For a reaction,
\(4\mathrm{M(s)}+n\mathrm{O_2(g)}\rightarrow2\mathrm{M_2O_n(s)}\)
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 :
\(4\mathrm{M(s)}+n\mathrm{O_2(g)}\rightarrow2\mathrm{M_2O_n(s)}\)
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 :
Show Hint
- 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
The Correct Option is A
Solution and Explanation
At low temperature \(\Delta G \lt 0\): the oxide is stable (formation is spontaneous). As T rises, \(\Delta G\) becomes less negative and eventually crosses zero (negative to positive); above this temperature the oxide decomposes.
Hence the temperature below which the oxide is stable is the point where \(\Delta G\) changes from negative to positive value. Changes of slope only mark phase changes (melting/boiling) of M or the oxide, not stability.
Answer: (1) the free energy change shows a change from negative to positive value
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