Answer:
a. First order.
b. [tex]k=3.7x10^{-3}s ^{-1}[/tex]
c. [tex]t_{1/2}=187.3s[/tex]
d. [tex][A]=0.111M[/tex]
Explanation:
Hello.
a. In this case since the slope is in s⁻¹ we can infer that the lineal form of the rate law is:
[tex]ln[A]=-kt+ln[A]_0[/tex]
Which is also:
[tex]y=mx+b[/tex]
It means that the reaction is first-order as slope equals the negative of the rate constant which also has units of first-oder reaction.
b. Since the slope is −3.7 x 10−3 s−1 the rate constant is:
[tex]k=-m\\\\k=-(-3.7x10^{-3}s ^{-1})\\\\k=3.7x10^{-3}s ^{-1}[/tex]
c. For first-order reactions, the half-life is:
[tex]t_{1/2}=\frac{ln(2)}{k}\\ \\t_{1/2}=\frac{ln(2)}{3.7x10^{-3}s^{-1}}\\\\t_{1/2}=187.3s[/tex]
d. In this case, since the integrated first-order rate law is:
[tex][A]=[A]_0exp(-kt)[/tex]
The concentration once 220 seconds have passed is:
[tex][A]=0.250Mexp(-3.7x10^{-3}s^{-1}*220s)\\\\[/tex]
[tex][A]=0.111M[/tex]
Regards.
