Question

Concrete is produced from a mixture of cement, water and small stones. Small amount of gypsum, $CaS{O}_4\cdot 2H_{2}O$ 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 $CaS{O}_4\cdot \frac{1}{2}{H}_{2}O$ according to reaction.
$CaS{O}_4\cdot 2H_{2}O\left(s\right)\to CaS{O}_4\cdot \frac{1}{2}{H}_{2}O\left(s\right)+\frac{3}{2}{H}_{2}O\left(g\right)$
The ${\Delta }_f{H}^{\ominus }$ of $CaS{O}_4\cdot 2H_{2}O\left(s\right),CaS{O}_4\frac{1}{2}{H}_{2}O\left(s\right),H_{2}O\left(g\right)$ are $-2021.0kJmo{l}^{-1}$, $-1575.0kJmo{l}^{-1}$ and $-241.8kJmo{l}^{-1}$ respectively. The respective values of their standard entropies are $194.0$, $130.0$ and $188.0J{K}^{-1}mo{l}^{-1}.$

$R=8.314J{K}^{-1}mo{l}^{-1}=0.0831Lbarmo{l}^{-1}{K}^{-1}$
Answer the follwoing questions on the basis of above information.

The value of equilibrium constant for reaction is:

• A. 0
• B. <1
• C. >1
• D. =1

Answer 1

The correct option is B <1

$\Delta H^{\ominus }=\Delta H_P^{\ominus }-\Delta H_R^{\ominus };\left(for1mol\right)$
$=[-1575.0kJmo{l}^{-1}-\frac{3}{2}\times 241.8]-[-2021.0kJmo{l}^{-1}]$
$=+83.3kJmo{l}^{-1}$
For 1 Kg $CaS{O}_4\cdot 2H_{2}O$
Numebr of moles $=\frac{1000}{172}$
$=5.81$
$\therefore$ Heat change for $5.81molofCaS{O}_4\cdot 2H_{2}O=5.81\times 83.3kJmo{l}^{-1}$
$=484kJmo{l}^{-1}$
$\Delta G=\Delta H-T\Delta S$
$=17.92kJ$ ........ $(\Delta S=S_P-S_R)$
$\Delta G=-nRTlnK$
$\therefore$ $K=e^{-\Delta GlnRT}<1$