For a reaction,
4M(s) + nO₂(g) → 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
For a reaction,
4M(s) + nO₂(g) → 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
Option 1 - <p>The slope changes from negative to positive</p>
Option 2 - <p>The free energy change shows a change from negative to positive value.</p>
Option 3 - <p>The slope changes from positive to negative.</p>
Option 4 - <p>The slope changes from positive to zero.</p>
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5 months ago
Correct Option - 2
Detailed Solution:
For oxide to be stable its ΔG value should be negative.
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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Chemistry Ncert Solutions Class 11th 2023
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