Home
Class 12
CHEMISTRY
For a reversible first-order reaction, ...

For a reversible first-order reaction,
`A underset(K_(2))overset(K_(1))(hArr)B, K_(f)=10^(-2) s^(-1) and B_(eq.)/A_(eq.)=4,`
If `A_(0)=0.01 ML^(-1)` and `B_(0)=0`, what will be concentration of `B` after `30 sec`?

Text Solution

Verified by Experts

The correct Answer is:
`2.50xx10^(-3) M;`
Promotional Banner

Topper's Solved these Questions

  • CHEMICAL KINETICS

    P BAHADUR|Exercise Exercise 8|6 Videos

  • DILUTE SOLUTION AND COLLIGATIVE PROPERTIES

    P BAHADUR|Exercise Exercise 9 Advanced Numerical|12 Videos

Similar Questions

Explore conceptually related problems

For a reversible first order reaction, K_(f) =10^(-2) sec^(-1) and (B_(eq))/(A_(eq))=4," If "A_(0) =0.01 ML^(-1) and B_(0) =0 , what will be concentration of B after 30 sec ?

For an elementary chemical reaction, A_(2) underset(k_(-1))overset(k_(1))(hArr) 2A , the expression for (d[A])/(dt) is

Consider the following reversible first - order reaction of X at an initial concentration [X]_(0) . The values of the rate constants are k_(f)= 2 s^(-1) and k_(b)= 1 s^(-1) X underset(k_(b))overset(k_(r))iff y A plot of concentration of X and Y as function of time is

For a zero order reaction : A rarr 2B, initial rate of a reaction is 10^(-1)"M min"^(-1) . If concentration of 'A' is 0.1 after 120 sec then what would be concentration of B after 60 sec.

Consider a reaction that is first order in both direction A overset(k_f)underset(k_b)hArr B Initially only A is present , and its concentration is A_0 . Assume A_t and A_(eq) are the concentrations of A at time 't' and at equilibrium, respectively. The time 't at which A_t = (A_0 + A_(eq))//2 is ,

For the reversible reaction A underset(k_(b))overset(k_(f))(hArr)B (having Ist order in both direction) the concentration as a function of time are given for a certain experimantal run. Calculate [B] at time t=1/5 min. ["Given" : k_(b) =3 "min"^(-1) "and"1/e=0.37]

Write a stoichiometric equation for the reaction between A_(2) and C whose mechanism is given below. Determine the value of equilibrium constant for the first step. Write a rate law equation for the over all reaction in terms of its initial reactants. ( i ) A_(2)overset(K_(1))underset(K_(2))hArr2A K_(1)=10^(10)s^(-1) and K_(2)=10^(10)M^(-1)s^(-1) ( ii ) A+CtoAC K=10^(-4)M^(-1)S^(-1)