Question

Standard entropy of $X_2,Y_2$ and $XY$ 3 are $60,40$ and $50$ $J{K}^{-1}$ mo $1^{-1}$, respectively.

For the reaction, $\frac{1}{2}{X}_2+\frac{3}{2}{Y}_2\to XY$ $3$, $\Delta H=-30kJ$, to be at equilibrium, the temperature will be______.

• A. $750$ $K$
• B. $1000$ $K$
• C. $1250$ $K$
• D. $500$ $K$

Answer 1

The correct option is A $750$ $K$

from the $T.dS$ relation we know that in terms of gibb's free energy and enthaply.

$G=H-T.dS$

we know that at equilibrium condition gibb's free energy will be $zero$

now from the given reaction we will find the value of entropy change $S$

so,$dS=(\frac{1}{2}\times 60)+(\frac{3}{2}\times 40)-50=40$

now put these value in above equation to find the temperature$T$

$T.dS=H$

$T=\frac{H}{dS}=\frac{30000}{40}=750K$