8. What did the Dust Bowl teach us?
1 answer:
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The integrated rate law for a second-order reaction is given by:
where, [A]t= the concentration of A at time t,
[A]0= the concentration of A at time t=0
<span>k =</span> the rate constant for the reaction
<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min
Hence,
<span> = 4.858</span>
<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>
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</span>
<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M