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

107.2 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

391.6 k J m o l^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

303.0 k J m o l^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $303.0 k J m o l^{- 1}$
$O_{2} (g) + O_{2} (g) \to O_{3} Δ H = - 107.2$ KJ/mol

To calculate average bond enthalpy in $O_{3}$

The equation of the bond enthalpy is
the change in enthalpy of reaction $=$ (the sum of the bonding energy of reactants)-(the sum of the bonding energy of the products)

The change of enthalpy is given as $= -$ $107.2$ KJ/mol

- $107.2$ KJ/mol =(bond enthalpy sum of reactants)-2(bond enthalpy sum of products)

- $107.2$ KJ/mol = $498.7$ KJ/mol-2(bond enthalpy of products)

$498.7$ KJ/mol is the BE of $O_{2}$

its 2 times the bonding enthalpy products because of their 2 kinds of bonds in $O_{3}$ - a single and a double bond

$605.9$ KJ/mol = 2BE

BE of products = $303.0$ KJ/mol

Suggest Corrections

Similar questions

For the reaction,

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)$ ; $Δ H = - 571 k J$ .

Bond energies: $(H - H) = 435 k J$ , $(O = O) = 498 k J$ .

Calculate the average bond energy of $(O - H)$ using the above data.

N_{2} O

△ H^{\circ} (N_{2} O) = 82 k J m o l^{- 1}

N^{Θ} = N^{⨁} = O

(N \equiv N) b o n d = 950 k J m o l^{- 1}

(N = N) b o n d = 420 k J

^{- 1}

O = O) b o n d = 500 k J m o l^{- 1}

(O = N) b o n d = 610 k J m o l^{- 1}

m o l^{- 1}

m o l^{- 1}

m o l^{- 1}

m o l^{- 1}

Δ H

2 C l_{2} + 2 H_{2} O \to 4 H C l + O_{2}

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

Δ H

H_{2} (g) + O_{2} (g) ⟶ H_{2} O_{2} (g)

H - H

436 k J

{m o l}^{- 1}

O - O

142 k J

{m o l}^{- 1}

O = O

499 k J

{m o l}^{- 1}

H - O

460 k J

{m o l}^{- 1}

Join BYJU'S Learning Program