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  • If the heat of dissolution of anhydrous <img alt="\mathrm{\mathrm{CuSO}_4 \: and \: \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\: is -60 \mathrm{KJ} / \mathrm{mol} \: and -11.2 \mathrm{KJ} / \m

If the heat of dissolution of anhydrous \mathrm{\mathrm{CuSO}_4 \: and \: \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\: is -60 \mathrm{KJ} / \mathrm{mol} \: and -11.2 \mathrm{KJ} / \mathrm{mol},} respectively, then heat of hydration \mathrm{(in\; \mathrm{kJ} / \mathrm{mol} ) \; of\; \mathrm{CuSO}_4} to form \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} is

Option: 1

71.2


Option: 2

-71.2


Option: 3

48.8


Option: 4

-48.8


Answers (1)

best_answer

Let's consider the following reaction:
\mathrm{CuSO}_4+5 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}

The heat of dissolution of \mathrm{\mathrm{CuSO}_4\: is -60 \mathrm{~kJ} / \mathrm{mol}}, which means that when 1 mole of \mathrm{CuSO}_4 dissolves in water, 60 \mathrm{~kJ} of heat is released.
The heat of dissolution of \mathrm{\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}\: is -11.2 \mathrm{~kJ} / \mathrm{mol}}, which means that when 1 mole of \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} dissolves in water, 11.2 \mathrm{~kJ} of heat is released.
Now, let's consider the hydration of \mathrm{\mathrm{CuSO}_4 \: to\: form \: \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}}. This can be represented as follows:

\mathrm{CuSO} 4+5 \mathrm{H}_ 2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CuSO} 4 \cdot 5 \mathrm{H}_2 \mathrm{O}(\mathrm{s})

We can break this reaction into two steps:

1. Hydration of \mathrm{CuSO_4: CuSO_4(\mathrm{s})+\mathrm{nH}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CuSO}_4 \cdot \mathrm{nH}_2 \mathrm{O}(\mathrm{s})}
2. Dissolution of \mathrm{CuSO4 \cdot \mathrm{nH}_2 \mathrm{O}} in excess water: \mathrm{\mathrm{CuSO} _4 \cdot \mathrm{nH}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{CuSO}_4 \cdot \mathrm{nH}_2 \mathrm{O}(\mathrm{aq})}

The heat of hydration of C u S O_4 can be calculated by subtracting the enthalpy change of the second step from the enthalpy change of the overall reaction:

\Delta \mathrm{H}=\Delta \mathrm{H}_1+\Delta \mathrm{H}_2
where \Delta \mathrm{H}_1 is the enthalpy change of the hydration step and \Delta \mathrm{H}_2 is the enthalpy change of the dissolution step.
Since \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} is formed, \mathrm{n}=5

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manish

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