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?

Rate = k[NO]^(2) [O_(2)]^(1) . By how many times does the rate of reaction change when the volume of the reaction vessel is reduced to 1//3^(rd) of its original volume ? Will ther he any change in the order of the reaction.

For the elementary reaction 2A + B_(2) rarr 2AB . Calculate how much the rate of reaction will change if the volume of the vessel is reduced to one third of its original volume?

The following reaction takes place is one step 2NO(g) + O(g) harr 2NO(g) How will the rate of the above reaction change if the volume of the reaction vessel is reduced to one-third of its original volume? Will there be any change in the order of reaction with the reduced volume?

How will the rate of reaction 2SO_(2)(g) + O_(2)(g) rarr 2SO_(3)(g) change if the volume of the reaction vessel is halved?

How will the rate of reaction 2SO_(2)(g) + O_(2)(g) rarr 2SO_(3)(g) change if the volume of the reaction vessel is halved?

Using rate law, how much rate of reaction: 2NO(g) + O_(2)(g) to 2NO_(2)(g) will change if the volume of vessel is reduced to 1/3 rd of its initial value ?