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Hess' Law

Heat of neutr...

Question

Heat of neutralisation is measured by mixing 1 equivalent of acid with 1 equivalent of base in a thermos flask. The values obtained are :

Δ H

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Δ U

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P Δ V

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Δ G

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Solution

The correct option is B $Δ H$
Heat of neutralisation is measured by mixing 1 equivalent of acid with 1 equivalent of base in a thermos flask.
Reaction is made at constant pressure in thermos flask.
The values obtained are enthalpy change $Δ H$ .

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\begin{matrix} List - I & List - II (|Enthalpy change in kcal|) & (Neutralisation) (I) < 13.7 kcal & (P) H C l 1 mol + N a O H 1 mol (II) = 13.7 kcal & (Q) H F 1 mol + N a O H 1 mol (III) > 13.7 kcal & (R) H C l 1 mol + N H_{4} O H 1 mol (IV) = 27.4 kcal & (S) N a O H 2 mol + H_{2} S O_{4} 1 mol (T) N a O H 1 mol + C H_{3} C O O H 1 mol \end{matrix}

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