Question

What is the heat evolved during the formation of $24.95$ grams of hydrated copper sulfate from anhydrous copper sulfate whose molecular weight is $159.5$ grams?
The equation for this reaction is $CuS{O_4} + 5{H_2}O \to CuS{O_4}.5{H_2}O$ and $\Delta H = - 78.2KJ$
(A) $- 78.21KJ$
(B) $78.21KJ$
(C) $7.82KJ$
(D) $- 7.82KJ$

Answer 1

It is given to us that the enthalpy change or the heat evolved in the formation of one mole of hydrated copper sulfate. Calculate the weight of hydrated copper sulfate present in that one mole so that the heat evolved in the formation of $24.95$ grams of it can be calculated.

Complete step-by-step solution: From the given reaction $CuS{O_4} + 5{H_2}O \to CuS{O_4}.5{H_2}O$ it is clear to us that the heat evolved in the formation of one mole of hydrated copper sulfate is $\Delta H = - 78.2KJ$
In order to find the weight of hydrated copper sulfate in one mole, we have to calculate the molecular weight of it
The molecular weight of hydrated copper sulfate is $249.5$ grams.
The heat evolved in forming $249.5$ grams of hydrated copper sulfate is $- 78.2KJ$
Let us assume that the heat evolved in forming $24.95$ grams is $X$
By cross multiplying, we get $249.5 \times X = - 78.2 \times 24.95$
$X = \dfrac{{ - 78.2 \times 24.95}}{{249.5}}$
By solving, we get the value of $X$ as $\dfrac{{1}}{{10}}th$ of the heat evolved during the formation of one mole of hydrated copper sulfate.
Therefore, the amount of heat evolved or the enthalpy change during the formation of $24.95$ grams of hydrated copper sulfate is $X = \dfrac{{ - 78.2}}{{10}} = - 7.82KJ$

Therefore the answer is option (D).

Note: The amount of heat evolved in the formation of one mole of hydrated copper sulfate is given to us so in order to calculate the amount of heat evolved for $24.95$ grams of it, we first find the weight of hydrated copper sulfate in one mole.