Question

Standard heat of formation of HgO(s) at 298K and at constant pressure is $- 90.8kJmo{l^{ - 1}}$. If excess of HgO(s) absorbs 41.84KJ 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

To solve the question; firstly write the equation of Hg with Oxygen(O) whose enthalpy is given in question and as we get HgO but the products we need are Hg and O so reverse the reaction and change the sign of enthalpy and keep the same magnitude. Now using the formula to calculate the heat of formation of Hg. As we are given heat absorbed and we have heat of formation now using these quantities calculate the moles of Hg formed and so weight.

Complete step by step answer:
Equation for formation of HgO
$Hg(l) + \dfrac{1}{2}O_2(g) \to HgO(s);\Delta H ^\circ = - 90.8kJ$
After reversing it:
$HgO(s) \to Hg(l) + \dfrac{1}{2}O_2(g);\Delta H ^\circ = + 90.8kJ$ ------(1)
Formula to calculate heat of formation of Hg(l) from HgO(S):
$\Delta {H^o} = \Delta {U^o} + \Delta nRT$
Where:
$\Delta {U^o}$ is the heat of formation
$\Delta {H^o}$ is the change in enthalpy
Now putting the given values of enthalpy and $\Delta nRT$ in the equation:
We get:
$90.8 = \Delta {U^o} + \dfrac{1}{2} \times 8.314 \times {10^{ - 3}} \times 298$
As in the equation 1 (change in mole) $\Delta n = \dfrac{1}{2}$
Solving the above equation we get:
$\Delta {U^o} = 89.56kJmo{l^{ - 1}}$
Now we are given that 41.84 kJ heat is absorbed by HgO
Thus, the number of moles formed is equal to:
$\dfrac{{41.84}}{{89.56}} = 0.4672mol$
As we have got number of moles formed, now multiply number of moles with the atomic mass of Hg we will get the weight of Hg formed:
${W_{Hg}} = 0.4672 \times 200 = 93.4g$
Therefore, option A is the correct answer.

Note:
Heat of formation, also called standard heat of formation, enthalpy of formation, or standard enthalpy of formation, the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements, each substance being in its normal physical state (gas, liquid, or solid). Usually the conditions at which the compound is formed are taken to be at a temperature of $25{\text{ }}^\circ C{\text{ }}\left( {298K} \right)$ and a pressure of 1 atmosphere, in which case the heat of formation can be called the standard heat of formation.