Question

For combustion of one mole of magnesium in an open container at 300K and 1bar pressure, ΔCHΘ=-601.70kJ 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)

Answer 1

The correct answer is 600
$Mg(s)+\frac{1}{2}O2→MgO(s)$
$ΔH=ΔU+ΔngRT$
$Δng=−\frac{1}{2}$
$−601.70=ΔU−\frac{1}{2}(8.3)(300)×10^{−3}$
ΔU = –601.70 + 1.245
ΔU ≃ –600 kJ
Hence , Magnitude of change in internal energy is 600 kJ.