Answer:
* [tex]t_{1/2}=2.56s[/tex]
* [tex][I_2]=0.011M[/tex]
Explanation:
Hello,
In this case, considering the reaction:
[tex]I_2(g)\rightarrow 2I[/tex]
Which is first-order with respect to I₂, we can compute the half-life by:
[tex]t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.271s^{-1}}\\ \\t_{1/2}=2.56s[/tex]
Moreover, since the integrated rate law is:
[tex][I_2]=[I_2]_0exp(-kt)[/tex]
We can compute the concentration of iodine once 5.37 s have passed:
[tex][I_2]=0.048Mexp(-0.271s^{-1}*5.37s)\\\\[/tex]
[tex][I_2]=0.011M[/tex]
Best regards.
