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 = $8.3 J K^{- 1} m o l^{- 1}$ ]

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Detailed Solution

the gas performs isothermal irreversible work (W).
Where, $Δ U$ =0 (change in internal energy)
From, 1^st law of thermodynamics,
$\Rightarrow$ $Δ U$ = $Δ Q$ + W
0 = $Δ Q$ + W $\Rightarrow$ $Δ Q$ = -W
Now, W = - $P_{e x t} (V_{2} - V_{1})$
= $- P_{e x t} (\frac{n R T}{p_{2}} - \frac{n R T}{p_{1}})$
= $- P_{e x t} \times n R T (\frac{1}{p_{2}} - \frac{1}{p_{1}})$
Given, $P_{e x t} = 4.3 M P a$ ,
$P_{1} = 2.1 M P a, p_{2} = 1.3 M P a$
$n = 5 m o l, T = 293 K$
and $R = 8.314 J m o l^{- 1} K^{- 1}$
= $- 4.3 \times 5 \times 8.314 \times 293 (\frac{1}{1.3} - \frac{1}{2.1})$
= -15347.70 J $m o l^{- 1}$
= 15.347 kJ $m o l^{- 1}$
$\approx - 15 k J m o l^{- 1}$
$\Rightarrow Δ Q = 15 k J m o l^{- 1}$