Answer:
13. t = 1183 min and 0.50 M.
14. [tex]t_{1/2}=2.67x10^{-8}hr[/tex]
Explanation:
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13. In this case, according to the units, we infer this is a second-order reaction which has the following integrated rate law:
[tex]\frac{1}{[A]} =\frac{1}{[A]_0} +kt[/tex]
Which can be solved for the time as shown below:
[tex]t=\frac{ \frac{1}{[A]}-\frac{1}{[A]_0}}{k}[/tex]
Thus, we plug in the given concentrations and rate constant to obtain:
[tex]t=\frac{ \frac{1}{0.250M}-\frac{1}{0.850M}}{0.002387M^{-1}min^{-1}}\\\\t= 1183min[/tex]
For the second part, we proceed by using the same rate constant and the new initial concentration as follows:
[tex]\frac{1}{[A]} =\frac{1}{[A]_0} +kt\\\\\frac{1}{[A]} =\frac{1}{0.750M} +0.680M^{-1}min^{-1}*0.996min\\\\\frac{1}{[A]} =1.99,M[/tex]
[tex][A]=0.50M[/tex]
14. In this case, according to the units of the rate constant, we infer this is a zeroth-order reaction, therefore we compute the half-life has shown below:
[tex]t_{1/2}=\frac{[A]_0}{2k}[/tex]
Thus, we plug in to obtain:
[tex]t_{1/2}=\frac{2.696x10^{-6}M}{2*50.5M*hr^{-1}}[/tex]
[tex]t_{1/2}=2.67x10^{-8}hr[/tex]
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