The heat of neutralization of a strong acid by a strong base is a constant

57.0 k J m o l^{- 1}

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13.7 K c a l m o l^{- 1}

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5.7 \times 10^{4} J m o l^{- 1}

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13.7 \times 10^{3} c a l m o l^{- 1}

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Solution

The correct options are
A $57.0 k J m o l^{- 1}$
B $5.7 \times 10^{4} J m o l^{- 1}$
C $13.7 \times 10^{3} c a l m o l^{- 1}$
D $13.7 K c a l m o l^{- 1}$
The heat of neutralization for strong acid and base is equal to $- 57.3 k J m o l^{- 1}$ or $- 13.7 K c a l m o l^{- 1}$ .
Hence, all options are correct as they have same numerical values

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- 13.7 k c a l m o l^{- 1}

\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}

- 13.7 k c a l m o l^{- 1}

\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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