The half life period of a reaction is halved as the initial concentration of the reactants is doubled. The order of reaction is:

Correct Answer - A `(l n2)/(k) =(log_(e)2)/(k)=(2.303xxlog2)/(k)=(0.693)/(k)`

Step. 1. Calculation of rate constant `k=0.693/t_(1//2) = 0.693/(2min) = 0.3465 min^(-1)` 1st case, `a=100%`, (a-x) = `25%` `t=(2.303)kloga/(a-x)` `=(2.303)/(0.3465 min^(-1)) log 100/10 = 2.303 /(0.3465 min^(-1)) log 10` `=2.303/(0.3465...

Correct Answer - D d) `underset("Initial conc")(12M)overset(t_(1//2))to6Moverset(t_(1//2))tounderset("Final conc.")(3M)` `t_(1//2) = 10`min , k = `(0.693)/(10)=0.0693 min^(-1)` After 20 minutes, the concentration will be 3M. rate =k[3M] = `(0.0693 min^(-1) xx 3M)`s

Correct Answer - C `A + 2B to C` Since the reaction rate becomes double when the concentration of A is made twice order w.r.t A=1 Since the reaction rate becomes...

`t_(1//2) propto 1/([A_(0)])^(n-1)` `(t_(1//2))_(1)/(t_(1//2))_(2) = {([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)` `20/10 = ((2a)/(a))^(n-1)` or `(2)^(1) =(2)^(n-1)` or `(2)^(1) = (2)^(n-1)` `therefore n-1 = 1` or n=2 (second order)

Correct Answer - B `xto` product (zero order) Till the concentration of X changes from `X_0` to `X_0/2` , it is zero order with rate constant k. Time taken for concentration...

`((t_(1//2))_(1))/((t_(1//2))_(2))=((p_(2))/(p_1))^(n-1)` `(350)/(175)=((40)/(80))^(n-1)` `2=((1)/(2))^(n-1)` n-1=-1 n=0 (zero order reaction)

Correct Answer - C `k=(2.030)/(t)log.((a)/(a-x))` `k=(2.303)/(1) log. ((100)/(5))=2xx0.693` `t_(1//2)=(0.693)/(k)=(0.693)/(2xx0.693)=(1)/(2)hr`

Correct Answer - B `t_(1//2)prop(1)/(a^(n-1))` `t_(1//2)prop(1)/(a)` where,n = order of reaction for second order reaction

Correct Answer - D Consider a hypothetical change. `A+BhArrC+D` For the reaction. `K_(eq)=([C][D])/([A][B])` For the above reaction if concentration of reactions are doubled then the rate of forward reaction increases for...