Question

The enthalpy of solution of $BaC{l}_2\left(s\right)$ and $BaC{l}_2.2H_{2}O\left(s\right)$ are $-20.6$ and $8.8kJmo{l}^{-1}$ respectively. The enthalpy change for the rection $BaC{l}_2\left(s\right)+2H_{2}O\left(aq\right)\to BaC{l}_2\left(s\right).2H_{2}O\left(s\right)$ will be:

• A. $+29.4kJ$
• B. $-11.8kJ$
• C. $-20.6kJ$
• D. $-29.4kJ$

Answer 1

The correct option is D $-29.4kJ$

Solution enthalpy for $BaC{l}_2$

$BaC{l}_2\left(s\right)+H_{2}O\to BaC{l}_2.2H_{2}O\left(aq\right)$ ------- (i)

Solution enthalpy for $BaC{l}_2.2H_{2}O\left(s\right)$-

$BaC{l}_2.2H_{2}O\left(s\right)+H_{2}O\to BaC{l}_2.2H_{2}O\left(aq\right)$ ----- (ii)

To get the enthalphy change for this reaction,

$BaC{l}_2\left(s\right)+2H_{2}O\to BaC{l}_2.2H_{2}O\left(s\right)$,

Reverse the 2nd reaction and add (i) and (ii).

$\Delta H_{sol.}=-20.6-8.8=-29.4kJ$