# Tough Chem kinetics: The following reaction was monitored as a function of time: AB→A+B?

0

A plot of 1/[AB] versus time yields a straight line with slope 5.4×10−2 (M⋅s)−1 .

1. What is the value of the rate constant (k) for this reaction at this temperature?

2. Write the rate law for the reaction.

3. What is the half-life when the initial concentration is 0.56 M ?

4. If the initial concentration of AB is 0.290 M , and the reaction mixture initially contains no products, what are the concentrations of A and B after 80 s ?

Also Check This  How do I format my Japanese address for amazon.co.jp shipping info section? What is Address Line 1 etc.?

*********

because 1/[AB] vs t yields straight line, the rxn is 2nd order in [AB] and

.. k = slope = 5.4×10^-2 /(M x sec)

rate law is….

.. rate = k x [AB]²

.. rate = (5.4×10^-2 / (M x sec)) x [AB]²

the integrated rate equation is

.. 1/[AB t] = +kt + 1/[AB o]

half life occurs when [AB t] = ½[AB o]… ie.. concentration of AB drops in ½ so that

.. (1/(½[AB o]) – 1/[AB o] ) / k = t½

.. (2 / [AB o] – 1/[AB o]) / k = t½

.. (1/[AB o]) / k = t½

so that

.. half life = (1 / 0.56M) / (5.4×10^-2 / (M x sec)) = 33sec

from integrated rate equation

.. 1/[AB t] = +kt + 1/[AB o]

.. 1/[AB t] = +(5.4×10^-2 / (M x sec)) x 80sec + 1/(0.290M) = 7.768 / M

.. [AB t] = 0.129M

and

.. if we assume 1 L solution

.. . .amount of AB consumed = (0.290M – 0.129M)x1L = 0.161mole

….. amount of A formed = 0.161mol… . (from coefficients of balanced equation)

.. . .amount of B formed = 0.161mol… . (from coefficients of balanced equation)

..so that 1 L of solution would contain

.. . .0.161mol A

.. . .0.161mol B

and finally

.. [A] = 0.161M

.. [B] = 0.161M

*********

questions?

.. 