Question

The heats evolved and absorbed when $100g$ each of $\text{CuS}{\text{O}}_{\text{2}}$ and $\text{CuS}{\text{O}}_{\text{2}}$.$\text{5}{\text{H}}_{\text{2}}\text{O}$ are dissolved in water are $41.7$ and $4.7kJ/mol$ respectively.The heat of hydration of $\text{CuS}{\text{O}}_{\text{2}}$ is (in kJ/mol)

• A. $-66.59$
• B. $+11.72$
• C. $-78.2$
• D. $+78.2$

Answer 1

The correct option is C $-78.2$

100 gm of $CuS{O}_4$ = $\frac{100}{159.5}mole$

Heat of solution of $CuS{O}_4$ = $-41.7\times \frac{159.5}{100}=-66.51kJ/mol$

100 gm of $CuS{O}_4.5H_{2}O$ = $\frac{100}{249.5}mole$

Heat of solution of $CuS{O}_4.5H_{2}O$ = $4.7\times \frac{249.5}{100}=11.72kJ/mol$

$\therefore$ Now,

$CuS{O}_4+\left(aq\right)\to {CuS{O}_4}_{\left(aq\right)}$ $\Delta H=-66.51kJ/mol$

${CuS{O}_4}_{\left(aq\right)}\to CuS{O}_4.5H_{2}O+aq$ $\Delta H=-11.72kJ/mol$

Adding both equation we get heat of hydration of $CuS{O}_4$ = -78.2kJ/mol.