First Order Kinetics

A substance ''A'' decomposes in solution following the first order kinetics. Flask I contains 1L of 1 M solution of A and falsk II constains 100 mL of 0.6 M solution. After 8 hr , the concentration, of A in flask I becomes 0.25 M . What will be the time for concentration of A in flask II to become 0.3 M ?

The decay of a radioactive element follows first order kinetic. Thus,

The vapour pressure of two miscible liquids (A) and (B) are 300 and 500 mm of Hg respectively. In a flask 10 mole of (A) is mixed with 12 mole of (B). However, as soon as (B) is added, (A) starts polymerising into a completely insoluble solid. The polymerisation follows first-order kinetics. After 100 minute, 0.525 mole of a solute is dissolved whivh arrests the polymerisation completely. The final vapour pressure of the solution is 400 mm of Hg . Estimate the rate constant of the polymerisation reaction. Assume negligible volume change on mixing and polymerisation and ideal behaviour for the final solution.

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) Amount of radioactive elements (activity) decreases with passage of time as

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) Half life of .^(60)Co is 5.3 years, the time taken for 99.9% decay will be

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) The rate of radioactive decay is

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) Select the correct relation.

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) Which "is"//"are" true about the decay cosntant?

Under the same reaction conditions, intial concentration of 1.386 mol dm^(-1) of a substance becomes half in 40 s and 20 s through first order and zero order kinetics respectively. Ratio ((k_1)/(k_0)) of the rate constants for first order (k_1) and zero order (k_0) of the reaction is

Under the same reaction conditions, the intial concentration of 1.386 mol dm^(-3) of a substance becomes half in 40 s and 20 s theough first order and zero order kinetics, respectively. The ratio (k_(1)//k_(0)) of the rate constants for first order (k_(1)) and zero order (k_(0)) of the reaction is