# The reaction between propanone,iodine and hydrochloric acid is a first order with respect to H+ ions. When[H+]=0.1M ,the rate is 0.865M/s. What is the rate when [H+] = 0.5 M?

Dec 19, 2015

${\text{4.3 M s}}^{- 1}$

#### Explanation:

All you have to do to answer this question is use the definition of a first-order reaction.

As you know, a first-order reaction is characterized by the fact that its rate is directly proportional to the concentration of one reactant.

Simply put, the rate of a first-order reaction will have a linear dependence of the concentration of one reactant. If the concentration of that reactant increases, the rate of the reaction Increases by the same amount.

Likewise, if the concentration of that reactant decreases, the rate of the reaction decreases by the same amount.

In your case, the reaction is first-order in hydrogen ions, ${\text{H}}^{+}$. Notice that the concentration of hydrogen ions increases five-fold.

(["H"^(+)]_2)/(["H"^(+)]_1) = (0.5 color(red)(cancel(color(black)("M"))))/(0.1color(red)(cancel(color(black)("M")))) = 5

This means that the rate of reaction will also increase five-fold, so that

${\text{rate"_2 = 5 xx "rate}}_{1}$

${\text{rate"_2 = 5 xx "0.865 M s"^(-1) = "4.325 M s}}^{- 1}$

You can also show that this is the case by calculating the rate constant for this reaction

color(blue)("rate" = -(d["H"^(+)])/dt = k * ["H"^(+)])

k = "rate"_1/(["H"^(+)]_1)

k = (0.865 color(red)(cancel(color(black)("M"))) "s"^(-1))/(0.1 color(red)(cancel(color(black)("M")))) = "8.65 s"^(-1)

This means that the new concentration of hydrogen ions will correspond to a reaction rate of

"rate"_2 = k * ["H"^(+)]_2

$\text{rate"_2 = "8.65 s"^(-1) * "0.5 M}$

${\text{rate"_2 = "4.325 M s}}^{- 1}$

Now, you should round this off to one sig fig, the number of sig figs you have for the two concentration of hydrogen ions, but I'll leave the answer rounded to two sig figs

"rate"_2 = color(green)("4.3 M s"^(-1))