question_answer

The heat of hydration of solution anhydrous \[CuS{{O}_{4}}\] and hydrated \[CuS{{O}_{4}}\cdot 5{{H}_{2}}O\] are - 66.5 and 11.7 kJ \[mo{{l}^{-1}}\]respectively. Calculate the heat of hydration of \[CuS{{O}_{4}}\] to \[CuS{{O}_{4}}\cdot 5{{H}_{2}}O\]

A) - 78.2 kJ

B) - 81.2 kJ

C) - 68.2 kJ

D) - 60.1 kJ

Correct Answer: A

Solution :

[a] The required equation is :

\[CuS{{O}_{4}}\left( S \right)+5{{H}_{2}}O\left( l \right)\to \]\[CuS{{O}_{4}}.5{{H}_{2}}O\left( s \right)\Delta H=?\]

given that

\[CuS{{O}_{4}}.5{{H}_{2}}O\left( S \right)+aq.\to CuS{{O}_{4}}.\left( aq \right);\Delta {{H}_{2}}=+11.7kJ\]

The process of hydration may be expressed as

\[\underset{\Delta H}{\mathop{CuS{{O}_{4}}}}\,\left( s \right)\xrightarrow{\Delta {{H}_{1}}}\underset{\Delta {{H}_{2}}}{\mathop{CuS{{O}_{4}}\left( aq \right)}}\,\]\[CuS{{O}_{4}}.5{{H}_{2}}O\]

According to Hess's law

\[\Delta {{H}_{1}}=\Delta H+\Delta {{H}_{2}}\]

\[\Delta {{H}_{1}}=\Delta H-\Delta {{H}_{2}}\]

\[=-\,66.5-11.7\text{ }=-78.2\text{ }kJ\]