!! LONG ANSWER !!

Let’s start with point a). If you look at the graphs for zeroth, first, and second order reactions, you’ll notice that only one of them has #”1/concentration”# plotted against time with a resulting positive slope.

https://260h.pbworks.com/w/page/70917374/Rate%20Laws

This means that, because the #1/([A]_0)# graph you plotted has a positive slope, your reaction can only be second-order. Since there is only one reactant, #”A”#, the rate law for your reaction can be written as

#”rate” = k * [A]^(2)#, where

#”k”# – the rate constant.

The trick here is to realize that the slope of the graph is equal to the rate constant k for this reaction.

http://nonsibihighschool.org/acchonch19.php

This means that your rate constant will be

#k = “slope” = 3.6 * 10^(-2)”L” * “mol”^(-1) * “s”^(-1)#

The integrated rate law for a second-order reaction looks like this

#1/([A]) – 1/([A]_0) = k* t#, where

#[A]# – the concetration of #”A”# after a time t has passed;
#[A]_0# – the initial concentration of #”A”#;

Moving on to point b). To determine the half-life of this reaction, replace #”t”# with #”t”_(“1/2”)# and #[A]# with #[A]_0/2# in the above equation. (Remember that, by definition, the reaction’s half-life is the time needed for the reactant’s initial concentration to be halved).

#1/([A]_0/2) – 1/([A]_0) = k * t_(“1/2”)#

Solving this for #t_(“1/2”)# will give you

#t_(“1/2”) = 1/(k * [A]_0)#

Plug in the values you have for k and #[A]_0# and you’ll get

#t_(“1/2″) = 1/(3.6 * 10^(-2)”L” * “mol”^(-1) * “s”^(-1) * 2.8 * 10^(-3)”mol” * “L”^(-1))#

#t_(“1/2”) = 9.92 * 10^(3)”s”#

Finally, point c). Use the integrated rate law to solve for #”t”#, since you know that the value of #[A]# must now be #7.00 * 10^(-4)”M”#. So,

#1/(7.00 * 10^(-4)”M”) – 1/(2.8 * 10^(-3)”M”) = 3.6 * 18^(-2)”M”^(-1)”s”^(-1) * t#

#1071.43 “M”^(-1) = 3.6 * 18^(-2)”M”^(-1)”s”^(-1) * t#

#t = 1071.43/(3.6 * 10^(-2)”s”^(-1)) = “29761.94 s”#, or

#t = 2.976 * 10^(4)”s”#

Rounded to two sig figs, the answer will be

#t = 3.0 * 10^(4)”s”#

SIDE NOTE I didn’t derive the integrated rate law for this second-order reaction because the answer would have been even longer, so I’ll leave that to you as practice.

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