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Compounds of Pnictogens-Trends

Enthalpy of n...

Question

Enthalpy of neutralization of the reaction between $C H_{3} C O O H_{(a q)}$ and $N a O H_{(a q)}$ is $- 13.2 K c a l e q^{- 1}$ and that of reaction between $H_{2} S O_{4 (a q)}$ and $K O H_{(a q)}$ is $- 13.7 K c a l e q^{- 1}$ . The enthalpy of dissociation of $C H_{3} C O O H_{(a q)}$ is:

A

- 0.5 k c a l e q^{- 1}

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B

+ 0.5 k c a l e q^{- 1}

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C

- 26.9 k c a l e q^{- 1}

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D

+ 13.45 k c a l e q^{- 1}

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Solution

The correct option is B $+ 0.5 k c a l e q^{- 1}$

Dissociation enthalpy of $C H_{3} C O O H$ = $13.7 - 13.2$ = $0.5 k c a l e q^{- 1}$

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Similar questions

Q. Enthalpy of neutralization of the reaction between

C H_{3} C O O H_{(a q)}

N a O H_{(a q)}

- 13.2 K c a l e q^{- 1}

and that of reaction between

H_{2} S O_{4 (a q)}

K O H_{(a q)}

- 13.7 K c a l e q^{- 1}

. The enthalpy of dissociation of

C H_{3} C O O H_{(a q)}

Q. Enthalpy of neutralization of the reaction between

C H_{3} C O O H_{(a q)}

N a O H_{(a q)}

- 13.2 K c a l e q^{- 1}

and that of reaction between

H_{2} S O_{4 (a q)}

K O H_{(a q)}

- 13.7 K c a l e q^{- 1}

. The enthalpy of dissociation of

C H_{3} C O O H_{(a q)}

Q. Enthalpy of neutralization of the reaction between

C H_{3} C O O H (a q)

N a O H (a q)

k c a l e q^{- 1}

and that of the reaction between

H_{2} S O_{4} (a q)

K O H (a q)

k c a l e q^{- 1}

. The enthalpy of dissociation of

C H_{3} C O O H (a q)

Q. Enthalpy of neutralisation of the reaction between

{C H}_{3} C O O H (a q .)

N a O H (a q .)

- 13.2 k c a l {e q}^{- 1}

and that of the reaction between

H_{2} {O S}_{4} (a q .)

K O H (a q .)

13.7 k c a l {e q}^{- 1}

. The enthalpy of dissociation of

{C H}_{3} C O O H (a q .)

Q. Enthalpy of neutralisation of the reaction between

C H_{3} C O O H (a q .) a n d N a O H (a q .) i s - 13.2 k c a l e q^{- 1}

and that of the reaction between

H_{2} S O_{4} (a q .)

and KOH (aq.) is - 13.7 kcal

e q^{- 1} .

The enthalpy of dissociation of

C H_{3} C O O H (a q .)

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