The following reaction was monitored as a function of time: AB-->A+B A plot of 1/AB versus time yields a straight line with slope 5.5 * 10^-2 (M?s)^-1.

What is the half-life when the initial concentration is 0.58 M ?

If the initial concentration of AB is 0.240 M , and the reaction mixture initially contains no products, what are the concentrations of A and B after 80s ?

half-life = 31.3 s

0.123 M A, 0.123 M B

Explanation:

When they tell us that a graph of 1 /[AB] versus time yields a straight line, they are telling us that the reaction is first order repect to AB.

A first order rection has a form:

rate = - ΔA/Δt = - k[A]²

The integrated rate law for this equation from calculus is:

1/[A]t = kt+ 1/[A]₀

which we see is the equation of a line with slope k and y intercept 1/[A]₀

Therefore k = 5.5 10⁻² /Ms

The above equation can rewritten as:

1/ (1/2 [A]₀) = k t1/2 + 1/[A]₀

2/[A]₀ = k t1/2 + 1/[A]₀

and the half life will be given by:

t 1/2 =  1 / k[A]₀

t 1/2  = 1 / [( 5.5 x 10⁻² /Ms ) x 0.58 M]

t 1/2  = 31.3 s

For the second part we make use of the equation from above:

1/[A]t = kt+ 1/[A]₀

to determine [A]t, and from the stoichiometry of the reaction we will calculate how much of A and B has been produced.

1/[A]t =  ( 5.5 x 10⁻²/Ms) x 80s + 1/0.240 M

1/[A]t = 4.40 / M +  4.167 / M = 8.56 / M

⇒ [A]t = 0.117 M

If after 80 seconds we have 0.117 M of AB, this means  (0.240 - 0.117)  of AB reacted to produce 0.123 M of A and .123 M of B.

It maybe a bit confusing that we almost have half of our original concentration of AB, and from the first part we know the half-life was 31.3 s.

But, you have to realize that the half-life for second order reactions depend on the initial concentration ( different from first order ). Calculating the half life in this part with an original concentration of 0.240 M gives us a half-life of 75.8 s which makes sense with our result.

yes they ca   is ur ansswerrr

b. is the answer hoped i