The enthalpies of neutralisation of HClO 4 and Cl 3 C - COOH are -13-5 kcal - g - equivalent and -14-7 kcal - g - equivalent respectively when 40 gmof solid NaOH is added to a mix of 1 g mol- HClO 4 and 1 g mol- Cl 3 C - COOH sodium perchlorate and sodium trichloroacetate are formed in molar ratio of 3- 1- Then-A- After reaction the total number of moles of acid remained are 0-5 is HClO 4 and -Cl 3 C - COOH B- H for the reaction of NaOH with the mix- is 6-45 kcal-C- H for the reaction of NaOH with the mix- is 13-8 kcal-D- After reaction the total weight of moles of acid remained are 147-7 g is HClO 4 and -Cl 3 C - COOH and H for the reaction of NaOH with mix- is -13-8 kcal

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Enthalpy of Neutralisation

The enthalpie...

Question

The enthalpies of neutralisation of $H C l O_{4}$ and $C l_{3} C - C O O H$ are $- 13.5 k c a l / g - e q u i v a l e n t$ and $- 14.7 k c a l / g - e q u i v a l e n t$ respectively when $40 g m$ of solid NaOH is added to a mix of $1 g m o l$ . $H C l O_{4}$ and $1 g m o l$ . $C l_{3} C - C O O H$ sodium perchlorate and sodium trichloroacetate are formed in molar ratio of 3 : 1. Then:

△ H

for the reaction of NaOH with the mix. is

6.45 k c a l

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△ H

for the reaction of NaOH with the mix. is

13.8 k c a l

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After reaction the total number of moles of acid remained are 0.5 is

H C l O_{4} a n d C l_{3} C - C O O H)

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After reaction the total weight of moles of acid remained are

147.7 g

H C l O_{4} a n d C l_{3} C - C O O H)

and

△ H

for the reaction of NaOH with mix. is

- 13.8 k c a l

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Solution

The correct option is D After reaction the total weight of moles of acid remained are $147.7 g$ is $H C l O_{4} a n d C l_{3} C - C O O H)$ and $△ H$ for the reaction of NaOH with mix. is $- 13.8 k c a l$
$H C l O_{4} + N a O H \to N a C l O_{4} + H_{2} O$
$C l_{3} C - C O O H + N a O H \to C l_{3} C - C O O N a + H_{2} O$
Let x mol of NaOH are used in 1st reaction and (1 - x) mol of NaOH are used in 2nd reaction.
Given- sodium perchlorate and sodium trichloroacetate are formed in molar ratio of 3 : 1 = moles ratio of $N a O H$ consumed for the reaction i.e. $\frac{\frac{x}{40}}{\frac{(1 - x)}{40}} = 3 \Rightarrow x = \frac{3}{4}$
$△ H = - 13.5 \times \frac{3}{4} + (- 14.7) \times \frac{1}{4}$
$= \frac{- 40.5 + 14.7}{4} = - 13.8 k c a l$
Weak acids $= \frac{1}{4} (1 + 35.5 + 64) + \frac{3}{4} (35.5 \times 3 + 24 + 32 + 1)$
$= 147.75 g$

Standard Enthalpy of Neutralization $(Δ_{n e u t} H^{o})$ :

Heat released when one mole of $H^{+}$ in dilute solution combines with one mole of $O H^{-}$ to give rise to undissociated water at $1 b a r, 298 K$ .

The amount of heat released in formation of one mole of water when an acid is neutralised by a base.

For example

$H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l)$

$Δ_{n e u t} H^{o} = - 57.1 k J / m o l = - 13.7 k c a l / m o l$

Always negative

Exothermic

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Similar questions

Q. The enthalpies of neutralisation of

H C l O_{4}

and

C l_{3} C - C O O H

are

- 13.5 k c a l / g - e q u i v a l e n t

and

- 14.7 k c a l / g - e q u i v a l e n t

respectively when

40 g m

of solid NaOH is added to a mix of

1 g m o l

H C l O_{4}

and

1 g m o l

C l_{3} C - C O O H

sodium perchlorate and sodium trichloroacetate are formed in molar ratio of 3 : 1. Then:

Q. Assertion :The heat of neutralisation of perchloric acid,

H C l O_{4}

, with

N a O H

is same as that of

H C l

with

N a O H

. Reason: Both

H C l

and

H C l O_{4}

are strong acids.

Solutions of sodium hydroxide and acetic acid are mixed by the reaction: $N a O H + C H_{3} C O O H ⟶ C H_{3} C O O N a + H_{2} O$

Acetic acid is a weak acid. If equal volumes of equal molar solutions are mixed, the pH of the solution will be:

Q. The heat of neutralisation of perchloric acid,

H C l O_{4}

, with

N a O H

is same as that of

H C l

with

N a O H

.
If true..................1
else.....................0

Q. The enthalpies of neutralisation of a strong acid

H A

and weak acid

H B

by NaOH are -13.68 kcal/equivalent and -2.9 kcal/equivalent respectively. If one equivalent of

H A

and one equivalent of

H B

are mixed with one equivalent of NaOH, the enthalpy change was found to be - 6.9 kcal/equivalent calculate the distribution of NaOH to neutralise

H A

and

H B

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Enthalpy of Neutralisation

Standard XII Chemistry

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