The data below were collected for the following reaction at a certain temperature: #X_2Y→2X+Y# (Data found as picture in answer box). What is the concentration of #X# after 12 hours?

2 Answers
Jul 22, 2015

Answer:

#[X]=0.15"M"#

Explanation:

enter image source here

If you plot a concentration time graph you get an exponential curve like this:

chemwiki.ucdavis.edu

This suggest a first order reaction. I plotted the graph in Excel and estimated the half-life. This is the time taken for the concentration to fall by one half of its initial value.

In this case I estimated the time taken for the concentration to fall from 0.1M to 0.05M. You need to extrapolate the graph to get this.

This gives #t_(1/2)=6min#

So we can see that 12mins = 2 half-lifes

After 1 half life the concentration is 0.05M

So after 2 half-lifes #[XY]=0.05/2 = 0.025M#

So in 1L of solution no. moles XY used up = 0.1 - 0.025 = 0.075

Since 2 moles of X form from 1 mole XY the no. moles X formed =0.075 x 2 = 0.15.

So #[X]=0.15"M"#

Jul 22, 2015

Answer:

The concentration of #X# will be equal to 0.134 M.

Explanation:

The values given to you are

enter image source here

In order to be able to determine what the concentration of #X# will be after 12 hours, you need to first determine two things

  • the order of the reaction
  • the rate constant

In order to determine the order of the reaction, you need to plot three graphs

  • #[X_2Y]# versus time, which looks like this

enter image source here

https://plot.ly/~stefan_zdre/3/col2/?share_key=vyrVdbciO8gLbNV6mmucNZ

  • #ln([X_2Y])# versus time, which looks like this

enter image source here

https://plot.ly/~stefan_zdre/17/col2/?share_key=gnsvMoGLJ2NDpZF0dN2B3p

  • #1/([X_2Y])# versus time, which looks like this

enter image source here

https://plot.ly/~stefan_zdre/7/col2/?share_key=M7By0sY6Wvq0W59uTv8Tv6

Now, the graph that fits a straight line will determine the reaction's rate order. As you can see, the third graph fits this patter, which means that the reaction will be second-order.

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

#1/([A]) = 1/([A_0]) + k * t#, where

#k# - the rate constant for the given reaction.
#t# - the time needed for the concentration to go from #[A_0]# to #[A]#.

In order to determine the value of #k#, you need to pick two sets of values from your table.

To make the calculations easier, I'll pick the first and second values. So, the concentration of #X_2Y# starts off at 0.100 M and, after 1 hour, drops to 0.0856 M. This means that you have

#1/([X_2Y]) = 1/([X_2Y]) + k * t#

#1/"0.0856 M" = 1/"0.100 M" + k * (1-0)"h"#

#"11.6822 M"^(-1)= "10.0 M"^(-1) + k * "1 h"#

#k = ((11.6822 - 10.0)"M"^(-1))/("1 h") = color(green)("1.68 M"^(-1)"h"^(-1)#

Use the same equation to determine what the concentration of #X_2Y# will be after 12 hours.

#1/([X_2Y]_12) = 1/("0.100 M") + 1.68 "M"^(-1) cancel("h"^(-1)) * (12 - 0)cancel("h")#

#1/([X_2Y]_12) = "10.0 M"^(-1) + "20.16 M"^(-1) = "30.16 M"^(-1)#

Therefore,

#[X_2Y]_12 = 1/("30.16 M"^(-1)) = color(green)("0.0332 M")#

To get the concentration fo #X#, use the mole ratio that exists between the two species in the chemical equation

#X_2Y -> color(red)(2)X + Y#

You know that every 1 mole of #X_2Y# will produce #color(red)(2)# moles of #X#. Assuming that you have a liter of solution (again, this is just to make thecalculations easier), the number of moles of #X_2Y# that reacted is

#[X_2Y]_"rct" = [X_2Y]_0 - [X_2Y]_12#

#[X_2Y] = 0.100 - 0.0332 = "0.0668 M"#

This is equivalent to

#n_(X_2Y) = "0.0668 moles"#

The number of moles of #X# produced will be equal to

#0.0668cancel("moles"X_2Y) * (color(red)(2)" moles "X)/(1cancel("mole"X_2Y)) = "0.1336 moles"# #X#

For your 1-L sample, this is equivalent to a molarity of

#[X]_12 = color(green)("0.134 M")#