The reaction `A+OH^(-)rarr` Products, obeys rate law expression as : `(-d[A])/(dt)=k[A][OH^(-)]` If initial concentrations of [A] and `[OH^(-)]` are 0

rate = `k[A][B]^(2)` By reducing the volume to `1/3` of its original volume, the concentration of the reactants will become three times. `therefore` rate = `k[3A][3B]^(2)=27k[A][B]^(2)` ltbr. Hence, the reaction...

a) The rate law epression is: Rate(r) = `k[A]^(1//3)[B]^(2)therefore` Order of reaction `=1/3 + 2=2(1/3)` b) Rate constant, k =`("rate")/([A]^(1//3)[B]^(2))` `therefore` Units of rate constant = `(moldm^(-3)s^(-1))/((mol dm^(-3^(1//3))(mol dm^(-3))^(2)))` `=(moldm^(-3))^(1-7//3)s^(-1)=(mol...

In an elementary reaction, the number of atoms or ions which are colliding to react is known as molecularity. Had the reaction been of elementary nature (single step reaction) the...

In order to make reaction rate positive.

Correct Answer - A a) The value of rate constant depends upon temperature and not upon the initial concentration of reactants and products.

Correct Answer - C c) In flask A: `k=(2.303)/8 log 1.0/(0.25) =2.303/4 log2` In flask B: `t=2.303/k log 0.6/0.3 = (2.303 xx 4)/(2.303) = 4` Log 2 cancels out

Correct Answer - A,B,D Unit of `k=("mol"/"litre")^(1-n) xx s^(-1)` for n=1 is `sec^(-1)`, n=2 is litre `mol^(-1) sec^(-1)`, n=3 is `"litre"^2 mol^(-2) sec^(-1)`

From expts. (2) and (3) , it is clear that when concentration of A is kept constant and that of B is doubled, the rate increasefour times. This shows that...

From the slow rate determining step `=-(d[O_(3)])/(dt)=k_(1)[O][O_(3)]` From the fast reaction., `K_(eq)=([O_(2)][O])/([O_(3)])` or `[O]=(K_(eq)[O_(3)])/([O_(2)])` Substituting the value of [O] in the above expression `r=-(d[O_(3)])/(dt)=(k_(1)K_(eq)[O_(3)]^(2))/([O_(2)])=(k[O_(3)]^(2))/([O_(2)])`

Correct Answer - D `0.028=k[0.01]^(2)[0.040]`