2 mol of Hg ( g ) is combusted in a fixed volume bomb calorimeter with excess of O 2 at 298 K and 1 atm into 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 ) are 20.00 kJ K -1 and 61.32 kJ mol -1 at 298 K, respectively, the calculated standard molar enthalpy of formation of HgO ( s ) at 298 K is X kJ mol -1. The value of | X | is [Given: Gas constant R =8.3 J K -1 mol -1 ]

Question

2 mol of Hg ( g ) is combusted in a fixed volume bomb calorimeter with excess of O 2 at 298 K and 1 atm into 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 ) are 20.00 kJ K -1 and 61.32 kJ mol -1 at 298 K, respectively, the calculated standard molar enthalpy of formation of HgO ( s ) at 298 K is X kJ mol -1. The value of | X | is [Given: Gas constant R =8.3 J K -1 mol -1 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Q rxn = C Δ T |Δ U | × 2=20 × 14.8 |Δ U |=148 kJ / mol Δ U =-148 kJ / mol Hg ( g )+(1/2) O 2( g ) longrightarrow HgO ( s ): Δ U =-148 kJ / mol Δ H =Δ U +Δ n g RT =-148-(3/2) × (8.3/1000) × 298=-151.7101 Hg (l)+(1/2) O 2( g ) longrightarrow HgO ( s ) Δ H =-151.7101+61.32=-90.39 kJ / mol

Qrxn​=CΔT ∣ΔU∣×2=20×14.8 ∣ΔU∣=148kJ/mol ΔU=−148kJ/mol Hg(g)+21​O2​(g)⟶HgO(s):ΔU=−148kJ/mol ΔH=ΔU+Δng​RT =−148−23​×10008.3​×298=−151.7101 Hg(l)+21​O2​(g)⟶HgO(s) ΔH=−151.7101+61.32=−90.39kJ/mol