Question

The enthalpy of solution of anhydrous $CuS{O}_4$ is $-16kCal$ and that of $CuS{O}_4.5H_{2}O$ is $3kCal$. Calculate $\Delta H$ for the following reaction at $500K$ and constant pressure.

Answer 1

$Cu{SO}_4\left(s\right)+aq\to Cu{SO}_4\left(aq\right)$

$x(-1)(Cu{SO}_4\cdot 5H_{2}O+aq)\to Cu{SO}_4\left(aq\right)+3H_{2}0$

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$Cu{SO}_4\left(s\right)+5H_{2}O\to Cu{SO}_4\cdot 5H_{2}O$

$\Delta H=3Kcal$

$=16-3$

$=-18Kcal$