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?

Jul 22, 2015

$\left[X\right] = 0.15 \text{M}$

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 ${t}_{\frac{1}{2}} = 6 \min$

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

After 1 half life the concentration is 0.05M

So after 2 half-lifes $\left[X Y\right] = \frac{0.05}{2} = 0.025 M$

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 $\left[X\right] = 0.15 \text{M}$

Jul 22, 2015

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

• $\left[{X}_{2} Y\right]$ versus time, which looks like this

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

• $\ln \left(\left[{X}_{2} Y\right]\right)$ versus time, which looks like this

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

• $\frac{1}{\left[{X}_{2} Y\right]}$ 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

$\frac{1}{\left[A\right]} = \frac{1}{\left[{A}_{0}\right]} + k \cdot t$, where

$k$ - the rate constant for the given reaction.
$t$ - the time needed for the concentration to go from $\left[{A}_{0}\right]$ to $\left[A\right]$.

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 ${X}_{2} Y$ starts off at 0.100 M and, after 1 hour, drops to 0.0856 M. This means that you have

$\frac{1}{\left[{X}_{2} Y\right]} = \frac{1}{\left[{X}_{2} Y\right]} + k \cdot t$

$\frac{1}{\text{0.0856 M" = 1/"0.100 M" + k * (1-0)"h}}$

$\text{11.6822 M"^(-1)= "10.0 M"^(-1) + k * "1 h}$

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 ${X}_{2} Y$ will be after 12 hours.

$\frac{1}{{\left[{X}_{2} Y\right]}_{12}} = \frac{1}{\text{0.100 M") + 1.68 "M"^(-1) cancel("h"^(-1)) * (12 - 0)cancel("h}}$

$\frac{1}{{\left[{X}_{2} Y\right]}_{12}} = {\text{10.0 M"^(-1) + "20.16 M"^(-1) = "30.16 M}}^{- 1}$

Therefore,

${\left[{X}_{2} Y\right]}_{12} = \frac{1}{\text{30.16 M"^(-1)) = color(green)("0.0332 M}}$

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

${X}_{2} Y \to \textcolor{red}{2} X + Y$

You know that every 1 mole of ${X}_{2} Y$ will produce $\textcolor{red}{2}$ moles of $X$. Assuming that you have a liter of solution (again, this is just to make thecalculations easier), the number of moles of ${X}_{2} Y$ that reacted is

${\left[{X}_{2} Y\right]}_{\text{rct}} = {\left[{X}_{2} Y\right]}_{0} - {\left[{X}_{2} Y\right]}_{12}$

$\left[{X}_{2} Y\right] = 0.100 - 0.0332 = \text{0.0668 M}$

This is equivalent to

${n}_{{X}_{2} Y} = \text{0.0668 moles}$

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

${\left[X\right]}_{12} = \textcolor{g r e e n}{\text{0.134 M}}$