For a reaction,
4M(s) + nO2(g) $\to$ 2M2On(s)
the free energy change is plotted as a function
of temperature. The temperature below which
the oxide is stable could be inferred from the
plot as the point at which :
Solution
$\Delta$G = $\Delta$H – T$\Delta$S
<br><br>$\Delta$G = –ve (stable oxide)
<br><br>$\Delta$G = +ve (unstable oxide)
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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