Question

The enthalpies of neutralisation of $HCl{O}_4$ and $C{l}_{3}C-COOH$ are $-13.5kcal/g-equivalent$ and $-14.7kcal/g-equivalent$respectively when $40gm$of solid NaOH is added to a mix of $1gmol$. $HCl{O}_4$ and $1gmol$. $C{l}_{3}C-COOH$ sodium perchlorate and sodium trichloroacetate are formed in molar ratio of 3 : 1. Then:

• A. $△H$ for the reaction of NaOH with the mix. is $6.45kcal$.
• B. $△H$ for the reaction of NaOH with the mix. is $13.8kcal$.
• C. After reaction the total number of moles of acid remained are 0.5 is $HCl{O}_{4}andC{l}_{3}C-COOH)$
• D. After reaction the total weight of moles of acid remained are $147.7g$ is $HCl{O}_{4}andC{l}_{3}C-COOH)$ and $△H$ for the reaction of NaOH with mix. is $-13.8kcal$

Answer 1

The correct option is D After reaction the total weight of moles of acid remained are $147.7g$ is $HCl{O}_{4}andC{l}_{3}C-COOH)$ and $△H$ for the reaction of NaOH with mix. is $-13.8kcal$

$HCl{O}_4+NaOH\to NaCl{O}_4+H_{2}O$
$C{l}_{3}C-COOH+NaOH\to C{l}_{3}C-COONa+H_{2}O$
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 $NaOH$ consumed for the reaction i.e. $\frac{\frac{x}{40}}{\frac{(1-x)}{40}}=3\Rightarrow x=\frac{3}{4}$
$△H=-13.5\times \frac{3}{4}+(-14.7)\times \frac{1}{4}$
$=\frac{-40.5+14.7}{4}=-13.8kcal$
Weak acids $=\frac{1}{4}(1+35.5+64)+\frac{3}{4}(35.5\times 3+24+32+1)$
$=147.75g$

Standard Enthalpy of Neutralization $\left({\Delta }_{neut}{H}^o\right)$:

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

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

For example

$H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\to H_{2}O\left(l\right)$

${\Delta }_{neut}{H}^o=-57.1kJ/mol=-13.7kcal/mol$

• Always negative

• Exothermic