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Question

Calculate the mass of mercury which can be liberated from HgO at 25C by the treatment of excess HgO with 41.84 kJ of heat at
(i) constant pressure (ii) constant volume conditions.
Given: Hf(HgO,s)=90 kJ mol1 and M(Hg)=200.6 g mol1.

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Solution

12O2(g)+Hg(l)HgO(s)ΔH=90 kJ/mol

(i) 1 mol of Hg90 kJ

x mol of Hg41.84 kJ

x=41.8490=0.465 mol

Mass of mercury =0.465×200.6=93.28 g

(ii) At constant volume, q=ΔV=ΔHΔngRT

q=90ΔngRT

Here, Δng= Difference between gaseous moles of reactants and products

Δng=012=12

q=90+12×8.314×298×103=88.76 kJ

Moles of Hg=41.8488.76=0.471 mol

Mass of Hg=0.471×200.694.5 g


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