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 8.8 kJ $mo{l}^{-1}$ respectively, the enthalpy change for the hydration of $BaC{l}_2\left(s\right)$ is:

• A. 29.4 kJ
• B. – 11.8 kJ
• C. – 29.4 kJ
• D. + 11.8 kJ

Answer 1

The correct option is C – 29.4 kJ

$\left(i\right)BaC{l}_2\left(s\right)\to BaC{l}_2\left(aq\right);\Delta H_1=-20.6kJ\left(ii\right)BaC{l}_{2}2H_{2}O\left(s\right)\to BaC{l}_2\left(aq\right);\Delta H_2=8.8kJ\text{Subtracting (ii) from (i) we get}BaC{l}_2\left(s\right)\to BaC{l}_2.2H_{2}O\left(s\right);\Delta H_1-\Delta H_2=-29.4kJ$