Question #23eaf

1 Answer
Mar 5, 2016

Answer:

Here's what I got.

Explanation:

Start by taking a look at the balanced chemical equation for this reaction

#"H"_2"SeO"_text(3(aq]) + color(red)(6)"I"_text((aq])^(-) + color(blue)(4)"H"_text((aq])^(+) -> 2"I"_text(3(aq])^(-) + 3"H"_2"O"_text((l]) + "Se"_text((s])#

Notice that you have a #color(red)(6):color(blue)(4)# mole ratio between the iodide anions and the hydrogen cations.

What does this tell you in terms of the rate of change of their respective concentrations as the reaction proceeds?

Well, since the reaction consumes #color(red)(3)# moles of iodide anions for every #color(blue)(2)# moles of hydrogen cations, the concentration of the iodide cations will decrease at a faster rate than the concentration of the hydrogen cations.

But keep in mind that the rate of the reaction must be the same for all the chemical species that take part in the reaction.

This means that if you want to express the rate of the reaction in terms of the rate of change in the concentration of each chemical species that takes part in the reaction, you're going to have to take into account stoichiometric coefficients.

The problem tells you that the rate of the reaction is defined as

#"rate"_"old" = -(Delta["H"^(+)])/(Deltat)#

Here the minus sign is used because the concentrations of reactants decrease as the reaction proceeds.

This is equivalent to looking at the rate of the reaction from the perspective of a single mole of hydrogen cation.

What happens to the concentration of iodide anions when one mole of hydrogen cations is consumed by the reaction?

Use the aforementioned mole ratio to find

#1color(red)(cancel(color(black)("mole H"^(+)))) * (color(red)(3)" moles I"^(-))/(color(blue)(2)color(red)(cancel(color(black)("moles H"^(+))))) = 3/2"moles I"^(-)#

When #1# mole of hydrogen cations is consumed, #1.5# moles of iodide anions are also consumed.

This means that you can write the rate of the reaction as

#"rate"_"old" = -(Delta["H"^(+)])/(Deltat) = -1/(3/2) (Delta["I"^(-)])/(Deltat)#

#"rate"_"old" = -(Delta["H"^(+)])/(Deltat) = - 2/3 (Delta["I"^(-)])/(Deltat)" " " "color(orange)("(*)")#

Now, If you were to define the rate of the reaction as

#"rate"_"new" = -(Delta["I"^(-)])/(Deltat)#

then you can say, using equation #color(orange)("(*)")#, that

# (Delta["I"^(-)])/(Deltat) = - 3/2 * "rate"_"old"#

and therefore

#color(green)(|bar(ul("rate"_"new" = 3/2 * "rate"_"old"))|)#

The rate of the reaction would increase by a factor of #3/2 = color(red)(6)/color(blue)(3)#.