Question

When $1g$ equivalent of strong acid reacts with strong base heat released is $13.5Kcal$. When $1g$ equivalent $H_{2}A$ is completely neutralised against strong base $13Kcal$ is released. When $1g$ equivalent $B(OH{)}_2$ is completely neutralised against strong acid 10 Kcal heat is released. Calculate enthalpy change when 1 mole $H_{2}A$ is completely neutralized by $B(OH{)}_2$.

• A. $-27Kcal$
• B. $-10Kcal$
• C. $-20Kcal$
• D. $-19Kcal$

Answer 1

The correct option is D $-19Kcal$

$MH+NOH\to MN+H_{2}O$

$\Delta H_1=13.5KCal$

$H_{2}A+2NOH\to N_{2}A+2H_{2}O$

$\Delta H_2=13kcal\times 2=26kcal$

$B(OH{)}_2+2MH\to B{M}_2+2H_{2}O$

$\Delta H_3=10\times 2=20Kcal$

$H_{2}A+B(OH{)}_2\to AB+2H_{2}O$

$2H_{2}O+2MN\to 2MH+2NOH$

$-{\Delta }_1=-13.5\times 2$

$H_{2}A+2NOH\to N_{2}A+2H_{2}O$

$\Delta H_2=26Kcal$

$B(OH{)}_2+2MH\to B{M}_2+2H_{2}O$

$\Delta H_3=20Kcal$

$-\Delta H_1+\Delta H_2+\Delta H_3$

$26+20-27=19Kcal$