2 moles 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 $H g (g)$ are 20.00 kJ $K^{- 1}$ and 61.32 kJ $m o l^{- 1}$ at 298 K, respectively, the calculated standard molar enthalpy of formation of HgO(s) at 298 K is X kJ $m o l^{- 1}$ . The value of |X| is…………
[Given, gas constant R = $8.3 J K^{- 1} m o l^{- 1}$ ]

Question

2 moles 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 $H g (g)$ are 20.00 kJ $K^{- 1}$ and 61.32 kJ $m o l^{- 1}$ at 298 K, respectively, the calculated standard molar enthalpy of formation of HgO(s) at 298 K is X kJ $m o l^{- 1}$ . The value of |X| is…………
[Given, gas constant R = $8.3 J K^{- 1} m o l^{- 1}$ ]

Infinity Learn · Accepted Answer

the gas performs isothermal irreversible work (W).Where, ΔU =0 (change in internal energy)From, 1st law of thermodynamics,⇒ΔU = ΔQ + W0 = ΔQ + W ⇒ ΔQ = -WNow, W = - Pext(V2−V1) = −Pext(nRTp2−nRTp1) = −Pext×nRT(1p2−1p1)Given, Pext=4.3MPa , P1=2.1MPa,p2=1.3MPa n=5mol,T=293Kand R=8.314J mol−1K−1= −4.3×5×8.314×293(11.3−12.1)= -15347.70 J mol−1= 15.347 kJ mol−1 ≈−15 kJ mol−1 ⇒ΔQ=15 kJ mol−1

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