When 1 gm equivalent of strong acid reacts with strong base heat … - Vidyadip

MCQ

When $1\ gm$ equivalent of strong acid reacts with strong base heat released is $13.5\ Kcal$ . when $1\ gm$ equivalent $H_2A$ is completely neutralised against strong base $13\ Kcal$ is released. When $1\ gm$ equivalent $B(OH)_2$ is completely neutralised against strong acid $10\ Kcal$ heat is released. Calculate enthalpy change when $1\ mole\ H_2A$ is completely neutralised by $B(OH)_2$........$K\,cal$

A
$-27$
B
$-10$
C
$-20$
✓
$-19$

✓

Answer

Correct option: D.

$-19$

d
$MH + NOH \rightarrow MN + H _{2} O$

$\Delta H _{1}=13.5 KCal$

$H _{2} A +2 NOH \rightarrow N _{2} A +2 H _{2} O$

$\Delta H _{2}=13 kcal \times 2=26 kcal$

$B ( OH )_{2}+2 MH \rightarrow BM _{2}+2 H _{2} O$

$\Delta H _{3}=10 \times 2=20 Kcal$

$H _{2} A + B ( OH )_{2} \rightarrow AB +2 H _{2} O$

$2 H _{2} O +2 MN \rightarrow 2 MH +2 NOH$

$-\Delta_{1}=-13.5 \times 2$

$H _{2} A +2 N OH \rightarrow N _{2} A +2 H _{2} O$

$\Delta H _{2}=26 K cal$

$B ( OH )_{2}+2 MH \rightarrow BM _{2}+2 H _{2} O$

$\Delta H _{3}=20 K cal$

$-\Delta H _{1}+\Delta H _{2}+\Delta H _{3}$

$26+20-27= - 19 K cal$

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