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 ?
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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