The heat of neutralisation of a strong acid and a strong alkali is $- 57.0 k J m o l^{- 1} .$ The heat released when $0.5 m o l e$ of $H N O_{3}$ solution is mixed with $0.2 m o l e$ of $K O H$ is:

57.0 k J

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11.4 k J

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28.5 k J

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34.9 k J

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Solution

The correct option is B $11.4 k J$
The reaction for neutralisation is given as follows:

$H N O_{3} + K O H \to K N O_{3} + H_{2} O$
This neutralisation releases $57 k J$ of energy.
We can cleary see that the molar ratio in which $H N O_{3} a n d K O H$ are required is 1:1
So, although we have, $0.5 m o l e s o f H N O_{3} a n d 0.2 m o l e s o f K O H$ ,
From the given data we can say,
$K O H$ will act as the limiting reagent and $0.2 m o l e s$ of each one will be neutralised.

So, given here is that,
$1 m o l$ of neutralisation of each releases $57.0 k J$
$∴$ $0.2 m o l$ neutralisation will release an energy of
$= 57.0 \times 0.2 k J = 11.4 k J$

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