Question

The correct descending order of the heat liberated (in $kJ$) during the neutralisation of the acids $C{H}_{3}COOH\left(W\right),HF\left(X\right),HCOOH\left(Y\right)$ and $HCN\left(Z\right)$ under identical conditions is:

($K_a$ of $C{H}_{3}COOH=1.8\times {10}^{-5},HCOOH=1.8\times {10}^{-4},HCN=4.9\times {10}^{-10}$ and $HF=3.2\times {10}^{-4}$)

• A. $Y>X>Z>W$
• B. $X>Y>W>Z$
• C. $W>X>Y>Z$
• D. $Z>W>Y>X$
• E. $Z>Y>X>W$

Answer 1

Answer: (B)

$X>Y>W>Z$

Given,
$K_a$ for $C{H}_{3}COOH=1.8\times {10}^{-5}\left(W\right)$

$K_a$ for $HCOOH=1.8\times {10}^{-4}\left(Y\right)$

$K_a$ for $HCN=4.9\times {10}^{-10}\left(Z\right)$ and

$K_a$ for $HF=3.2\times {10}^{-4}\left(X\right)$

Now,

$\because K_a\propto$ Strength of acid

or, $K_a\propto$ Heat liberated during neutralisation

Thus, the order of heat liberated during neutralisation is:

$HF>HCOOH>C{H}_{3}COOH>HCN$

i.e. $X>Y>W>Z$