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Enthalpy of Combustion

The enthalpy ...

Question

The enthalpy of combustion of $H_{2} (g)$ at 298 K to give $H_{2} O (g)$ is $- 249 k J m o l^{- 1}$ and bond enthalpies of H - H and O = O are $433 k J m o l^{- 1}$ and $492 k J m o l^{- 1},$ respectively. The bond enthalpy of O - H is

464 k J m o l^{- 1}

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928 k J m o l^{- 1}

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232 k J m o l^{- 1}

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

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Solution

The correct option is A $464 k J m o l^{- 1}$
Combustion of $H_{2}$ gives $H_{2} O$ (g),
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$
$Enthalpy of combustion of reaction, Δ H = - 249 k J m o l^{- 1}$
$Enthalpy of reaction = Bond enthalpy of reactant - Bond enthalpy of product$
$Δ H = B . E . o f H_{2} + \frac{1}{2} B . E . o f O_{2} - B . E . o f H_{2} O$
$H_{2} O$ has two OH bond hence,
$Δ H = B . E . o f H_{2} + \frac{1}{2} B . E . o f O_{2} - 2 \times B . E . o f O H$
$- 249 = 433 + \frac{1}{2} \times 492 - 2 \times B . E . o f O H$
$- 928 = - 2 \times B . E . o f O H$
$Bond ethalpy of OH = \frac{928}{2}$
$Bond enthalpy of OH is 464 k J m o l^{- 1}$

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Enthalpy of Combustion

Standard XIII Chemistry

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The enthalpy of combustion of H2(g) at 298 K to give H2O(g) is −249 kJ mol−1 and bond enthalpies of H - H and O = O are 433 kJ mol−1 and 492 kJ mol−1, respectively. The bond enthalpy of O - H is

The enthalpy of combustion of $H_{2} (g)$ at 298 K to give $H_{2} O (g)$ is $- 249 k J m o l^{- 1}$ and bond enthalpies of H - H and O = O are $433 k J m o l^{- 1}$ and $492 k J m o l^{- 1},$ respectively. The bond enthalpy of O - H is