If the rate constant is 0.0693 seat
the time
taken for 99.9%
complexion
of
the reaction is ?
(b) 0.693 sec
(d) 10 sec.
(a) 69.3 sec
c)100 sec
Answers
ANSWER WILL BE OPTION (B)...
As we know,
As we know,for a first order reaction,
As we know,for a first order reaction,t
As we know,for a first order reaction,t 1/2
As we know,for a first order reaction,t 1/2
As we know,for a first order reaction,t 1/2 =
As we know,for a first order reaction,t 1/2 = k
As we know,for a first order reaction,t 1/2 = k0.693
As we know,for a first order reaction,t 1/2 = k0.693
As we know,for a first order reaction,t 1/2 = k0.693 =
As we know,for a first order reaction,t 1/2 = k0.693 = 0.0693
As we know,for a first order reaction,t 1/2 = k0.693 = 0.06930.693
As we know,for a first order reaction,t 1/2 = k0.693 = 0.06930.693
As we know,for a first order reaction,t 1/2 = k0.693 = 0.06930.693 =10min
As we know,for a first order reaction,t 1/2 = k0.693 = 0.06930.693 =10minReactant after 10 min = 5 mol
As we know,for a first order reaction,t 1/2 = k0.693 = 0.06930.693 =10minReactant after 10 min = 5 molRate (
As we know,for a first order reaction,t 1/2 = k0.693 = 0.06930.693 =10minReactant after 10 min = 5 molRate ( dt
As we know,for a first order reaction,t 1/2 = k0.693 = 0.06930.693 =10minReactant after 10 min = 5 molRate ( dtdx
As we know,for a first order reaction,t 1/2 = k0.693 = 0.06930.693 =10minReactant after 10 min = 5 molRate ( dtdx
As we know,for a first order reaction,t 1/2 = k0.693 = 0.06930.693 =10minReactant after 10 min = 5 molRate ( dtdx )=k[A]=0.0693×5molmin
As we know,for a first order reaction,t 1/2 = k0.693 = 0.06930.693 =10minReactant after 10 min = 5 molRate ( dtdx )=k[A]=0.0693×5molmin −1
Answer:
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