$\frac{1}{2}{X_2} + \frac{3}{2}{Y_2} \to X{Y_3},\,\Delta H = - 30\,kJ$ , to be at equilibrium, the temperature will be ............... $\mathrm{K}$
since $\Delta G=\Delta H-T \Delta S$
so at equilibrium $\Delta H-T \Delta S=0$
or $\quad \Delta H=T \Delta S$
For the reaction
$\frac{1}{2} X_{2}+\frac{3}{2} Y_{2} \rightarrow X Y_{3}$
$\Delta H=-30 \,k J$ (given)
Calculating $\Delta S$ for the above reaction, we get
$\Delta S=50-\left[\frac{1}{2} \times 60+\frac{3}{2} \times 40\right] \,J K^{-1}$
$=50-(30+60) \,J K^{-1}$
$=-40 \,J K^{-1}$
At equilibrium, $T \Delta S=\Delta H$
$[\text { as } \quad \Delta G=0]$
$\therefore \quad T \times(-40)=-30 \times 1000$
$[\text { as } 1 \,k J=1000\, J]$
$\quad T=\frac{-30 \times 1000}{-40}=\quad 750 \,K$
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${A_2}\left( g \right) + {B_2}\left( g \right) \rightleftharpoons {C_2}\left( g \right) + {D_2}\left( g \right)$
If we take $1\ mole$ of each of the four gases in a $10\ litre$ container, what would be equilibrium concentration of $A_2(g)$?
$I.$ $1$ molecule of oxygen
$II.$ $1$ atom of nitrogen
$III.$ $1 \times {10^{ - 10}}$ $g$ molecular weight of oxygen
$IV.$ $1 \times {10^{ - 10}}$ $g$ atomic weight of copper
