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 halflife 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 ?
Please explain
1 Answer

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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
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questions?
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