Question

Standard heat of formation of $HgO\left(s\right)$ at 298 K and at constant pressure is $-90.8kJmo{l}^{-1}$. If excess of $HgO\left(s\right)$ absorbs 41.84 kJ of heat at constant volume, the mass of Hg that can be obtained at constant volume and 298 K is:
Note : Atomic mass of $Hg=200$.

• A. $93.4g$
• B. $46.7g$
• C. $85.56g$
• D. $75.56g$

Answer 1

The correct option is A $93.4g$

Given that,

$Hg\left(l\right)+\frac{1}{2}{O}_2\left(g\right)⟶HgO\left(s\right);\Delta H^{\circ }=-90.8kJ$

$\therefore HgO\left(s\right)⟶Hg\left(l\right)+\frac{1}{2}{O}_2\left(g\right);\Delta H^{\circ }=+90.8kJ$

$\Delta H^{\circ }=\Delta U^{\circ }+\Delta nRT$

$90.8=\Delta U^{\circ }+\frac{1}{2}\times 8.314\times {10}^{-3}\times 298[\because \Delta n=\frac{1}{2}]$

$\therefore \Delta U^{\circ }=89.56kJmo{l}^{-1}$

$\therefore \Delta U^{\circ }$ is heat of formation of $Hg\left(l\right)$ from $HgO\left(s\right)$.

Now, $41.84kJ$ heat is absorbed by HgO,

Thus, mole of Hg formed

$=\frac{41.84}{89.56}=0.4672mol$

$\therefore W_{Hg}=0.4672\times 200=93.4g$