R-I : - 2NaNO3 → 2NaNO2 + O2R-II : - 2H2O2 → 2H2O + O2Identify disproportionation reaction from the above two reactions.

I) 2NaN+5O3−2→2NaNO2+3+O20‘N’ undergoes reduction‘O’ undergoes oxidation∴R-I is not a disproportionation reactionII) 2H2O2−1→2H2O−2+O20‘O’ undergoes both oxidation and reduction.∴R-II is  a disproportionation reaction.