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JEE Main 2022 · 28 Jun · Shift 2, Chemistry question 82

For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, ΔCH⊖=−601.70\Delta_CH^\ominus=-601.70 kJ mol−1^{-1}, the magnitude of change in internal energy for the reaction is ______ kJ. (Nearest integer)

(Given : R = 8.3 J K−1^{-1} mol−1^{-1})

Answer and worked solution

Official answer: 600

Reaction. Mg(s)+12O2(g)→MgO(s)\mathrm{Mg(s)+\frac12O_2(g)\rightarrow MgO(s)}, Δng=−12\Delta n_g=-\frac12.

ΔU\Delta U =ΔH−ΔngRT=−601.70+12×8.3×300×10−3≈−600.5=\Delta H-\Delta n_gRT=-601.70+\frac12\times8.3\times300\times10^{-3}\approx-600.5 kJ. Magnitude 600.

Answer: 600

Topics:First law and enthalpy
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