The rate law for the reaction `O_(3) + O rarr 2O_(2)` is rate `= k[O_(3)][NO]`. Then which is/are correct? (more than one correct)

The rate law of the reaction 2N_2O_5 rarr4NO_2+O_2 is

The rate law expression for the reaction, 2NO+O_(2) rarr 2NO_(2) is: rate =K[NO][O_(2)] Suggest the possible mechanism.

For the reaction 2NO_(2)rarr2NO+O_(2) , rate expressed as :

The rate of the reaction 2NO+O_(2)rarr2NO_(2),"at" 25^(@)C is r=k[NO]^(2)[O_(2)] If the initial concentrations of the reactant are O_(2)=0.040 "mol"L^(-1) and NO==0.01 "mol" L^(-1) , the rate constant of the reaction is :

The rate law for the reaction, 2Cl_(2)Orarr2Cl_(2)+O_(2) at 200^(@)C is found to be : rate = k[Cl_(2)O]^(2) (a) How would the rate change if [Cl_(2)O] is reduced to one- third of its original value (b) How should the [Cl_(2)O] be chabged in order to double the rate (c) How would the rate change is [Cl_(2)O] is raised to threefold of its original value ?

Statement: For the reaction 2O_(3) rarr 3O_(2) , the rate =K[O_(3)]^(2)[O_(2)]^(-1) Explanation: The reaction has -ve order for O_(2)

The rate law of a chemical reaction given below: 2NO+O_(2)rarr 2NO_(2) is given as rate =K[NO]^(2)[O_(2)] . How will the rate of reaction change if the volume of reaction vessel is reduced to 1//4th of its original valur?