Answer:
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.
