`T_(50)` of first order reaction is 10 min. starting with 10 `"mol"^(-1)` rate after 20 min is :

Correct option is: (c) 5 : 00 pm

`A+2B to 3C + 2D` `Rate = (-d[A])/(dt) = -1/2(d[B])/(dt) = 1/3(d[C])/(dt) = 1/2(d[D])/(dt)` `(d[A])/(dt) = 1/2(d[B])/(dt) = 1/2 xx (1 xx 10^(-2))=0.5 xx 10^(2) mol L^(-1)s^(-1)` `(d[C])/(dt) = 3/2...

Step. I. Calculation of rate constant (k). For first order reaction, rate( r) `propto C` or kC Rate `(r_(1)) = kC_(1)` and Rate `(r_(2)) =kC_(2)` `therefore` Rate `(r_(1))= kC_(1)` and...

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 c) Correct reason. Rate constant remains constant for the reaction and is independent of the initial concentration of the reason.

Correct Answer - A Unit of rate and rrte constant are same for zero order reaction.

Correct Answer - B `(k_(T_(2)))/(k_(T_(1)))=(mu)^(DeltaT//10)` `(10^(-3))/(k_(T_(1)))=(mu)^(10//10)=2` `k_(T_(1))=(10^(-3))/(2)=0.5xx10^(-3)=5xx10^(-4)"min"^(-1)`

`(r_(CH_(4)))/(r_(gas))=sqrt((M_(gas))/(16))=(20)/(40)=(1)/(2)` `M_(gas)=4gm`

Correct Answer - 3.98

(a) `0.5+2=2.5` (b) For the first order reaction, `t_(1//2)=(0.693)/(K)=(0.693)/(5.5xx10^(-14)S^(-1))=1.26xx10^(13)S.`