For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, $Δ_{C} H^{⊖}$ = -601.70 kJ mol^-1, the magnitude of change in internal energy for the reaction is __________ kJ. (Nearest integer)
(Given : R = 8.3 J K^-1 mol^-1)

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answer is 600.

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Detailed Solution

$\begin{array}{l} Mg (s) + \frac{1}{2} O_{2} (g) \to MgO (s) \\ ΔH = ΔU + {Δn}_{g} RT \\ - 601 .70 \times 10^{3} = ΔU - \frac{1}{2} \times 8 .3 \times 300 \\ - 601 .70 kJ = ΔU - 1 .245 kJ \\ ΔU = - 600 .455 kJ \end{array}$

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