Question

Concrete is produced from a mixture of cement, water and small stones. A small amount of gypsum, $CaS{O_4} \cdot 2{H_2}O$ is added in cement production to improve the subsequent hardening of concrete.
The elevated temperature during the production of cement may lead to the formation of unwanted hemihydrate $CaS{O_4} \cdot 21{H_2}O$ according to reaction.
$CaS{O_4} \cdot 2{H_2}O \to CaS{O_4} \cdot 21{H_2}O(s) + 23{H_2}O(g)$
The ${\Delta _f}{H^\Theta }$ of $CaS{O_4} \cdot 2{H_2}O(s)$, $CaS{O_4} \cdot 21{H_2}O(s),{H_2}$ ​
$O(g)$ are $- 2021.0$ kJ/mol, $- 1575.0$ kJ/mol and $- 241.8$ kJ/mol respectively. The respective values of their standard entropies are $194.0,130.0$ and $188.0$ /JKmol.
R = $8.314 JK^{−1}mol^{−1}$ = $0.0831 L bar mol^{−1}K^{−1}$
Answer the following questions on the basis of above information.
Heat change occurring during conversion of $1$ Kg of $CaS{O_4} \cdot 2{H_2}O(s)$ (molar mass 172 g/ mol) of $CaS{O_4} \cdot \dfrac{1}{2}{H_2}O(s),O(s)$ ​O(s) is equal to:
a) $484$
b) $400$
c) $- 484$
d) $- 1000$

Answer 1

The given question is from the topic Thermodynamics. The answer is simple as we will use the equation for the change in heat to find the value of heat change in the formation of $CaS{O_4} \cdot 2{H_2}O$.

Complete Step by step answer:
The equation of heat of formation is given by
$\Delta {H^\Theta } = \Delta {H_P}^{^\Theta } - \Delta {H_R}^{^\Theta }$ for 1 mole
In the question we have $\Delta {H_P}^{^\Theta } = - 1575.0 - \dfrac{3}{2} \times 241.8,\Delta {H_R}^{^\Theta } = - 2021.0$ ( the units are in KJ/mol)

Putting these together we get:
$\Delta {H^\Theta } = [ - 1575.0$ KJ/mol $- \dfrac{3}{2} \times 241.8]$ $- [ - 2021.0]$ KJ/mol
$\Rightarrow \Delta {H^\Theta } = - 1575.0 - \dfrac{3}{2} \times 241.8 + 2021.0$ KJ/mol
On simplification of the above equation we have:
$\Delta {H^\Theta } = - 1575.0 - 362.7 + 2021.0$ KJ/mol
$\Rightarrow \Delta {H^\Theta } = - 1937.7 + 2021.0$ KJ/mol
$\Rightarrow \Delta {H^\Theta } = 83.3$ KJ/mol
The above answer is for one mole.
Now for 1 kg of $CaS{O_4} \cdot 2{H_2}O$, number of moles is $\dfrac{{1000}}{{172}}$
Therefore, the heat change for $\dfrac{{1000}}{{172}}$ moles of $CaS{O_4} \cdot 2{H_2}O = \dfrac{{1000}}{{172}} \times 83.3 = 484.302 \approx 484$.

The answer we found out is identical to option ‘a’, so it is the correct option.

Note: Always remember the process of conversion of moles into Kg and other units.
Here in the solution we have to find out the heat change for $\dfrac{{1000}}{{172}}$ which is equal to 5.81 mole so here in this process instead of finding directly for 5.81 mole it is better to find the heat change for one mole and the calculate for 5.81 moles.