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Formation of ozone takes place as O(2)(g...

Formation of ozone takes place as `O_(2)(g) + O(g) rarr O_(3)(g) , DeltaH^(@) =- 107 . 2kJ` . Assuming `O = O` bond energy as `498.8 kJ mol^(-1)`, the average bond energy of ozone is

A

`391.6 kJmol^(-1)`

B

` 606 . 0 kJmol^(-1)`

C

` 107.2kJ mol^(-1)`

D

` 302 .6 kJ mol^(-1)`

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The correct Answer is:
To find the average bond energy of ozone (O₃), we can use the given reaction and the bond energies involved. Let's break it down step by step. ### Step 1: Write the reaction and identify the given values The formation of ozone is represented by the reaction: \[ O_2(g) + O(g) \rightarrow O_3(g) \] Given: - \(\Delta H^\circ = -107.2 \, \text{kJ}\) - Bond energy of \(O=O\) (double bond) = \(498.8 \, \text{kJ/mol}\) ### Step 2: Understand the bond energies involved In the reaction, we have: - One \(O=O\) bond in \(O_2\) - One \(O\) atom that will form bonds in ozone - Ozone (O₃) has a resonance structure, which means it has equivalent bonds that can be considered when calculating average bond energy. ### Step 3: Write the expression for the change in enthalpy The change in enthalpy for the reaction can be expressed in terms of bond energies: \[ \Delta H^\circ = \text{Bond energies of reactants} - \text{Bond energies of products} \] For this reaction: - Bond energy of \(O_2\) (1 O=O bond) = \(498.8 \, \text{kJ/mol}\) - Bond energy of \(O\) = \(0 \, \text{kJ/mol}\) (since it is a single atom) - Bond energy of \(O_3\) can be represented as \(3 \times BE_{O}\) (since there are three bonds in ozone). ### Step 4: Set up the equation Substituting the values into the equation: \[ -107.2 \, \text{kJ} = (498.8 + 0) - 3 \times BE_{O} \] This simplifies to: \[ -107.2 = 498.8 - 3 \times BE_{O} \] ### Step 5: Rearranging the equation Rearranging gives: \[ 3 \times BE_{O} = 498.8 + 107.2 \] \[ 3 \times BE_{O} = 606 \, \text{kJ} \] ### Step 6: Solve for the average bond energy of ozone Now, divide by 3 to find the average bond energy: \[ BE_{O} = \frac{606}{3} = 202 \, \text{kJ/mol} \] ### Conclusion The average bond energy of ozone (O₃) is: \[ \boxed{202 \, \text{kJ/mol}} \]

To find the average bond energy of ozone (O₃), we can use the given reaction and the bond energies involved. Let's break it down step by step. ### Step 1: Write the reaction and identify the given values The formation of ozone is represented by the reaction: \[ O_2(g) + O(g) \rightarrow O_3(g) \] Given: - \(\Delta H^\circ = -107.2 \, \text{kJ}\) - Bond energy of \(O=O\) (double bond) = \(498.8 \, \text{kJ/mol}\) ...
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H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g) DeltaH =- 242 kJ mol^(-1) Bond energy of H_(2) and O_(2) is 436 and 500 kJ mol^(-1) , respectively. What is bond energy of O-H bond?

For the reaction: 2H_(2)(g) +O_(2)(g) rarr 2H_(2)O(g), DeltaH =- 571 kJ bond enegry of (H-H) = 435 kJ and of (O=O) = 498 kJ . Then, calculate the average bond enegry of (O-H) bond using the above data.

Knowledge Check

  • For the reaction 2 H_(2) + O_(2) rarr 2H_(2)O, DeltaH = - 571 . Bond energy of H-H =435 and O=O=498 . Then the average bond energy of O-H bond will be

    A
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    B
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    C
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    D
    `-271`
  • The enthalpy changes at 298 kJ in successive breaking of O-H bonds of water, are H_(2)O(g) rarr H(g)+OH(g),DeltaH=498kJ mol^(-1) OH(g) rarr H(g)+O(g), DeltaH =428 kJ mol^(-1) The bond energy of the O-H bond is

    A
    `498 kJ mol^(-1)`
    B
    `463 kJ mol^(-1)`
    C
    `428 kJ mol^(-1)`
    D
    `70 kJ mol^(-1)`
  • For the reaction 2H_(2)+O_(2)rarr2H_(2)O, DeltaH=-571 . Bond energy of H-H = 435, O=O = 498, then calculate the average bond energy of O-H bond using the above data

    A
    484
    B
    `- 484`
    C
    271
    D
    `- 271`
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