Question

Standard heat of formation of $HgO\left( s \right)$ at $298K$ and at constant pressure is $-90.8kJmo{{l}^{-1}}$ . If the excess of $HgO\left( s \right)$absorbs $41.8kJ$ of heat at constant volume the mass of $Hg$ that can be obtained at constant volume and $298K$is:
Note: atomic mass of$Hg=200$.
a.) $93.4g$
b.) $46.7g$
c.) $85.56g$
d.) $75.56g$

Answer 1

Hint:. We need to find the heat of formation of $Hg$, using that find the number of moles $Hg$ formed, since we have given the excess of $HgO\left( s \right)$ absorbs $41.8kJ$ of heat at constant volume from which the weight of the formation of $Hg$ can be obtained.

Complete step by step answer:
Given data is as follows
Standard heat of formation of $HgO\left( s \right)$ at $298K$ and at constant pressure is $-90.8kJmo{{l}^{-1}}$
That is $Hg\left( l \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)\to HgO\left( s \right);\Delta H = -90.8kJ$
By reversing the above equation we can have the heat of formation of $Hg$ as shown below
$HgO\left( s \right)\to Hg\left( l \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right);\Delta H = +90.8kJ$

We know that the change of enthalpy is the sum of change of internal energy and the work done
That is $\Delta H=\Delta U+\Delta nRT$
Since we know that the work done $\Delta PV$ can be written as $\Delta PV=\Delta nRT$
Where $\Delta H=\text{heat of formation}$
$\Delta U=\text{internal energy}$
$\Delta n=\text{gaseous moles of products - gaseous moles of reactants}$
$R=\text{real gas constant}$
$T=\text{ Temperature}$

By substituting the above values in the enthalpy equation we have
$\Rightarrow 90.8=\Delta U+\dfrac{1}{2}\times 0.008314\times 298$
Therefore $\Delta U=89.56kJmo{{l}^{-1}}$

Given that the excess of $HgO\left( s \right)$ absorbs $41.8kJ$ of heat at constant volume
Therefore the number of moles of formation of $Hg$ from $HgO$ is
\[\text{Number of moles = }\dfrac{41.84}{89.56} = 0.467 mol\]
We know that weight is the product of moles and molecular weight
\[\text{Weight }\!\!~\!\!\text{ is = 0}\text{.467}\times \text{200 = 93}\text{.4g}\]
Therefore the required weight of $Hg$ formed from $HgO$ is $93.4g$
So, the correct answer is “Option A”.

Note: The standard heat of formation or the standard enthalpy can be defined as the change of enthalpy during the formation of 1 mole of a substance from its constituent elements. We need to take care that the reactants should be in their standard form to find the standard heat of formation of the product formed from those reactants.