1.

Ozone is thermodynamically unstable. Give reasons.

Answer»

Solution :`O_(3)` on decomposition gives. `O_(2)` i.e., `2 O_(3) rarr 3 O_(2) , DELTA H^(@) (298 K) = - 142 kJ mol^(-)`.
Since `DeltaH` is negative (because `O_(3)` is an endothermic compound) and `DELTAS` is positive (because degree of RANDOMNESS increases on decomposition), therefore, according to Gibbs equation, `Delta G = Delta H - T Delta S, Delta G` is -ve and HENCE decomposition of `O_(3)` is spontaneous. In other words, ozone is themodynamically UNSTABLE.


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