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Chemistry · Thermodynamics & Thermochemistry

JEE Advanced 2022 — Paper 1 — Question 35

2 mol2 \mathrm{~mol} of Hg(g)\mathrm{Hg}(g) is combusted in a fixed volume bomb calorimeter with excess of

O2\mathrm{O}_{2} at 298 K and 1 atm into HgO(s)\mathrm{HgO}(s). During the reaction, temperature increases from 298.0 K to 312.8 K .

If heat capacity of the bomb calorimeter and enthalpy of formation of Hg(g)\mathrm{Hg}(g) are 20.00 kJ K−120.00 \mathrm{~kJ} \mathrm{~K}^{-1}

and 61.32 kJ mol−161.32 \mathrm{~kJ} \mathrm{~mol}^{-1} at 298 K , respectively, the calculated standard molar enthalpy of formation of

HgO(s)\mathrm{HgO}(s) at 298 Kis XkJmol−1298 \mathrm{~K}^{\text {is }} \mathrm{X} \mathrm{kJ} \mathrm{mol}^{-1}. The value of ∣X∣|\mathrm{X}| is

Answer: 90.39

Numerical answer — enter this value.

Step-by-step solution

ΔT=312.8−298=14.8\Delta \mathrm{T}=312.8-298=14.8 Molar heat capacity of calorimeter =20 kJ K−1=20 \mathrm{~kJ} \mathrm{~K}^{-1}

Heat released by combustion of 2 moles of Hg(g)\mathrm{Hg}(\mathrm{g}) =−20×14.8=-20 \times 14.8 =−296 kJ=-296 \mathrm{~kJ}

ΔUcombustion =−2962=−148 kJ mol−1\Delta \mathrm{U}_{\text {combustion }}=-\frac{296}{2}=-148 \mathrm{~kJ} \mathrm{~mol}^{-1}

Hg(g)+12O2( g)⟶HgO(s)\mathrm{Hg}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HgO}(\mathrm{s})

Δng=0−(1+12)=−32\Delta \mathrm{n}_{\mathrm{g}}=0-\left(1+\frac{1}{2}\right)=-\frac{3}{2}

ΔHcombustion =ΔUcombustion +ΔngRT\Delta \mathrm{H}_{\text {combustion }}=\Delta \mathrm{U}_{\text {combustion }}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}

=−148+(−32)×8.3×10−3×298=-148+\left(-\frac{3}{2}\right) \times 8.3 \times 10^{-3} \times 298 =−148−3.710=-148-3.710

=−151.710 kJ mol−1=-151.710 \mathrm{~kJ} \mathrm{~mol}^{-1}

Hg(ℓ)+12O2( g)⟶HgO(s)\mathrm{Hg}(\ell)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HgO}(\mathrm{s})

ΔHcombustion ∘=ΔHf∘(HgO)−ΔHHg(ℓ)⟶Hg(g)∘+12ΔHf∘O2\Delta \mathrm{H}_{\text {combustion }}^{\circ}=\Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HgO})-\Delta \mathrm{H}_{\mathrm{Hg}(\ell) \longrightarrow \mathrm{Hg}(\mathrm{g})}^{\circ}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{f}}^{\circ} \mathrm{O}_{2}

−151.710=ΔHf∘(HgO)−61.32+0-151.710=\Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HgO})-61.32+0 ΔHf∘(HgO)=−151.710+61.32=90.39\Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HgO})=-151.710+61.32=90.39 ΔHf∘=90.39\Delta \mathrm{H}_{\mathrm{f}}{ }^{\circ}=\mathbf{9 0 . 3 9}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2022
Paper
Paper 1
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Thermochemistry and Enthalpy Changes
2 mol of Hg (g) is combusted in a fixed volume bomb calorimeter with… | JEE Advanced 2022 PYQ with Solution · DhiX AI