Question

* Please show work step by step explaining how you got to the solution* Consider the...

* Please show work step by step explaining how you got to the solution*

Consider the following data.

Time (s) 0.0 5.0 10.0 15.0 20.0
[A] (M) 0.20 0.14 0.10 0.071 0.050

Graph the [A] vs. time, ln[A] vs time and 1/[A] vs. time.  Determine the rate law and rate constant (including units) for the reaction. Determine the time at which the concentration of A at will be 0.12 M.

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Answer #1

The plots are shown below.

[A] vs time 0.250 0.200 0.150 0.050 0.000 0.0 5.0 10.0 15.0 20.0 25.0

They are characteristic of first order reaction.

The rate law is rate = k [A]

The rate constant k=-slope=-(-0.069) = 0.069 \: s

Here slope is the slope of the graph of ln [A] vs time.

t=\frac {1}{k}ln \frac {a}{a-x}= \frac {1}{0.069} \times ln \frac {0.20}{0.12} =7.4 \: s

The concentration will reduce to 0.12 M in 7.4 s.

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