Question

The value of $△H_{sol.}$ of ${BaCl}_2\left(s\right)$ and ${BaCl}_2$. ${2H}_{2}O\left(s\right)$ are $a$ kJ and $b$ kJ respectively. The value of $△H_{Hydration}$ of ${BaCl}_2\left(s\right)$ is:

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

Answer 1

Answer: (C)

$-a-b$

$\left(i\right)-{BaCl}_{2\left(s\right)}+aq⟶{BaCl}_{2\left(aq\right)},{△H}_{sol}=-aKJ$

$\left(ii\right)-{BaCl}_2.2H_2{O}_{\left(s\right)}+aq⟶{BaCl}_{2\left(aq\right)},{△H}_{sol}=bKJ$

Equation$\left(ii\right)$ can be written in two steps:-

$\left(iii\right)-{BaCl}_{2\left(s\right)}+2H_{2}O⟶{BaCl}_2.2H_{2}O\left(s\right),△H_1$

$\left(iv\right)-{BaCl}_2.2H_{2}O\left(s\right)+aq⟶{BaCl}_2\left(aq\right),△r{H}_2=+bKJ$

${△}_r{H}_1+{△}_r{H}_2=-a$

$-a-b={△}_r{H}_1$

But we want,

$\left(v\right)=\left(iii\right)$

$\therefore {△H}_{hydration}=-a-b={△}_r{H}_1$