Question

The value of $\Delta H_{sol.}$ of $BaC{l}_2\left(s\right)$ and $BaC{l}_2.2H_{2}O\left(s\right)$ are - a kJ and b kJ respectively. The value of $\Delta H_{Hydration}$ of $BaC{l}_2\left(s\right)$ is:

• A. b - a
• B. a + b
• C. - a - b
• D. a - b

Answer 1

Answer: (C)

- a - b

Solution:- (C) $-a-b$

${Ba{Cl}_2}_{\left(s\right)}+aq.⟶{Ba{Cl}_2}_{aq.}\Delta H=-akJ.....\left(1\right)$

${Ba{Cl}_2.2H_{2}O}_{\left(s\right)}+aq.⟶{Ba{Cl}_2}_{(aq.)}+2H_{2}O\Delta H=bkJ$

${Ba{Cl}_2}_{(aq.)}+2H_{2}O⟶{Ba{Cl}_2.2H_{2}O}_{\left(s\right)}+aq.\Delta H=-bkJ.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

${Ba{Cl}_2}_{\left(s\right)}+aq.+{Ba{Cl}_2}_{(aq.)}+2{H_{2}O}_{\left(l\right)}⟶{Ba{Cl}_2}_{aq.}+{Ba{Cl}_2.2H_{2}O}_{\left(s\right)}+aq.\Delta H=(-a)+(-b)$

${Ba{Cl}_2}_{\left(s\right)}+2{H_{2}O}_{\left(l\right)}⟶Ba{Cl}_2.2H_{2}O\Delta H=-a-bkJ$

Hence the heat of hydration of anhydrous $Ba{Cl}_2$ is $-a-bkJ$.