If $x_{1}$ , $x_{2}$ and $x_{3}$ are enthalpies $H - H, O = O$ and $O - H$ bonds respectively, and $x_{4}$ is the enthalpy of vaporization of water. Estimate the standard enthalpy of combustion of hydrogen.

x_{1} + \frac{x_{2}}{2} - 2 x_{3} + x_{4}

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x_{1} + \frac{x_{2}}{2} - 2 x_{3} - x_{4}

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x_{1} + \frac{x_{2}}{2} - x_{3} + x_{4}

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2 x_{3} - x_{1} - \frac{x_{2}}{2} - x_{4}

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Solution

The correct option is B $x_{1} + \frac{x_{2}}{2} - 2 x_{3} - x_{4}$
Combustion of hydrogen:
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O (H - O - H)$
As water contains two $O - H$ bonds.
So, combustion enthalpy of hydrogen is:
$Δ H =$ Bond energy of reactant $-$ Bond energy of the product $-$ Enthalpy of vapourisation.

$Δ H = x_{1} + \frac{x_{2}}{2} - 2 x_{3} - x_{4}$

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If x1, x2 and x3 are enthalpies H−H,O=O and O−H bonds respectively, and x4 is the enthalpy of vaporization of water. Estimate the standard enthalpy of combustion of hydrogen.

The correct option is B x1+x22−2x3−x4Combustion of hydrogen:H2+12O2→H2O (H−O−H)As water contains two O−H bonds.So, combustion enthalpy of hydrogen is:ΔH= Bond energy of reactant − Bond energy of the product − Enthalpy of vapourisation. ΔH=x1+x22−2x3−x4

If $x_{1}$ , $x_{2}$ and $x_{3}$ are enthalpies $H - H, O = O$ and $O - H$ bonds respectively, and $x_{4}$ is the enthalpy of vaporization of water. Estimate the standard enthalpy of combustion of hydrogen.