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Difference between Internal Energy and Enthalpy

Enthalpy of n...

Question

Enthalpy of neutralization of $H_{3} {P O}_{3}$ acid is $- 106.68 k J / m o l$ using $N a O H$ . If the enthalpy of neutralization of $H C l$ by $N a O H$ is $- 55.84 k J / m o l$ . Calculate $Δ H_{i o n i z a t i o n}$ of $H_{3} {P O}_{3}$ into its ions:

50.84 k J / m o l

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5 k J / m o l

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2.5 k J / m o l

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none of these

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Solution

The correct option is B $5 k J / m o l$

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Similar questions

H_{3} P O_{3}

N a O H

H C l

N a O H

△ H_{i o n i z a t i o n} o f H_{3} P O_{3}

H_{3} P O_{3}

N a O H

- 106.68 {kJ.mol}^{- 1}

H C l

N a O H

- 55.84 {kJ.mol}^{- 1}

Δ H_{ionization}

H_{3} P O_{3}

{C H}_{3} C O O H

N a O H

- 49.86 k J / m o l

{C H}_{3} C O O H

H^{+} (a q) + {O H}^{-} (a q) ⟶ H_{2} O (l); Δ_{r} H^{o} = - 55.84 k J / m o l

N H_{4} O H

N H_{4} O H

C H_{3} C O O H

N a O H

- 50.6 k J / m o l

N a O H

- 55.9 k J / m o l

Δ H

C H_{3} C O O H

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