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 :

For the reaction , 2NO_(2) rarr 2NO+ O_(2) rate is expressed as :

The rate expression for the chemical reaction, 2NO_2Cl rarr 2NO_3+Cl_2 , is given as : Rate= k[NO_2Cl] Propose a possible mechanism for the above reaction and give the order of the reaction.

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

The order of the reaction 2NO_(g)+O_(2(g))rarr 2NO_(2(g)) having Rate = k[NO_2]^2[O_2] is

Express the rate of the following reaction : 2NO_2 rarr 2NO+ 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?

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