Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after
is 0.00345 mol/L.
Explanation:

Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t = 
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after
is 0.00345 mol/L.
The answer would be , a Chemical change!
Answer:
See explanation
Explanation:
For this question, we have to remember the effect of an atom with high <u>electronegativity</u> as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an <u>inductive effect</u>. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be <u>weaker</u> and the compound will be more acid (because is easier to produce the hydronium ion
).
With this in mind, for A in the last compound, we have <u>2 Br atoms</u> near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have <u>more acidity</u>. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.
In B, the difference between the molecules is the <u>position</u> of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a <u>higher inductive effect</u> and more <u>acidity</u>.
See figure 1
I hope it helps!