JEE MAIN - Chemistry (2020 - 6th September Evening Slot - No. 5)
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 :
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 :
the free energy change shows a change
from negative to positive value
the slope changes from positive to negative
the slope changes from negative to positive
the slope changes from positive to zero
Explanation
$$\Delta $$G = $$\Delta $$H – T$$\Delta $$S
$$\Delta $$G = –ve (stable oxide)
$$\Delta $$G = +ve (unstable oxide)
$$\Delta $$G = –ve (stable oxide)
$$\Delta $$G = +ve (unstable oxide)
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