Question

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?

Answer 1

`[X]=0.15"M"`

Explanation:

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

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"`

Answer 2

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

Explanation:

The values given to you are

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

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

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

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

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

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")`