In the reaction, 2no+o2 2no2 when 1 mole of no and 1 mole of o2 are made to react to completion then

A reaction is second order with respect to a reactant. How is the rate of rection affected if the concentration of the reactant is
(i) Doubled
(ii) Reduced to 1/2?

Let the reaction A → B is a 2nd order reaction w.r.t A and conc. of A is ‘a’ mol/L, then rate of reaction can be written as:

dxdt=k[A]2 =ka2

(i) When conc. [A] is doubled

(ii) Similarly, when conc. of A is reduced to 12 i.e., [A] is a/2 then new rate of reaction,

d'(x)dt = ka22 = 14ka2 = 14dxdt

Now the stoichiometry CLEARLY indicates, that for each mole, each #30*g# of #NO(g)#, a half-mole quantity, #16*g#, of dioxygen gas reacts, to give a #46*g# mass of #NO_2(g)#.

Do you agree? I take it that you can calculate the molecular masses of each reactant and product. All I am doing is taking the masses from the Periodic Table, and adding them appropriately.

And because the products and reactants are GASEOUS we could also say that each #2*L# of #NO(g)# reacts with #1*L# dioxygen to gives a #2*L# volume of #NO_2(g)#.....(Because under the same conditions of temperature and pressure, volume is proportional to the molar quantity, i.e. #Vpropn#, which is old Avogadro's law.)

All of this relates to the given stoichiometry of the given reaction, which is stoichiometrically balanced with respect to mass and charge:

#2NO(g) + O_2(g) rarr 2NO_2(g)#

i.e. as written there are #2xx30*g+32*g=92*g# of reactant, and necessarily #92*g# of product.

If you don't follow what I say, I am certainly willing to have another spray.