You're dealing with a second order reaction that has the rate constant equal to #"0.15 M"^(-1)"min"^(-1)# and the initial concentration concentration of your reactant, #[B]_0#, equal to **0.5 M**.

The generic form of your reaction is

#B -> P#

The rate of reaction for a second order reaction is

#"rate" = k[B]^(2)#

The integrated rate law for this reaction will be

#1/([B]) - 1/([B]_0) = k*t# #color(blue)((1))#, where

#[B]# - the concentration of #B# after the passing of a certain amount of time, #t#;

#[B]_0# - the initial concentration of #B#;

#k# - the rate constant - #"0.15 M"^(-1)"min"^(-1)#;

#t# - time.

So, you have to calculate the *half-life* of the reaction, which expresses the amount of time needed for the **concentration of B to be halved**. Mathematically, this can be written as

#[B] = ([B]_0)/2#

Plug this equation into equation #color(blue)((1))# and solve for #t#

#1/(([B]_0)/2) - 1/([B]_0) = k * t_"1/2"#

#2/([B]_0) - 1/([B]_0) = k * t_"1/2" => t_"1/2" = 1/(k * [B]_0)#

#t_"1/2" = 1/(0.15cancel("M"^(-1))"s"^(-1) * "0.5"cancel("M")) = color(green)("13 min")#

Now you have to calculate how much time must pass for the concentration of #B# to drop to **45%** of the initial concentration.

Even before doing any calculations, you can predict that *more* than the half-life time must pass, since **45%** of #[B]_0# is a smaller amount than **50%** of #[B]_0#.

The same approach can be used in this case as well

#[B] = 45/100 * [B]_0#

Once again, plug this into equation #color(blue)((1))# and solve for #t#

#1/(45/100 * [B]_0) - 1/([B]_0) = k * t_"45%"#

#t_"45%" = (1/(45/100 * 0.5) - 1/0.5)cancel("M"^(-1)) * 1/(0.15cancel(M^(-1))"s"^(-1))#

#t_"45%" = color(green)("16 min")#

To get the graph of a second-order integrated rate law, rearrange equation #color(blue)((1))# to get

#underbrace(1/([B]))_text(y) = underbrace(k)_text(slope) * t + underbrace(1/([B]_0))_text(y-intercept)#

Your graph will look like this